Separation of Variables

Separation of Variables is a special method to solve some Differential Equations

A Differential Equation is an equation with a function and one or more of its derivatives :

When Can I Use it?

Separation of Variables can be used when:

All the y terms (including dy) can be moved to one side of the equation, and

All the x terms (including dx) to the other side.

Three Steps:

  • Step 1 Move all the y terms (including dy) to one side of the equation and all the x terms (including dx) to the other side.
  • Step 2 Integrate one side with respect to y and the other side with respect to x . Don't forget "+ C" (the constant of integration).
  • Step 3 Simplify

Example: Solve this (k is a constant):

  dy dx = ky

Step 1 Separate the variables by moving all the y terms to one side of the equation and all the x terms to the other side:

Step 2 Integrate both sides of the equation separately:

C is the constant of integration. And we use D for the other, as it is a different constant.

Step 3 Simplify:

We have solved it:

This is a general type of first order differential equation which turns up in all sorts of unexpected places in real world examples.

We used y and x , but the same method works for other variable names, like this:

rabbits

Example: Rabbits!

The more rabbits you have the more baby rabbits you will get. Then those rabbits grow up and have babies too! The population will grow faster and faster.

The important parts of this are:

  • the population N at any time t
  • the growth rate r
  • the population's rate of change dN dt

The rate of change at any time equals the growth rate times the population:

But hey! This is the same as the equation we just solved! It just has different letters:

  • N instead of y
  • t instead of x
  • r instead of k

So we can jump to a solution:

And here is an example, the graph of N = 0.3e 2t :

There are other equations that follow this pattern such as continuous compound interest .

More Examples

OK, on to some different examples of separating the variables:

Example: Solve this:

dy dx = 1 y

We integrated both sides in the one line.

We also used a shortcut of just one constant of integration C. This is perfectly OK as we could have +D on one, +E on the other and just say that C = E−D.

Note: This is not the same as y = √(2x) + C, because the C was added before we took the square root. This happens a lot with differential equations. We cannot just add the C at the end of the process. It is added when doing the integration.

y = ±√(2(x + C))

A harder example:

dy dx = 2xy 1+x 2

Step 1 Separate the variables:

Multiply both sides by dx, divide both sides by y:

1 y dy = 2x 1+x 2 dx

∫ 1 y dy = ∫ 2x 1+x 2 dx

The left side is a simple logarithm, the right side can be integrated using substitution:

It is already as simple as can be. We have solved it:

y = k(1 + x 2 )

An even harder example: the famous Verhulst Equation

Example: Rabbits Again!

Remember our growth Differential Equation:

Well, that growth can't go on forever as they will soon run out of available food.

A guy called Verhulst included k (the maximum population the food can support) to get:

dN dt = rN(1−N/k)

The Verhulst Equation

Can this be solved?

Yes, with the help of one trick ...

Step 2 Integrate:

∫ 1 N(1−N/k) dN = ∫ r dt

Hmmm... the left side looks hard to integrate. In fact it can be done with a little trick from Partial Fractions ... we rearrange it like this:

Now it is a lot easier to solve. We can integrate each term separately, like this:

(Why did that become minus ln(k−N)? Because we are integrating with respect to N.)

We are getting close! Just a little more algebra to get N on its own:

And we have our solution:

N = k 1 + Ae −rt

Here is an example , the graph of 40 1 + 5e −2t

Module 4: Differential Equations

Separation of variables, learning outcomes.

  • Use separation of variables to solve a differential equation
  • Solve applications using separation of variables

We start with a definition and some examples.

A separable differential equation is any equation that can be written in the form

The term ‘separable’ refers to the fact that the right-hand side of the equation can be separated into a function of [latex]x[/latex] times a function of [latex]y[/latex]. Examples of separable differential equations include

The second equation is separable with [latex]f\left(x\right)=6{x}^{2}+4x[/latex] and [latex]g\left(y\right)=1[/latex], the third equation is separable with [latex]f\left(x\right)=1[/latex] and [latex]g\left(y\right)=\sec{y}+\tan{y}[/latex], and the right-hand side of the fourth equation can be factored as [latex]\left(x+3\right)\left(y - 2\right)[/latex], so it is separable as well. The third equation is also called an autonomous differential equation because the right-hand side of the equation is a function of [latex]y[/latex] alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables .

Problem-Solving Strategy: Separation of Variables

  • Check for any values of [latex]y[/latex] that make [latex]g\left(y\right)=0[/latex]. These correspond to constant solutions.
  • Rewrite the differential equation in the form [latex]\frac{dy}{g\left(y\right)}=f\left(x\right)dx[/latex].
  • Integrate both sides of the equation.
  • Solve the resulting equation for [latex]y[/latex] if possible.
  • If an initial condition exists, substitute the appropriate values for [latex]x[/latex] and [latex]y[/latex] into the equation and solve for the constant.

Note that Step 4. states “Solve the resulting equation for [latex]y[/latex] if possible.” It is not always possible to obtain [latex]y[/latex] as an explicit function of [latex]x[/latex]. Quite often we have to be satisfied with finding [latex]y[/latex] as an implicit function of [latex]x[/latex].

Example: Using Separation of Variables

Find a general solution to the differential equation [latex]y^{\prime} =\left({x}^{2}-4\right)\left(3y+2\right)[/latex] using the method of separation of variables.

Follow the five-step method of separation of variables.

  • In this example, [latex]f\left(x\right)={x}^{2}-4[/latex] and [latex]g\left(y\right)=3y+2[/latex]. Setting [latex]g\left(y\right)=0[/latex] gives [latex]y=-\frac{2}{3}[/latex] as a constant solution.

Let [latex]u=3y+2[/latex]. Then [latex]du=3\frac{dy}{dx}dx[/latex], so the equation becomes

Now we use some logic in dealing with the constant [latex]C[/latex]. Since [latex]C[/latex] represents an arbitrary constant, [latex]3C[/latex] also represents an arbitrary constant. If we call the second arbitrary constant [latex]{C}_{1}[/latex], the equation becomes

Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base [latex]e[/latex]).

Again define a new constant [latex]{C}_{2}={e}^{{c}_{1}}[/latex] (note that [latex]{C}_{2}>0[/latex]):

This corresponds to two separate equations: [latex]3y+2={C}_{2}{e}^{{x}^{3}-12x}[/latex] and [latex]3y+2=\text{-}{C}_{2}{e}^{{x}^{3}-12x}[/latex]. The solution to either equation can be written in the form [latex]y=\frac{-2\pm {C}_{2}{e}^{{x}^{3}-12x}}{3}[/latex]. Since [latex]{C}_{2}>0[/latex], it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant [latex]C[/latex] is entirely arbitrary, and can be dropped. Therefore the solution can be written as

  • No initial condition is imposed, so we are finished.

Watch the following video to see the worked solution to Example: Using Separation of Variables

For closed captioning, open the video on its original page by clicking the Youtube logo in the lower right-hand corner of the video display. In YouTube, the video will begin at the same starting point as this clip, but will continue playing until the very end.

You can view the transcript for this segmented clip of “4.3.1” here (opens in new window) .

Use the method of separation of variables to find a general solution to the differential equation [latex]y^{\prime} =2xy+3y - 4x - 6[/latex].

First factor the right-hand side of the equation by grouping, then use the five-step strategy of separation of variables.

[latex]y=2+C{e}^{{x}^{2}+3x}[/latex]

Example: Solving an Initial-Value Problem

Using the method of separation of variables, solve the initial-value problem

  • In this example, [latex]f\left(x\right)=2x+3[/latex] and [latex]g\left(y\right)={y}^{2}-4[/latex]. Setting [latex]g\left(y\right)=0[/latex] gives [latex]y=\pm 2[/latex] as constant solutions.

To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity

Then integration becomes

Multiplying both sides of this equation by [latex]4[/latex] and replacing [latex]4C[/latex] with [latex]{C}_{1}[/latex] gives

Next we can remove the absolute value and let [latex]{C}_{2}[/latex] be either positive or negative. Then multiply both sides by [latex]y+2[/latex].

Now collect all terms involving y on one side of the equation, and solve for [latex]y\text{:}[/latex]

Therefore the solution to the initial-value problem is

A graph of this solution appears in Figure 1.

A graph of the solution over [-5, 3] for x and [-3, 2] for y. It begins as a horizontal line at y = -2 from x = -5 to just before -3, almost immediately steps up to y = 2 from just after x = -3 to just before x = 0, and almost immediately steps back down to y = -2 just after x = 0 to x = 3.

Figure 1. Graph of the solution to the initial-value problem [latex]y^{\prime} =\left(2x+3\right)\left({y}^{2}-4\right),y\left(0\right)=-3[/latex].

Find the solution to the initial-value problem

using the method of separation of variables.

Follow the steps for separation of variables to solve the initial-value problem.

[latex]y=\frac{4+14{e}^{{x}^{2}+x}}{1 - 7{e}^{{x}^{2}+x}}[/latex]

Applications of Separation of Variables

Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.

Solution concentrations

Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations .

Example: Determining Salt Concentration over Time

A tank containing [latex]100\text{L}[/latex] of a brine solution initially has [latex]4\text{kg}[/latex] of salt dissolved in the solution. At time [latex]t=0[/latex], another brine solution flows into the tank at a rate of [latex]2\text{L/min}\text{.}[/latex] This brine solution contains a concentration of [latex]0.5\text{kg/L}[/latex] of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of [latex]2\text{L/min}[/latex], so that the level of liquid in the tank remains constant (Figure 2). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.

A diagram of a cylinder filled with water with input and output. It is a 100 liter tank which initially contains 4 kg of salt. The input is 0.5 kg salt / liter and 2 liters / minute. The output is 2 liters / minute.

Figure 2. A brine tank with an initial amount of salt solution accepts an input flow and delivers an output flow. How does the amount of salt change with time?

First we define a function [latex]u\left(t\right)[/latex] that represents the amount of salt in kilograms in the tank as a function of time. Then [latex]\frac{du}{dt}[/latex] represents the rate at which the amount of salt in the tank changes as a function of time. Also, [latex]u\left(0\right)[/latex] represents the amount of salt in the tank at time [latex]t=0[/latex], which is [latex]4[/latex] kilograms.

The general setup for the differential equation we will solve is of the form

INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of [latex]2[/latex] L/min, and each liter of solution contains [latex]0.5[/latex] kilogram of salt, every minute [latex]2\left(0.5\right)=1\text{kilogram}[/latex] of salt enters the tank. Therefore INFLOW RATE = [latex]1[/latex].

To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time [latex]t[/latex] is equal to [latex]u\left(t\right)[/latex]. Thus, the concentration of salt is [latex]\frac{u\left(t\right)}{100}[/latex] kg/L, and the solution leaves the tank at a rate of [latex]2[/latex] L/min. Therefore salt leaves the tank at a rate of [latex]\frac{u\left(t\right)}{100}\cdot 2=\frac{u\left(t\right)}{50}[/latex] kg/min, and OUTFLOW RATE is equal to [latex]\frac{u\left(t\right)}{50}[/latex]. Therefore the differential equation becomes [latex]\frac{du}{dt}=1-\frac{u}{50}[/latex], and the initial condition is [latex]u\left(0\right)=4[/latex]. The initial-value problem to be solved is

The differential equation is a separable equation, so we can apply the five-step strategy for solution.

Step 1. Setting [latex]1-\frac{u}{50}=0[/latex] gives [latex]u=50[/latex] as a constant solution. Since the initial amount of salt in the tank is [latex]4[/latex] kilograms, this solution does not apply.

Step 2. Rewrite the equation as

Then multiply both sides by [latex]dt[/latex] and divide both sides by [latex]50-u\text{:}[/latex]

Step 3. Integrate both sides:

Step 4. Solve for [latex]u\left(t\right)\text{:}[/latex]

Eliminate the absolute value by allowing the constant to be either positive or negative:

Finally, solve for [latex]u\left(t\right)\text{:}[/latex]

Step 5. Solve for [latex]{C}_{1}\text{:}[/latex]

The solution to the initial value problem is [latex]u\left(t\right)=50 - 46{e}^{\frac{\text{-}t}{50}}[/latex]. To find the limiting amount of salt in the tank, take the limit as [latex]t[/latex] approaches infinity:

Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is [latex]50[/latex] kilograms, then it remains constant. If it starts at less than 50 kilograms, then it approaches 50 kilograms over time.

Watch the following video to see the worked solution to Example: Determining Salt Concentration over Time

You can view the transcript for this segmented clip of “4.3.2” here (opens in new window) .

A tank contains [latex]3[/latex] kilograms of salt dissolved in [latex]75[/latex] liters of water. A salt solution of [latex]0.4\text{kg salt/L}[/latex] is pumped into the tank at a rate of [latex]6\text{L/min}[/latex] and is drained at the same rate. Solve for the salt concentration at time [latex]t[/latex]. Assume the tank is well mixed at all times.

Follow the steps in the example: Determining Salt Concentration over Time and determine an expression for INFLOW and OUTFLOW. Formulate an initial-value problem, and then solve it.

Initial value problem:

[latex]\frac{du}{dt}=2.4-\frac{2u}{25},u\left(0\right)=3[/latex]

[latex]\text{Solution:}u\left(t\right)=30 - 27{e}^{\frac{\text{-}t}{50}}[/latex]

Newton’s law of cooling

Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let [latex]T\left(t\right)[/latex] represent the temperature of an object as a function of time, then [latex]\frac{dT}{dt}[/latex] represents the rate at which that temperature changes. The temperature of the object’s surroundings can be represented by [latex]{T}_{s}[/latex]. Then Newton’s law of cooling can be written in the form

The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We call this temperature [latex]{T}_{0}[/latex]. Therefore the initial-value problem that needs to be solved takes the form

where [latex]k[/latex] is a constant that needs to be either given or determined in the context of the problem. We use these equations in the next example.

Example: Waiting for a Pizza to Cool

A pizza is removed from the oven after baking thoroughly, and the temperature of the oven is [latex]350^\circ\text{F}\text{.}[/latex] The temperature of the kitchen is [latex]75^\circ\text{F}[/latex], and after [latex]5[/latex] minutes the temperature of the pizza is [latex]340^\circ\text{F}\text{.}[/latex] We would like to wait until the temperature of the pizza reaches [latex]300^\circ\text{F}[/latex] before cutting and serving it (Figure 3). How much longer will we have to wait?

A diagram of a pizza pie. The room temperature is 75 degrees, and the pizza temperature is 350 degrees.

Figure 3. From Newton’s law of cooling, if the pizza cools [latex]10^\circ\text{F}[/latex] in [latex]5[/latex] minutes, how long before it cools to [latex]300^\circ\text{F?}[/latex]

The ambient temperature (surrounding temperature) is [latex]75^\circ\text{F}[/latex], so [latex]{T}_{s}=75[/latex]. The temperature of the pizza when it comes out of the oven is [latex]350^\circ\text{F}[/latex], which is the initial temperature (i.e., initial value), so [latex]{T}_{0}=350[/latex]. Therefore our equation becomes

To solve the differential equation, we use the five-step technique for solving separable equations.

  • Setting the right-hand side equal to zero gives [latex]T=75[/latex] as a constant solution. Since the pizza starts at [latex]350^\circ\text{F}[/latex], this is not the solution we are seeking.

To determine the value of [latex]k[/latex], we need to use the fact that after [latex]5[/latex] minutes the temperature of the pizza is [latex]340^\circ\text{F}\text{.}[/latex] Therefore [latex]T\left(5\right)=340[/latex]. Substituting this information into the solution to the initial-value problem, we have

So now we have [latex]T\left(t\right)=75+275{e}^{-0.007408t}[/latex]. When is the temperature [latex]300^\circ\text{F?}[/latex] Solving for [latex]t[/latex], we find

Watch the following video to see the worked solution to Example: Waiting for a Pizza to Cool

A cake is removed from the oven after baking thoroughly, and the temperature of the oven is [latex]450^\circ\text{F}\text{.}[/latex] The temperature of the kitchen is [latex]70^\circ\text{F}[/latex], and after [latex]10[/latex] minutes the temperature of the cake is [latex]430^\circ\text{F}\text{.}[/latex]

  • Write the appropriate initial-value problem to describe this situation.
  • Solve the initial-value problem for [latex]T\left(t\right)[/latex].
  • How long will it take until the temperature of the cake is within [latex]5^\circ\text{F}[/latex] of room temperature?

Determine the values of [latex]{T}_{s}[/latex] and [latex]{T}_{0}[/latex] then use the example: Solving an Initial-Value Problem.

  • Initial-value problem [latex]\frac{dT}{dt}=k\left(T - 70\right),T\left(0\right)=450[/latex]
  • [latex]T\left(t\right)=70+380{e}^{kt}[/latex]
  • Approximately [latex]114[/latex] minutes.
  • 4.3.1. Authored by : Ryan Melton. License : CC BY: Attribution
  • 4.3.2. Authored by : Ryan Melton. License : CC BY: Attribution
  • Calculus Volume 2. Authored by : Gilbert Strang, Edwin (Jed) Herman. Provided by : OpenStax. Located at : https://openstax.org/books/calculus-volume-2/pages/1-introduction . License : CC BY-NC-SA: Attribution-NonCommercial-ShareAlike . License Terms : Access for free at https://openstax.org/books/calculus-volume-2/pages/1-introduction

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On this page

  • Differential Equations
  • Predicting AIDS - a DEs example
  • 1. Solving Differential Equations
  • 2. Separation of Variables
  • 3. Integrable Combinations
  • 4. Linear DEs of Order 1
  • 5. Application: RL Circuits
  • 6. Application: RC Circuits
  • 7. Second Order DEs - Homogeneous
  • 8. Second Order DEs - Damping - RLC
  • 9. Second Order DEs - Forced Response
  • 10. Second Order DEs - Solve Using SNB
  • 11. Euler's Method - a numerical solution for Differential Equations
  • 12. Runge-Kutta (RK4) numerical solution for Differential Equations

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Particular solutions RL circuit Terminal velocity

Some differential equations can be solved by the method of separation of variables (or "variables separable") . This method is only possible if we can write the differential equation in the form

A ( x ) dx + B ( y ) dy = 0,

where A ( x ) is a function of x only and B ( y ) is a function of y only.

Once we can write it in the above form, all we do is integrate throughout, to obtain our general solution.

NOTE: In this variables separable section we only deal with first order, first degree differential equations.

Example 1 - Separation of Variables form

a) The differential equation (which we saw earlier in Solutions of Differential Equations ):

`(dy)/(dx)ln\ x-y/x=0`

can be expressed in the required form, A ( x ) dx + B ( y ) dy = 0 , after some algebraic juggling:

` (dy)/(dx)ln\ x-y/x=0` `dy\ ln\ x-(y\ dx)/x=0` `dy-(y\ dx)/(x\ ln\ x)=0` `(dy)/y-(dx)/(x\ ln\ x)=0` `1/ydy-1/(x\ ln\ x)dx=0`

Here, `A(x) = -1/(x\ ln\ x)` and `B(y) = 1/y`.

[To solve the equation, we would then integrate throughout].

b) The following differential equation cannot be expressed in the required form, so it cannot be solved using separation of variables:

`(dy)/(dx)=(3(x+y))/(x(y-2))`

Solve the differential equation:

y 2 dy + x 3 dx = 0

This is already in the required form (since the x -terms are together with dx terms, and y -terms are together with dy terms), so we simply integrate:

`inty^2dy+intx^3dx=0`
`y^3/3+x^4/4=K`

This is the general solution for the differential equation.

We can continue on to solve this as an explicit function in x , as follows:

`y^3=3(K-x^4/4)` `y=root(3)(3(K-x^4/4))`

Taking a typical constant value `K=5`, we have this solution graph:

Typical solution graph `y=root(3)(3(5-x^4/4))`.

Solve the differential equation: `2(dy)/(dx)=(y(x+1))/x`

First we must separate the variables:

`(2\ dy)/y=((x+1)dx)/x`

This gives us: `(2\ dy)/y=(1+1/x)dx`

We now integrate:

`int(2 dy)/y=int(1+1/x)dx` `2 ln y=x+ln x+K`

Go back to Integration: Basic Logarithm Form if you are rusty on this integration.

We could continue with our solution and express `y` as an explicit function of `x`, as follows:

`ln y=(x+ln x+K)/2` `y=e^((x+ln x+K)//2)`

Taking a typical constant value `K=1`, we have this solution graph:

Typical solution graph `y=e^((x+ln x+1)//2)`.

Solve `t(dx)/(dt)-x=3`

`t(dx)/(dt)-x=3`

We want all x 's on one side, all the terms in t on the other.

`t(dx)/(dt)=x+3`

Divide both sides by `(x + 3)` and multiply both sides by `dt`:

`t(dx)/(x+3)=dt`

Next, divide both sides by t .

`(dx)/(x+3)=(dt)/t`

Integrate both sides.

`int(dx)/(x+3)=int(dt)/t`

Important: dx and dt must be on top of the fraction (i.e. in the numerator).

`ln|x+3|=ln|t|+C`

Take " e to power of both sides":

`x+3=e^(ln|t|+C)=e^(ln|t|)e^C=Kt`

(We let `e^C=K`, constant)

So `x = Kt - 3`.

Typical solution and graph

We let `K = 7`, for illustration):

Typical solution graph `x=7t-3`.

Solve `sqrt(1+4x^2)\ dy=y^3x\ dx`

Separating variables gives us:

`(dy)/y^3=(x\ dx)/(sqrt(1+4x^2))`

Integrating gives us:

`int(dy)/y^3=int(x\ dx)/(sqrt(1+4x^2)`

We now proceed to integrate the 2 sides separately. That is, we integrate the left side in y only (since after separating the variables we have terms in y and a dy on the left) and we work on the right side in x only (since we have terms in x and a dx only on the right).

For the right hand side involving x , let u = (1 + 4 x 2 ), so du = 8 x dx and du /8 = x dx.

`int(dy)/y^3=1/8int(du)/sqrtu`

`=(-1)/(2y^2)` `=1/8(2)u^(1"/"2)+K` `=1/4sqrt(1+4x^2)+K`

So the solution is given by:

`(-1)/(2y^2)=1/4sqrt(1+4x^2)+K`

We could go on to solve this in y , as follows:

Multiply both sides by `−2`:

`1/y^2=-1/2sqrt(1+4x^2)-2K`

For convenience, introduce a new variable `K_1 = -4K`, so that we'll have `-2K=K_1/2`. Our solution becomes:

`1/y^2=-1/2sqrt(1+4x^2)+K_1/2 =` ` (K_1 - sqrt(1+4x^2))/2`

Take the reciprocal of both sides:

`y^2=2/(K_1-sqrt(1+4x^2))`

Then solve for y :

`y=(+-sqrt2)/sqrt(K_1-sqrt(1+4x^2))`

(The constant K 1 can be chosen so that the expression in the denominator is real.)

Here is the graph of a typical solution for Example 5 where we have taken `K=50`:

Typical solution graph `y=(+-sqrt2)/sqrt(50-sqrt(1+4x^2))`.

Particular Solutions

Our examples so far in this section have involved some constant of integration, K .

We now move on to see particular solutions, where we know some boundary conditions and we substitute those into our general solution to give a particular solution .

Find the particular solution for

`(dy)/(dx)+2y=6`

given that x = 0 when y = 1.

We solve as before, and then use the given information to find the value of the unknown K.

Separating variables:

`(dy)/(dx)+2y=6` `(dy)/(dx)=6-2y` `dy=(6-2y)dx` `(dy)/(6-2y)=dx`

Integrating gives:

`int(dy)/(6-2y)=intdx` `(-1)/2ln|6-2y|=x+K`

[For the integral involving `y`, we put `u = 6 - 2y` giving `du = - 2\ dy`. This means we'll replace `dy` with `(-1/2)du` and integrate `(1/u)du` giving `ln\ u`.]

Now, when `x = 0`, `y = 1`; so we have:

`(-1)/2ln|6-2(1)|=0+K` `K=(-1)/2ln|4|`

So, on substituting this back into our previous equation and doing some algebra, we obtain:

`(-1)/2ln|6-2y|=x+(-1)/2ln|4|`

`(-1)/2[ln|6-2y|-ln 4]=x`

`ln (6-2y)/4=-2x`

Taking "`e` to both sides":

`(6-2y)/4=e^(-2x)`

`6-2y=4e^(-2x)`

`y=3-2e^(-2x)`

Checking our solution: `(dy)/(dx)=4e^(-2x)`and so

`"LHS"=(dy)/(dx)+2y`

`=4e^(-2x)+2(3-2e^(-2x))`

`="RHS"`

Also, when `x = 0`, `y = 3 - 2e^0= 1`.

So the particular solution is given by: `y = 3 - 2e^(-2x)`

Here is the graph of our solution for Example 6:

Solution graph `y=3-2e^(-2x)`, showing the curve passing through `(0, 1)`.

`x\ dy = y\ ln\ y\ dx` ,
`x = 2` when `y = e`.
`x\ dy=y\ ln\ y\ dx` `(dy)/(y\ ln\ y)=(dx)/x`

Integrating: [For the y part, let u = ln y , then du = dy/y ].

`int(dy)/(y\ ln\ y)=int(dx)/x` `ln(ln\ y)=ln\ x+K`

Substituting x = 2 when y = e gives:

`ln(ln\ e)=ln\ 2+K` `ln(1)=ln\ 2+K` `0=ln\ 2+K` `K=-ln\ 2`

Substituting this in our general solution:

`ln(ln\ y)=ln\ x-ln\ 2` `=ln\ x/2`

This gives us:

`ln\ y=x/2`

So the particular solution is given by:

`y=e^(x"/"2)`

Here is the graph of our solution for Example 7:

Solution graph `y=e^(x//2)`, showing the curve passes through `(2, e)`.

Example 8 - RL Circuit Application

In an RL circuit, the differential equation formed using Kirchhoff's law, is

`Ri+L(di)/(dt)=V`

Solve this DE, using separation of variables, given that

R = 10 Ω, L = 3 H and V = 50 volts, and i (0) = 0.

Substituting R = 10, L = 3 and V = 50 gives:

`10i+3(di)/(dt)=50`

`3(di)/(dt)=50-10i`

First, we separate the variables.

`(di)/((50-10i))=(dt)/3`

`1/10int(di)/(5-i)=1/3intdt`

`-1/10ln(5-i)=t/3+K`

Since i (0) = 0,

`-1/10ln(5-0)=0+K`

`K=(-ln\ 5)/10`

So, substituting for K :

`-1/10ln(5-i)=t/3-(ln\ 5)/10`

Put log parts together.

`-t/3=1/10ln(5-i)-(ln\ 5)/10`

Multiply both sides by 10

`-(10t)/3=ln(5-i)-ln\ 5`

`-(10t)/3=ln\ (5-i)/5`

`e^(-10t"/"3)=(5-i)/5`

`5e^(-10t"/"3)=5-i`

`i=5-5e^(-10t"/"3)=5(1-e^(-10t"/"3))`

The graph shows that the current builds up and levels out at a maximum value of 5 A.

Solution graph `i=5(1-e^(-10t"/"3))`.

NOTE: We could have solved this for `i` another way. Here it is - you may find it easier.

`ln(5-i)=-(10t)/3+ln\ 5`

Raising both sides as a power of e :

`5-i=e^(-10t"/"3 + ln\ 5)` `=e^(-10t"/"3)e^(ln\ 5)` `=5e^(-10t"/"3)`

`i=5-5e^(-10t"/"3)` `=5(1-e^(-10t"/"3))`

Example 9 - Skydiver's Terminal Velocity

skydiver leaping from Cessna

When a skydiver jumps out of a (perfectly good) aeroplane, apart from experiencing sheer terror (and exhilaration), she experiences two forces:

Gravity , acting downwards Air resistance , acting upwards

Gravity is written g and on earth its acceleration has value 9.8 ms -2 .

Air resistance depends on the size and shape of the item that is dropping through the air. We use the constant k to represent the amount of air resistance. It is called the coefficient of drag .

The air resistance is proportional to the square of the velocity , so the upwards resistance force due to air resistance can be represented by

F air = kv 2 .

Now the force on the object due to gravity is:

F gravity = mg , where g is the acceleration due to gravity.

The total force acting on the body is

`F_("total") = F_("air") - F_("gravity") = kv^2 - mg`

Dividing throughout by m gives us a good model for the velocity v of an object falling through the air:

`a=k/mv^2-g`

We can write this as

`(dv)/(dt)=k/mv^2-g`

skydivers

a. Find an expression for the velocity of a sky diver at time t .

b. Find the velocity after 5 seconds for a sky diver of mass m = 80 kg and k = 0.2

c. Find the terminal velocity for any object for general values of g and k .

d. Find the terminal velocity our skydiver with mass m = 80 kg and k = 0.2.

a. General Formula for Velocity of an Object Falling in Air

We rewrite our differential equation:

in a more convenient form :

`(dv)/(dt)=g(k/(mg)v^2-1)`

Also for convenience, let c be defined as:

`c=sqrt((mg)/k`
`1/c^2=k/(mg)`

We can rewrite our differential equation as:

`(dv)/(dt)=g(v^2/c^2-1)`

This is the same as:

`(dv)/(dt)=g((v^2-c^2)/c^2)`

Separating the variables:

`(c^2dv)/(v^2-c^2)=g\ dt`

Integrating both sides:

`c^2int(dv)/(v^2-c^2)=intg dt`

We multiply the fraction by `-1`, thus reversing the order of `(v^2-c^2)`, and also at the front to compensate, so the substitution step coming up later is possible. (Otherwise, we would be trying to find the log of a negative number when finding `K`.)

`-c^2int(dv)/(c^2-v^2)=int g dt`

To perform this integration, we could either:

  • Factor the denominator, use partial fractions and then integrate (it needs the logarithm form), or
  • Use a table of integrals (integral #13), (easier); or
  • Use a computer algebra system, like Scientific Notebook . (easiest)

We integrate and obtain:

`(-c^2)1/(2c)ln((c+v)/(c-v))= g t +K`

" K " is the constant of integration.

At `t = 0`, `v = 0` and after substituting, we conclude `K = 0`. Therefore:

`-c/2ln((c+v)/(c-v))=g t`

Now we solve for v .

Multiply both sides by `-2` and divide by c :

`ln((c+v)/(c-v))=-(2g t)/c`

Take e to both sides to remove the logarithm:

`(c+v)/(c-v)=e^(-2g t"/"c)`

Multiply out and solve for v :

`c+v=(c-v)e^(-2g t"/"c)` `c+v=ce^(-2g t"/"c)-ve^(-2g t"/"c)` `ve^(-2g t"/"c)+v=ce^(-2g t"/"c)-c` `v=c(e^(-2g t"/"c)-1)/(e^(-2g t"/"c)+1)`

This is the velocity of our object at time t .

b. Velocity for Particular Sky Diver at Specific Time

First, we find c for our given situation. This was the expression for c :

`c=sqrt((mg)/k)`

We use the given mass and the coefficient of drag for the skydiver.

mass = m = 80 kg coefficient of drag = k = 0.2
`c=sqrt((mg)/k)` `=sqrt((80xx9.8)/0.2)` `=62.6`

Next, we substitute the given time value ( t = 5 ) and c = 62.6 to find the required velocity:

`v=c(e^(-2g t"/"c)-1)/(e^(-2g t"/"c)+1)` `=62.6(e^(-2(9.8)(5)"/"62.6)-1)/(e^(-2(9.8)(5)"/"62.6)+1)` `=-40.9580`

The units are ms -1 , so the velocity at time t = 5 s is approximately 147 km/h (1 ms -1 = 3.6 km/h) , in the downward direction.

c. Terminal velocity

The expression for velocity at time t we found earlier:

`v=c(e^(-2g t"/"c)-1)/(e^(-2g t"/"c)+1)`

As t → ∞, the value of the fraction approaches −1, since e -2 gt/c → 0, giving us the terminal velocity `v=-c`.

is the terminal velocity for the falling object (in the downward direction).

Note: We could have obtained the above expression without knowing the expression for velocity at time t , by simply noting the velocity of the object reaches terminal velocity when the acceleration is 0.

That is, solving the following acceleration expression to find the velocity:

`a=k/mv^2-g=0`

This gives terminal velocity

`v=sqrt((mg)/k)`

d. Terminal Velocity for Skydiver Example

We already found the expression for c (which is the terminal velocity) in Part (b)

`=sqrt((80xx9.8)/0.2)`

So the terminal velocity is approximately 225 km/h , since 1 ms -1 = 3.6 km/h.

The graph of the velocity against time shows that it takes around 15 seconds to reach (or "get very close to") the terminal velocity:

Velocity graph `v=225(e^(-2(9.8) t"/"62.6)-1)/(e^(-2(9.8) t"/"62.6)+1)`, showing the point `(5, -147)`.

  • The graph shows that the terminal velocity is never actually reached - the skydiver's velocity just gets closer and closer to that velocity.
  • The graph includes the point representing the velocity at time t = 5, found earlier.
  • Actually, the air resistance increases as the air gets more dense nearer the Earth's surface. We have assumed it remains constant for this problem.
  • The human sky diver can change k easily by either spreading their arms and legs (which will slow them down), or diving down with arms and legs tightly together (which will increase their speed)

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Solved example of separable differential equations

Rewrite the differential equation in the standard form $M(x,y)dx+N(x,y)dy=0$

The differential equation $3y^2dy-2xdx=0$ is exact, since it is written in the standard form $M(x,y)dx+N(x,y)dy=0$, where $M(x,y)$ and $N(x,y)$ are the partial derivatives of a two-variable function $f(x,y)$ and they satisfy the test for exactness: $\displaystyle\frac{\partial M}{\partial y}=\frac{\partial N}{\partial x}$. In other words, their second partial derivatives are equal. The general solution of the differential equation is of the form $f(x,y)=C$

 Intermediate steps

Find the derivative of $M(x,y)$ with respect to $y$

The derivative of the constant function ($-2x$) is equal to zero

Find the derivative of $N(x,y)$ with respect to $x$

The derivative of the constant function ($3y^2$) is equal to zero

Using the test for exactness, we check that the differential equation is exact

The integral of a function times a constant ($-2$) is equal to the constant times the integral of the function

Applying the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, in this case $n=1$

Since $y$ is treated as a constant, we add a function of $y$ as constant of integration

Integrate $M(x,y)$ with respect to $x$ to get

The derivative of the constant function ($-x^2$) is equal to zero

The derivative of $g(y)$ is $g'(y)$

Now take the partial derivative of $-x^2$ with respect to $y$ to get

Simplify and isolate $g'(y)$

$x+0=x$, where $x$ is any expression

Rearrange the equation

Set $3y^2$ and $0+g'(y)$ equal to each other and isolate $g'(y)$

Integrate both sides with respect to $y$

The integral of a function times a constant ($3$) is equal to the constant times the integral of the function

Apply the power rule for integration, $\displaystyle\int x^n dx=\frac{x^{n+1}}{n+1}$, where $n$ represents a number or constant function, such as $2$

Any expression multiplied by $1$ is equal to itself

Find $g(y)$ integrating both sides

We have found our $f(x,y)$ and it equals

Then, the solution to the differential equation is

Group the terms of the equation

Removing the variable's exponent raising both sides of the equation to the power of $\frac{1}{3}$

Divide $1$ by $3$

Simplify $\sqrt[3]{y^{3}}$ using the power of a power property: $\left(a^m\right)^n=a^{m\cdot n}$. In the expression, $m$ equals $3$ and $n$ equals $\frac{1}{3}$

Multiply $3$ times $\frac{1}{3}$

Find the explicit solution to the differential equation. We need to isolate the variable $y$

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AP®︎/College Calculus AB

Course: ap®︎/college calculus ab   >   unit 7.

  • Separable equations introduction
  • Addressing treating differentials algebraically

Separable differential equations

  • Separable differential equations: find the error
  • Worked example: separable differential equations
  • Worked example: identifying separable equations
  • Identifying separable equations
  • Identify separable equations
  • (Choice A)   y = A 9 x 2 2 3 + C ‍   A y = A 9 x 2 2 3 + C ‍  
  • (Choice B)   y = A 9 x 2 2 + C 3 ‍   B y = A 9 x 2 2 + C 3 ‍  
  • (Choice C)   y = A 9 ln ⁡ | x | + C 3 ‍   C y = A 9 ln ⁡ | x | + C 3 ‍  
  • (Choice D)   y = A 9 ln ⁡ | x + C | 3 ‍   D y = A 9 ln ⁡ | x + C | 3 ‍  

4.3 Separable Equations

Learning objectives.

  • 4.3.1 Use separation of variables to solve a differential equation.
  • 4.3.2 Solve applications using separation of variables.

We now examine a solution technique for finding exact solutions to a class of differential equations known as separable differential equations. These equations are common in a wide variety of disciplines, including physics, chemistry, and engineering. We illustrate a few applications at the end of the section.

Separation of Variables

We start with a definition and some examples.

A separable differential equation is any equation that can be written in the form

The term ‘separable’ refers to the fact that the right-hand side of the equation can be separated into a function of x x times a function of y . y . Examples of separable differential equations include

The second equation is separable with f ( x ) = 6 x 2 + 4 x f ( x ) = 6 x 2 + 4 x and g ( y ) = 1 , g ( y ) = 1 , the third equation is separable with f ( x ) = 1 f ( x ) = 1 and g ( y ) = sec y + tan y , g ( y ) = sec y + tan y , and the right-hand side of the fourth equation can be factored as ( x - 2 ) ( y + 3 ) , ( x - 2 ) ( y + 3 ) , so it is separable as well. The third equation is also called an autonomous differential equation because the right-hand side of the equation is a function of y y alone. If a differential equation is separable, then it is possible to solve the equation using the method of separation of variables .

Problem-Solving Strategy

Problem-solving strategy: separation of variables.

  • Check for any values of y y that make g ( y ) = 0 . g ( y ) = 0 . These correspond to constant solutions.
  • Rewrite the differential equation in the form d y g ( y ) = f ( x ) d x . d y g ( y ) = f ( x ) d x .
  • Integrate both sides of the equation.
  • Solve the resulting equation for y y if possible.
  • If an initial condition exists, substitute the appropriate values for x x and y y into the equation and solve for the constant.

Note that Step 4. states “Solve the resulting equation for y y if possible.” It is not always possible to obtain y y as an explicit function of x . x . Quite often we have to be satisfied with finding y y as an implicit function of x . x .

Example 4.10

Using separation of variables.

Find a general solution to the differential equation y ′ = ( x 2 − 4 ) ( 3 y + 2 ) y ′ = ( x 2 − 4 ) ( 3 y + 2 ) using the method of separation of variables.

Follow the five-step method of separation of variables.

  • In this example, f ( x ) = x 2 − 4 f ( x ) = x 2 − 4 and g ( y ) = 3 y + 2 . g ( y ) = 3 y + 2 . Setting g ( y ) = 0 g ( y ) = 0 gives y = − 2 3 y = − 2 3 as a constant solution.
  • Rewrite the differential equation in the form d y 3 y + 2 = ( x 2 − 4 ) d x . d y 3 y + 2 = ( x 2 − 4 ) d x .
  • Integrate both sides of the equation: ∫ d y 3 y + 2 = ∫ ( x 2 − 4 ) d x . ∫ d y 3 y + 2 = ∫ ( x 2 − 4 ) d x . Let u = 3 y + 2 . u = 3 y + 2 . Then d u = 3 d y d x d x , d u = 3 d y d x d x , so the equation becomes 1 3 ∫ 1 u d u = 1 3 x 3 − 4 x + C 1 3 ln | u | = 1 3 x 3 − 4 x + C 1 3 ln | 3 y + 2 | = 1 3 x 3 − 4 x + C . 1 3 ∫ 1 u d u = 1 3 x 3 − 4 x + C 1 3 ln | u | = 1 3 x 3 − 4 x + C 1 3 ln | 3 y + 2 | = 1 3 x 3 − 4 x + C .
  • To solve this equation for y , y , first multiply both sides of the equation by 3 . 3 . ln | 3 y + 2 | = x 3 − 12 x + 3 C ln | 3 y + 2 | = x 3 − 12 x + 3 C Now we use some logic in dealing with the constant C . C . Since C C represents an arbitrary constant, 3 C 3 C also represents an arbitrary constant. If we call the second arbitrary constant C 1 , C 1 , the equation becomes ln | 3 y + 2 | = x 3 − 12 x + C 1 . ln | 3 y + 2 | = x 3 − 12 x + C 1 . Now exponentiate both sides of the equation (i.e., make each side of the equation the exponent for the base e ) . e ) . e ln | 3 y + 2 | = e x 3 − 12 x + C 1 | 3 y + 2 | = e C 1 e x 3 − 12 x e ln | 3 y + 2 | = e x 3 − 12 x + C 1 | 3 y + 2 | = e C 1 e x 3 − 12 x Again define a new constant C 2 = e c 1 C 2 = e c 1 (note that C 2 > 0 ) : C 2 > 0 ) : | 3 y + 2 | = C 2 e x 3 − 12 x . | 3 y + 2 | = C 2 e x 3 − 12 x . This corresponds to two separate equations: 3 y + 2 = C 2 e x 3 − 12 x 3 y + 2 = C 2 e x 3 − 12 x and 3 y + 2 = − C 2 e x 3 − 12 x . 3 y + 2 = − C 2 e x 3 − 12 x . The solution to either equation can be written in the form y = −2 ± C 2 e x 3 − 12 x 3 . y = −2 ± C 2 e x 3 − 12 x 3 . Since C 2 > 0 , C 2 > 0 , it does not matter whether we use plus or minus, so the constant can actually have either sign. Furthermore, the subscript on the constant C C is entirely arbitrary, and can be dropped. Therefore the solution can be written as y = −2 + C e x 3 − 12 x 3 . y = −2 + C e x 3 − 12 x 3 .
  • No initial condition is imposed, so we are finished.

Checkpoint 4.10

Use the method of separation of variables to find a general solution to the differential equation y ′ = 2 x y + 3 y − 4 x − 6 . y ′ = 2 x y + 3 y − 4 x − 6 .

Example 4.11

Solving an initial-value problem.

Using the method of separation of variables, solve the initial-value problem

  • In this example, f ( x ) = 2 x + 3 f ( x ) = 2 x + 3 and g ( y ) = y 2 − 4 . g ( y ) = y 2 − 4 . Setting g ( y ) = 0 g ( y ) = 0 gives y = ± 2 y = ± 2 as constant solutions.
  • Divide both sides of the equation by y 2 − 4 y 2 − 4 and multiply by d x . d x . This gives the equation d y y 2 − 4 = ( 2 x + 3 ) d x . d y y 2 − 4 = ( 2 x + 3 ) d x .
  • Next integrate both sides: ∫ 1 y 2 − 4 d y = ∫ ( 2 x + 3 ) d x . ∫ 1 y 2 − 4 d y = ∫ ( 2 x + 3 ) d x . (4.4) To evaluate the left-hand side, use the method of partial fraction decomposition. This leads to the identity 1 y 2 − 4 = 1 4 ( 1 y − 2 − 1 y + 2 ) . 1 y 2 − 4 = 1 4 ( 1 y − 2 − 1 y + 2 ) . Then Equation 4.4 becomes 1 4 ∫ ( 1 y − 2 − 1 y + 2 ) d y = ∫ ( 2 x + 3 ) d x 1 4 ( ln | y − 2 | − ln | y + 2 | ) = x 2 + 3 x + C . 1 4 ∫ ( 1 y − 2 − 1 y + 2 ) d y = ∫ ( 2 x + 3 ) d x 1 4 ( ln | y − 2 | − ln | y + 2 | ) = x 2 + 3 x + C . Multiplying both sides of this equation by 4 4 and replacing 4 C 4 C with C 1 C 1 gives ln | y − 2 | − ln | y + 2 | = 4 x 2 + 12 x + C 1 ln | y − 2 y + 2 | = 4 x 2 + 12 x + C 1 . ln | y − 2 | − ln | y + 2 | = 4 x 2 + 12 x + C 1 ln | y − 2 y + 2 | = 4 x 2 + 12 x + C 1 .
  • It is possible to solve this equation for y . First exponentiate both sides of the equation and define C 2 = e C 1 : C 2 = e C 1 : | y − 2 y + 2 | = C 2 e 4 x 2 + 12 x . | y − 2 y + 2 | = C 2 e 4 x 2 + 12 x . Next we can remove the absolute value and let C 2 C 2 be either positive or negative. Then multiply both sides by y + 2 . y + 2 . y − 2 = C 2 ( y + 2 ) e 4 x 2 + 12 x y − 2 = C 2 y e 4 x 2 + 12 x + 2 C 2 e 4 x 2 + 12 x . y − 2 = C 2 ( y + 2 ) e 4 x 2 + 12 x y − 2 = C 2 y e 4 x 2 + 12 x + 2 C 2 e 4 x 2 + 12 x . Now collect all terms involving y on one side of the equation, and solve for y : y : y − C 2 y e 4 x 2 + 12 x = 2 + 2 C 2 e 4 x 2 + 12 x y ( 1 − C 2 e 4 x 2 + 12 x ) = 2 + 2 C 2 e 4 x 2 + 12 x y = 2 + 2 C 2 e 4 x 2 + 12 x 1 − C 2 e 4 x 2 + 12 x . y − C 2 y e 4 x 2 + 12 x = 2 + 2 C 2 e 4 x 2 + 12 x y ( 1 − C 2 e 4 x 2 + 12 x ) = 2 + 2 C 2 e 4 x 2 + 12 x y = 2 + 2 C 2 e 4 x 2 + 12 x 1 − C 2 e 4 x 2 + 12 x .

Checkpoint 4.11

Find the solution to the initial-value problem

using the method of separation of variables.

Applications of Separation of Variables

Many interesting problems can be described by separable equations. We illustrate two types of problems: solution concentrations and Newton’s law of cooling.

Solution concentrations

Consider a tank being filled with a salt solution. We would like to determine the amount of salt present in the tank as a function of time. We can apply the process of separation of variables to solve this problem and similar problems involving solution concentrations .

Example 4.12

Determining salt concentration over time.

A tank containing 100 L 100 L of a brine solution initially has 4 kg 4 kg of salt dissolved in the solution. At time t = 0 , t = 0 , another brine solution flows into the tank at a rate of 2 L/min . 2 L/min . This brine solution contains a concentration of 0.5 kg/L 0.5 kg/L of salt. At the same time, a stopcock is opened at the bottom of the tank, allowing the combined solution to flow out at a rate of 2 L/min , 2 L/min , so that the level of liquid in the tank remains constant ( Figure 4.16 ). Find the amount of salt in the tank as a function of time (measured in minutes), and find the limiting amount of salt in the tank, assuming that the solution in the tank is well mixed at all times.

First we define a function u ( t ) u ( t ) that represents the amount of salt in kilograms in the tank as a function of time. Then d u d t d u d t represents the rate at which the amount of salt in the tank changes as a function of time. Also, u ( 0 ) u ( 0 ) represents the amount of salt in the tank at time t = 0 , t = 0 , which is 4 4 kilograms.

The general setup for the differential equation we will solve is of the form

INFLOW RATE represents the rate at which salt enters the tank, and OUTFLOW RATE represents the rate at which salt leaves the tank. Because solution enters the tank at a rate of 2 2 L/min, and each liter of solution contains 0.5 0.5 kilogram of salt, every minute 2 ( 0.5 ) = 1 kilogram 2 ( 0.5 ) = 1 kilogram of salt enters the tank. Therefore INFLOW RATE = 1 . 1 .

To calculate the rate at which salt leaves the tank, we need the concentration of salt in the tank at any point in time. Since the actual amount of salt varies over time, so does the concentration of salt. However, the volume of the solution remains fixed at 100 liters. The number of kilograms of salt in the tank at time t t is equal to u ( t ) . u ( t ) . Thus, the concentration of salt is u ( t ) 100 u ( t ) 100 kg/L, and the solution leaves the tank at a rate of 2 2 L/min. Therefore salt leaves the tank at a rate of u ( t ) 100 · 2 = u ( t ) 50 u ( t ) 100 · 2 = u ( t ) 50 kg/min, and OUTFLOW RATE is equal to u ( t ) 50 . u ( t ) 50 . Therefore the differential equation becomes d u d t = 1 − u 50 , d u d t = 1 − u 50 , and the initial condition is u ( 0 ) = 4 . u ( 0 ) = 4 . The initial-value problem to be solved is

The differential equation is a separable equation, so we can apply the five-step strategy for solution.

Step 1. Setting 1 − u 50 = 0 1 − u 50 = 0 gives u = 50 u = 50 as a constant solution. Since the initial amount of salt in the tank is 4 4 kilograms, this solution does not apply.

Step 2. Rewrite the equation as

Then multiply both sides by d t d t and divide both sides by 50 − u : 50 − u :

Step 3. Integrate both sides:

Step 4. Solve for u ( t ) : u ( t ) :

Eliminate the absolute value by allowing the constant to be either positive or negative:

Finally, solve for u ( t ) : u ( t ) :

Step 5. Solve for C 1 : C 1 :

The solution to the initial value problem is u ( t ) = 50 − 46 e − t / 50 . u ( t ) = 50 − 46 e − t / 50 . To find the limiting amount of salt in the tank, take the limit as t t approaches infinity:

Note that this was the constant solution to the differential equation. If the initial amount of salt in the tank is 50 50 kilograms, then it remains constant. If it starts at less than 50 kilograms, then it approaches 50 kilograms over time.

Checkpoint 4.12

A tank contains 3 3 kilograms of salt dissolved in 75 75 liters of water. A salt solution of 0.4 kg salt/L 0.4 kg salt/L is pumped into the tank at a rate of 6 L/min 6 L/min and is drained at the same rate. Solve for the salt concentration at time t . t . Assume the tank is well mixed at all times.

Newton’s law of cooling

Newton’s law of cooling states that the rate of change of an object’s temperature is proportional to the difference between its own temperature and the ambient temperature (i.e., the temperature of its surroundings). If we let T ( t ) T ( t ) represent the temperature of an object as a function of time, then d T d t d T d t represents the rate at which that temperature changes. The temperature of the object’s surroundings can be represented by T s . T s . Then Newton’s law of cooling can be written in the form

The temperature of the object at the beginning of any experiment is the initial value for the initial-value problem. We call this temperature T 0 . T 0 . Therefore the initial-value problem that needs to be solved takes the form

where k k is a constant that needs to be either given or determined in the context of the problem. We use these equations in Example 4.13 .

Example 4.13

Waiting for a pizza to cool.

A pizza is removed from the oven after baking thoroughly, and the temperature of the pizza when it comes out of the oven is 200 ° F . 200 ° F . The temperature of the kitchen is 75 ° F , 75 ° F , and after 1 1 minute the temperature of the pizza is 190 ° F . 190 ° F . We would like to wait until the temperature of the pizza reaches 150 ° F 150 ° F before cutting and serving it ( Figure 4.17 ). How much longer will we have to wait?

The ambient temperature (surrounding temperature) is 75 ° F , 75 ° F , so T s = 75 . T s = 75 . The temperature of the pizza when it comes out of the oven is 200 ° F , 200 ° F , which is the initial temperature (i.e., initial value), so T 0 = 200 . T 0 = 200 . Therefore Equation 4.4 becomes

To solve the differential equation, we use the five-step technique for solving separable equations.

  • Setting the right-hand side equal to zero gives T = 75 T = 75 as a constant solution. Since the pizza starts at 200 ° F , 200 ° F , this is not the solution we are seeking.
  • Rewrite the differential equation by multiplying both sides by d t d t and dividing both sides by T − 75 : T − 75 : d T T − 75 = k d t . d T T − 75 = k d t .
  • Integrate both sides: ∫ d T T − 75 = ∫ k d t ln | T − 75 | = k t + C . ∫ d T T − 75 = ∫ k d t ln | T − 75 | = k t + C .
  • Solve for T T by first exponentiating both sides: e ln | T − 75 | = e k t + C | T − 75 | = C 1 e k t T − 75 = C 1 e k t T ( t ) = 75 + C 1 e k t . e ln | T − 75 | = e k t + C | T − 75 | = C 1 e k t T − 75 = C 1 e k t T ( t ) = 75 + C 1 e k t .
  • Solve for C 1 C 1 by using the initial condition T ( 0 ) = 200 : T ( 0 ) = 200 : T ( t ) = 75 + C 1 e k t T ( 0 ) = 75 + C 1 e k ( 0 ) 200 = 75 + C 1 C 1 = 125 . T ( t ) = 75 + C 1 e k t T ( 0 ) = 75 + C 1 e k ( 0 ) 200 = 75 + C 1 C 1 = 125 . Therefore the solution to the initial-value problem is T ( t ) = 75 + 125 e k t . T ( t ) = 75 + 125 e k t . To determine the value of k , k , we need to use the fact that after 1 1 minute the temperature of the pizza is 190 ° F . 190 ° F . Therefore T ( 1 ) = 190 . T ( 1 ) = 190 . Substituting this information into the solution to the initial-value problem, we have T ( t ) = 75 + 125 e k t T ( 1 ) = 190 = 75 + 125 e k 115 = 125 e k 115 125 = 23 25 = e k ln e k = ln ( 23 25 ) k = ln ( 23 25 ) ≈ − . 08338 T ( t ) = 75 + 125 e k t T ( 1 ) = 190 = 75 + 125 e k 115 = 125 e k 115 125 = 23 25 = e k ln e k = ln ( 23 25 ) k = ln ( 23 25 ) ≈ − . 08338 So now we have T ( t ) = 75 + 125 e - . 08338 t . T ( t ) = 75 + 125 e - . 08338 t . When is the temperature 150 ° F? 150 ° F? Solving for t , t , we find T ( t ) = 75 + 125 e − . 08338 t 150 = 75 + 125 e − . 08338 t 75 = 125 e − . 08338 t 75 125 = 3 5 = e - . 08338 t - . 08338 t = ln 3 5 t = ln 3 5 - . 08338 ≈ 6 . 12 . T ( t ) = 75 + 125 e − . 08338 t 150 = 75 + 125 e − . 08338 t 75 = 125 e − . 08338 t 75 125 = 3 5 = e - . 08338 t - . 08338 t = ln 3 5 t = ln 3 5 - . 08338 ≈ 6 . 12 . Therefore we need to wait an additional 6.12 6.12 minutes (after the temperature of the pizza reached 200 ° F ) . 200 ° F ) . That should be just enough time to finish this calculation.

Checkpoint 4.13

A cake is removed from the oven after baking thoroughly, and the temperature of the cake when it comes out of the oven is 450 ° F . 450 ° F . The temperature of the kitchen is 70 ° F , 70 ° F , and after 10 10 minutes the temperature of the cake is 330 ° F . 330 ° F .

  • Write the appropriate initial-value problem to describe this situation.
  • Solve the initial-value problem for T ( t ) . T ( t ) .
  • How long will it take until the temperature of the cake is within 5 ° F 5 ° F of room temperature?

Section 4.3 Exercises

Solve the following initial-value problems with the initial condition y 0 = 0 y 0 = 0 and graph the solution.

d y d t = y + 1 d y d t = y + 1

d y d t = y − 1 d y d t = y − 1

d y d t = –y + 1 d y d t = –y + 1

d y d t = − y − 1 d y d t = − y − 1

Find the general solution to the differential equation.

x 2 y ′ = ( x + 1 ) y x 2 y ′ = ( x + 1 ) y

y ′ = tan ( y ) x y ′ = tan ( y ) x

y ′ = 2 x y 2 y ′ = 2 x y 2

d y d t = y cos ( 3 t + 2 ) d y d t = y cos ( 3 t + 2 )

2 x d y d x = y 2 2 x d y d x = y 2

y ′ = e y x 2 y ′ = e y x 2

( 1 + x ) y ′ = ( x + 2 ) ( y − 1 ) ( 1 + x ) y ′ = ( x + 2 ) ( y − 1 )

d x d t = 3 t 2 ( x 2 + 4 ) d x d t = 3 t 2 ( x 2 + 4 )

t d y d t = 1 − y 2 t d y d t = 1 − y 2

y ′ = e x e y y ′ = e x e y

Find the solution to the initial-value problem.

y ′ = e y − x , y ( 0 ) = 0 y ′ = e y − x , y ( 0 ) = 0

y ′ = y 2 ( x + 1 ) , y ( 0 ) = 2 y ′ = y 2 ( x + 1 ) , y ( 0 ) = 2

d y d x = y 3 x e x 2 , y ( 0 ) = 1 d y d x = y 3 x e x 2 , y ( 0 ) = 1

d y d t = y 2 e x sin ( 3 x ) , y ( 0 ) = 1 d y d t = y 2 e x sin ( 3 x ) , y ( 0 ) = 1

y ′ = x sech 2 y , y ( 0 ) = 0 y ′ = x sech 2 y , y ( 0 ) = 0

y ′ = 2 x y ( 1 + 2 y ) , y ( 0 ) = −1 y ′ = 2 x y ( 1 + 2 y ) , y ( 0 ) = −1

d x d t = ln ( t ) 1 − x 2 , x ( 1 ) = 0 d x d t = ln ( t ) 1 − x 2 , x ( 1 ) = 0

y ′ = 3 x 2 ( y 2 + 4 ) , y ( 0 ) = 0 y ′ = 3 x 2 ( y 2 + 4 ) , y ( 0 ) = 0

y ′ = e y 5 x , y ( 0 ) = ln ( ln ( 5 ) ) y ′ = e y 5 x , y ( 0 ) = ln ( ln ( 5 ) )

y ′ = −2 x tan ( y ) , y ( 0 ) = π 2 y ′ = −2 x tan ( y ) , y ( 0 ) = π 2

For the following problems, use a software program or your calculator to generate the directional fields. Solve explicitly and draw solution curves for several initial conditions. Are there some critical initial conditions that change the behavior of the solution?

[T] y ′ = 1 − 2 y y ′ = 1 − 2 y

[T] y ′ = y 2 x 3 y ′ = y 2 x 3

[T] y ′ = y 3 e x y ′ = y 3 e x

[T] y ′ = e y y ′ = e y

[T] y ′ = y ln ( x ) y ′ = y ln ( x )

Most drugs in the bloodstream decay according to the equation y ′ = c y , y ′ = c y , where y y is the concentration of the drug in the bloodstream. If the half-life of a drug is 2 2 hours, what fraction of the initial dose remains after 6 6 hours?

A drug is administered intravenously to a patient at a rate r r mg/h and is cleared from the body at a rate proportional to the amount of drug still present in the body, d d Set up and solve the differential equation, assuming there is no drug initially present in the body.

[T] How often should a drug be taken if its dose is 3 3 mg, it is cleared at a rate c = 0.1 c = 0.1 mg/h, and 1 1 mg is required to be in the bloodstream at all times?

A tank contains 1 1 kilogram of salt dissolved in 100 100 liters of water. A salt solution of 0.1 0.1 kg salt/L is pumped into the tank at a rate of 2 2 L/min and is drained at the same rate. Solve for the salt concentration at time t . t . Assume the tank is well mixed.

A tank containing 10 10 kilograms of salt dissolved in 1000 1000 liters of water has two salt solutions pumped in. The first solution of 0.2 0.2 kg salt/L is pumped in at a rate of 20 20 L/min and the second solution of 0.05 0.05 kg salt/L is pumped in at a rate of 5 5 L/min. The tank drains at 25 25 L/min. Assume the tank is well mixed. Solve for the salt concentration at time t . t .

[T] For the preceding problem, find how much salt is in the tank 1 1 hour after the process begins.

Torricelli’s law states that for a water tank with a hole in the bottom that has a cross-sectional area of A T A T and with a height of water h h above the bottom of the tank, the rate of change of volume of water flowing from the tank is proportional to the square root of the height of water, according to d V d t = − A T 2 g h , d V d t = − A T 2 g h , where g g is the acceleration due to gravity. Note that d V d t = A d h d t , d V d t = A d h d t , where A T A T is the cross-sectional area of the tank. Solve the resulting initial-value problem for the height of water, assuming a tankof radius 24 2 24 2 with a circular hole of radius 2 2 ft.

For the preceding problem, determine how long it takes the tank to drain.

For the following problems, use Newton’s law of cooling.

The liquid base of an ice cream has an initial temperature of 200 ° F 200 ° F before it is placed in a freezer with a constant temperature of 0 ° F . 0 ° F . After 1 1 hour, the temperature of the ice-cream base has decreased to 140 ° F . 140 ° F . Formulate and solve the initial-value problem to determine the temperature of the ice cream.

[T] The liquid base of an ice cream has an initial temperature of 210 ° F 210 ° F before it is placed in a freezer with a constant temperature of 20 ° F . 20 ° F . After 2 2 hours, the temperature of the ice-cream base has decreased to 170 ° F . 170 ° F . At what time will the ice cream be ready to eat? (Assume 30 ° F 30 ° F is the optimal eating temperature.)

[T] You are organizing an ice cream social. The outside temperature is 80 ° F 80 ° F and the ice cream is at 10 ° F . 10 ° F . After 10 10 minutes, the ice cream temperature has risen by 10 ° F . 10 ° F . How much longer can you wait before the ice cream melts at 40 ° F? 40 ° F?

For Exercises 159—162, assume a cooling constant of k = - 0 . 125 k = - 0 . 125 and assume time t t is in minutes.

You have a cup of coffee at temperature 70 ° C 70 ° C and the ambient temperature in the room is 20 ° C . 20 ° C . Assuming a cooling rate k of 0.125 , k of 0.125 , write and solve the differential equation to describe the temperature of the coffee with respect to time.

[T] You have a cup of coffee at temperature 70 ° C 70 ° C that you put outside, where the ambient temperature is 0 ° C . 0 ° C . After 5 5 minutes, how much colder is the coffee?

You have a cup of coffee at temperature 70 ° C 70 ° C and you immediately pour in 1 1 part milk to 5 5 parts coffee. The milk is initially at temperature 1 ° C . 1 ° C . Write and solve the differential equation that governs the temperature of this coffee.

You have a cup of coffee at temperature 70 ° C , 70 ° C , which you let cool 10 10 minutes before you pour in the same amount of milk at 1 ° C 1 ° C as in the preceding problem. How does the temperature compare to the previous cup after 10 10 minutes?

Solve the generic problem y ′ = a y + b y ′ = a y + b with initial condition y ( 0 ) = c . y ( 0 ) = c .

Prove the basic continual compounded interest equation. Assuming an initial deposit of P 0 P 0 and an interest rate of r , r , set up and solve an equation for continually compounded interest.

Assume an initial nutrient amount of I I kilograms in a tank with L L liters. Assume a concentration of c c kg/L being pumped in at a rate of r r L/min. The tank is well mixed and is drained at a rate of r r L/min. Find the equation describing the amount of nutrient in the tank.

Leaves accumulate on the forest floor at a rate of 2 2 g/cm 2 /yr and also decompose at a rate of 90 % 90 % per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor, assuming at time 0 0 there is no leaf litter on the ground. Does this amount approach a steady value? What is that value?

Leaves accumulate on the forest floor at a rate of 4 4 g/cm 2 /yr. These leaves decompose at a rate of 10 % 10 % per year. Write a differential equation governing the number of grams of leaf litter per square centimeter of forest floor. Does this amount approach a steady value? What is that value?

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  • Authors: Gilbert Strang, Edwin “Jed” Herman
  • Publisher/website: OpenStax
  • Book title: Calculus Volume 2
  • Publication date: Mar 30, 2016
  • Location: Houston, Texas
  • Book URL: https://openstax.org/books/calculus-volume-2/pages/1-introduction
  • Section URL: https://openstax.org/books/calculus-volume-2/pages/4-3-separable-equations

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Mathematics LibreTexts

4.6: PDEs, Separation of Variables, and The Heat Equation

  • Last updated
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  • Page ID 345

  • Jiří Lebl
  • Oklahoma State University

Let us recall that a partial differential equation or PDE is an equation containing the partial derivatives with respect to several independent variables. Solving PDEs will be our main application of Fourier series.

A PDE is said to be linear if the dependent variable and its derivatives appear at most to the first power and in no functions. We will only talk about linear PDEs. Together with a PDE, we usually have specified some boundary conditions , where the value of the solution or its derivatives is specified along the boundary of a region, and/or some initial conditions where the value of the solution or its derivatives is specified for some initial time. Sometimes such conditions are mixed together and we will refer to them simply as side conditions .

We will study three specific partial differential equations, each one representing a more general class of equations. First, we will study the heat equation , which is an example of a parabolic PDE. Next, we will study the wave equation , which is an example of a hyperbolic PDE . Finally, we will study the Laplace equation , which is an example of an elliptic PDE . Each of our examples will illustrate behavior that is typical for the whole class.

Heat on an Insulated Wire

Let us first study the heat equation. Suppose that we have a wire (or a thin metal rod) of length \(L\) that is insulated except at the endpoints. Let \(x\) denote the position along the wire and let \(t\) denote time. See Figure \(\PageIndex{1}\).

heat-wire-tex4ht.png

Let \(u(x,t)\) denote the temperature at point \(x\) at time \(t\). The equation governing this setup is the so-called one-dimensional heat equation :

\[\frac{\partial u}{\partial t} = k \frac{\partial^2 u}{\partial x^2}, \nonumber \]

where \(k>0\) is a constant (the thermal conductivity of the material). That is, the change in heat at a specific point is proportional to the second derivative of the heat along the wire. This makes sense; if at a fixed \(t\) the graph of the heat distribution has a maximum (the graph is concave down), then heat flows away from the maximum. And vice-versa.

We will generally use a more convenient notation for partial derivatives. We will write \(u_t\) instead of \( \frac{\partial u}{\partial t}\), and we will write \(u_{xx}\) instead of \(\frac{\partial^2 u}{\partial x^2} \). With this notation the heat equation becomes

\[ u_t=ku_{xx}. \nonumber \]

For the heat equation, we must also have some boundary conditions. We assume that the ends of the wire are either exposed and touching some body of constant heat, or the ends are insulated. For example, if the ends of the wire are kept at temperature 0, then we must have the conditions

\[ u(0,t)=0 \quad\text{and}\quad u(L,t)=0. \nonumber \]

If, on the other hand, the ends are also insulated we get the conditions

\[ u_x(0,t)=0 \quad\text{and}\quad u_x(L,t)=0. \nonumber \]

Let us see why that is so. If \(u_{x}\) is positive at some point \(x_{0}\), then at a particular time, \(u\) is smaller to the left of \(x_{0}\), and higher to the right of \(x_{0}\). Heat is flowing from high heat to low heat, that is to the left. On the other hand if \(u_{x}\) is negative then heat is again flowing from high heat to low heat, that is to the right. So when \(u_{x}\) is zero, that is a point through which heat is not flowing. In other words, \(u_{x}(0,t)=0\) means no heat is flowing in or out of the wire at the point \(x=0\).

We have two conditions along the \(x\)-axis as there are two derivatives in the \(x\) direction. These side conditions are said to be homogeneous (i.e., \(u\) or a derivative of \(u\) is set to zero).

We also need an initial condition—the temperature distribution at time \(t=0\). That is,

\[ u(x,0)=f(x), \nonumber \]

for some known function \(f(x)\). This initial condition is not a homogeneous side condition.

Separation of Variables

The heat equation is linear as \(u\) and its derivatives do not appear to any powers or in any functions. Thus the principle of superposition still applies for the heat equation (without side conditions). If \(u_1\) and \(u_2\) are solutions and \(c_1,c_2\) are constants, then \( u= c_1u_1+c_2u_2\) is also a solution.

Exercise \(\PageIndex{1}\)

Verify the principle of superposition for the heat equation.

Superposition also preserves some of the side conditions. In particular, if \(u_1\) and \(u_2\) are solutions that satisfy \(u(0,t)=0\) and \(u_(L,t)=0\), and \(c_1,\: c_2\) are constants, then \( u= c_1u_1+c_2u_2\) is still a solution that satisfies \(u(0,t)=0\) and\(u_(L,t)=0\). Similarly for the side conditions \(u_x(0,t)=0\) and \(u_x(L,t)=0\). In general, superposition preserves all homogeneous side conditions.

The method of separation of variables is to try to find solutions that are sums or products of functions of one variable. For example, for the heat equation, we try to find solutions of the form

\[ u(x,t)=X(x)T(t). \nonumber \]

That the desired solution we are looking for is of this form is too much to hope for. What is perfectly reasonable to ask, however, is to find enough “building-block” solutions of the form \( u(x,t)=X(x)T(t)\) using this procedure so that the desired solution to the PDE is somehow constructed from these building blocks by the use of superposition.

Let us try to solve the heat equation

\[u_t=ku_{xx} \quad\text{with}\quad u(0,t)=0,\quad u(L,t)=0, \quad\text{and}\quad u(x,0)=f(x). \nonumber \]

Let us guess \(u(x,t)=X(x)T(t)\). We will try to make this guess satisfy the differential equation, \(u_{t}=ku_{xx}\), and the homogeneous side conditions, \(u(0,t)=0\) and \(u(L,t)=0\). Then, as superposition preserves the differential equation and the homogeneous side conditions, we will try to build up a solution from these building blocks to solve the nonhomogeneous initial condition \(u(x,0)=f(x)\).

First we plug \(u(x,t)=X(x)T(t)\) into the heat equation to obtain

\[ X(x)T'(t)=kX''(x)T(t). \nonumber \]

We rewrite as \[ \frac{T'(t)}{kT(t)}= \frac{X''(x)}{X(x)}. \nonumber \]

This equation must hold for all \(x\) and all \(t\). But the left hand side does not depend on \(x\) and the right hand side does not depend on \(t\). Hence, each side must be a constant. Let us call this constant \(- \lambda\) (the minus sign is for convenience later). We obtain the two equations

\[ \frac{T'(t)}{kT(t)}= - \lambda = \frac{X''(x)}{X(x)}. \nonumber \]

In other words

\[\begin{align}\begin{aligned} X''(x) + \lambda X(x) &=0, \\ T'(t) + \lambda k T(t)& =0.\end{aligned}\end{align} \nonumber \]

The boundary condition \(u(0,t)=0\) implies \( X(0)T(t)=0\). We are looking for a nontrivial solution and so we can assume that \(T(t)\) is not identically zero. Hence \(X(0)=0\). Similarly, \(u(L,t)=0\) implies \(X(L)=0\). We are looking for nontrivial solutions \(X\) of the eigenvalue problem \( X'' + \lambda X = 0, X(0)=0, X(L)=0\). We have previously found that the only eigenvalues are \( \lambda_n = \frac{n^2 \pi^2}{L^2}\), for integers \( n \geq 1\), where eigenfunctions are \( \sin \left( \frac{n \pi}{L}x \right)\). Hence, let us pick the solutions

\[ X_n(x)= \sin \left( \frac{n \pi}{L}x \right). \nonumber \]

The corresponding \(T_n\) must satisfy the equation

\[ T'_n(t) + \frac{n^2 \pi^2}{L^2}kT_n(t)=0. \nonumber \]

By the method of integrating factor, the solution of this problem is

\[T_n(t)=e^{\frac{-n^2 \pi^2}{L^2}kt}. \nonumber \]

It will be useful to note that \(T_n(0)=1\). Our building-block solutions are

\[u_n(x,t)=X_n(x)T_n(t)= \sin \left( \frac{n \pi}{L}x \right) e^{\frac{-n^2 \pi^2}{L^2}kt}. \nonumber \]

We note that \( u_n(x,0)= \sin \left( \frac{n \pi}{L}x \right)\). Let us write \(f(x)\) as the sine series

\[ f(x)= \sum_{n=1}^{\infty} b_n \sin \left( \frac{n \pi}{L}x \right). \nonumber \]

That is, we find the Fourier series of the odd periodic extension of \(f(x)\). We used the sine series as it corresponds to the eigenvalue problem for \(X(x)\) above. Finally, we use superposition to write the solution as

\[ u(x,t)= \sum^{\infty}_{n=1}b_n u_n (x,t)= \sum^{\infty}_{n=1}b_n \sin \left(\frac{n \pi}{L}x \right)e^{\frac{-n^2 \pi^2}{L^2}kt}. \nonumber \]

Why does this solution work? First note that it is a solution to the heat equation by superposition. It satisfies \(u(0,t)=0\) and \(u(L,t)=0\), because \(x=0\) or \(x=L\) makes all the sines vanish. Finally, plugging in \(t=0\), we notice that \(T_n(0)=1\) and so

\[ u(x,0)= \sum^{\infty}_{n=1}b_n u_n (x,0)= \sum^{\infty}_{n=1}b_n \sin \left(\frac{n \pi}{L}x \right)=f(x). \nonumber \]

Example \(\PageIndex{1}\)

Suppose that we have an insulated wire of length \(1\), such that the ends of the wire are embedded in ice (temperature 0). Let \(k=0.003\). Then suppose that initial heat distribution is \(u(x,0)=50x(1-x)\). See Figure \(\PageIndex{2}\).

clipboard_e967ebbe224347ae2e113ae6dc37980c4.png

We want to find the temperature function \(u(x,t)\). Let us suppose we also want to find when (at what \(t\) ) does the maximum temperature in the wire drop to one half of the initial maximum of \(12.5\).

We are solving the following PDE problem:

\[\begin{align}\begin{aligned} u_t &=0.003u_{xx}, \\ u(0,t) &= u(1,t)=0, \\ u(x,0) &= 50x(1-x) ~~~~ {\rm{for~}} 0<x<1.\end{aligned}\end{align} \nonumber \]

We write \(f(x)=50x(1-x)\) for \(0<x<1\) as a sine series. That is, \(f(x)= \sum^{\infty}_{n=1}b_n \sin(n \pi x)\), where

\[ b_n= 2 \int^1_0 50x(1-x) \sin(n \pi x)dx = \frac{200}{\pi^3 n^3}-\frac{200(-1)^n}{\pi^3 n^3}= \left\{ \begin{array}{cc} 0 & {\rm{if~}} n {\rm{~even,}} \\ \frac{400}{\pi^3 n^3} & {\rm{if~}} n {\rm{~odd.}} \end{array} \right. \nonumber \]

clipboard_ed6ee3470352b18dd3d391c1104f2c7a3.png

The solution \(u(x,t)\), plotted in Figure \(\PageIndex{3}\) for \( 0 \ leq t \ leq 100\), is given by the series:

\[ u(x,t)= \sum^{\infty}_{\underset{n~ {\rm{odd}} }{n=1}} \frac{400}{\pi^3 n^3} \sin(n \pi x) e^{-n^2 \pi^2 0.003t}. \nonumber \]

Finally, let us answer the question about the maximum temperature. It is relatively easy to see that the maximum temperature will always be at \(x=0.5\), in the middle of the wire. The plot of \(u(x,t)\) confirms this intuition.

If we plug in \(x=0.5\) we get

\[ u(0.5,t)= \sum^{\infty}_{\underset{n~ {\rm{odd}} }{n=1}} \frac{400}{\pi^3 n^3} \sin(n \pi 0.5) e^{-n^2 \pi^2 0.003t}. \nonumber \]

For \(n=3\) and higher (remember \(n\) is only odd), the terms of the series are insignificant compared to the first term. The first term in the series is already a very good approximation of the function. Hence

\[u(0.5,t) \approx \frac{400}{\pi^3}e^{-\pi^2 0.003t}. \nonumber \]

The approximation gets better and better as \(t\) gets larger as the other terms decay much faster. Let us plot the function \(0.5,t\), the temperature at the midpoint of the wire at time \(t\), in Figure \(\PageIndex{4}\). The figure also plots the approximation by the first term.

clipboard_e04300606d2604caf9fc4abe56178bbd3.png

After \(t=5\) or so it would be hard to tell the difference between the first term of the series for \(u(x,t)\) and the real solution \(u(x,t)\). This behavior is a general feature of solving the heat equation. If you are interested in behavior for large enough \(t\), only the first one or two terms may be necessary.

Let us get back to the question of when is the maximum temperature one half of the initial maximum temperature. That is, when is the temperature at the midpoint \(12.5/2=6.25\). We notice on the graph that if we use the approximation by the first term we will be close enough. We solve

\[ 6.25=\frac{400}{\pi^3}e^{-\pi^2 0.003t}. \nonumber \]

\[ t=\frac{\ln{\frac{6.25 \pi^3}{400}}}{-\pi^2 0.003} \approx 24.5. \nonumber \]

So the maximum temperature drops to half at about \(t=24.5\).

We mention an interesting behavior of the solution to the heat equation. The heat equation “smoothes” out the function \(f(x)\) as \(t\) grows. For a fixed \(t\), the solution is a Fourier series with coefficients \(b_n e^{\frac{-n^2 \pi^2}{L^2}kt}\). If \(t>0\), then these coefficients go to zero faster than any \(\frac{1}{n^P}\) for any power \(p\). In other words, the Fourier series has infinitely many derivatives everywhere. Thus even if the function \(f(x)\) has jumps and corners, then for a fixed \(t>0\), the solution \(u(x,t)\) as a function of \(x\) is as smooth as we want it to be.

Example \(\PageIndex{2}\)

When the initial condition is already a sine series, then there is no need to compute anything, you just need to plug in. Consider \[u_t = 0.3 \, u_{xx}, \qquad u(0,t)=u(1,t)=0, \qquad u(x,0) = 0.1 \sin(\pi t) + \sin(2\pi t) . \nonumber \] The solution is then \[u(x,t) = 0.1 \sin(\pi t) e^{- 0.3 \pi^2 t} + \sin(2 \pi t) e^{- 1.2 \pi^2 t} . \nonumber \]

Insulated Ends

Now suppose the ends of the wire are insulated. In this case, we are solving the equation

\[ u_t=ku_{xx}\quad\text{with}\quad u_x(0,t)=0,\quad u_x(L,t)=0,\quad\text{and}\quad u(x,0)=f(x). \nonumber \]

Yet again we try a solution of the form \(u(x,t)=X(x)T(t)\). By the same procedure as before we plug into the heat equation and arrive at the following two equations

\[\begin{align}\begin{aligned} X''(x)+\lambda X(x) &=0, \\ T'(t)+\lambda kT(t) &=0.\end{aligned}\end{align} \nonumber \]

At this point the story changes slightly. The boundary condition \(u_x(0,t)=0\) implies \(X'(0)T(t)=0\). Hence \(X'(0)=0\). Similarly, \(u_x(L,t)=0\) implies \(X'(L)=0\). We are looking for nontrivial solutions \(X\) of the eigenvalue problem \(X''+ \lambda X=0,\) \(X'(0)=0,\) \(X'(L)=0,\). We have previously found that the only eigenvalues are \(\lambda_n=\frac{n^2 \pi^2}{L^2}\), for integers \( n \geq 0\), where eigenfunctions are \(\cos(\frac{n \pi}{L})X\) (we include the constant eigenfunction). Hence, let us pick solutions

\[X_n(x)= \cos(\frac{n \pi}{L}x)\quad\text{and}\quad X_0(x)=1. \nonumber \]

\[T'_n(t)+ \frac{n^2 \pi^2}{L^2}kT_n(t)=0. \nonumber \]

For \(n \geq 1\), as before,

\[T_n(t)= e^{\frac{-n^2 \pi^2}{L^2}kt}. \nonumber \]

For \(n=0\), we have \(T'_0(t)=0\) and hence \(T_0(t)=1\). Our building-block solutions will be

\[u_n(x,t)=X_n(x)T_n(t)= \cos \left( \frac{n \pi}{L} x \right) e^{\frac{-n^2 \pi^2}{L^2}kt}, \nonumber \]

\[u_0(x,t)=1. \nonumber \]

We note that \(u_n(x,0) =\cos \left( \frac{n \pi}{L} x \right)\). Let us write \(f\) using the cosine series

\[f(x)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n \cos \left( \frac{n \pi}{L} x \right). \nonumber \]

That is, we find the Fourier series of the even periodic extension of \(f(x)\).

We use superposition to write the solution as

\[u(x,t)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n u_n(x,t)= \frac{a_0}{2} + \sum^{\infty}_{n=1} a_n \cos \left( \frac{n \pi}{L} x \right) e^{\frac{-n^2 \pi^2}{L^2}kt}. \nonumber \]

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Solving the variable coefficient nonlinear partial differential equations based on the bilinear residual network method

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  • Published: 29 March 2024

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  • Xue-Sha Wu 1 &
  • Jian-Guo Liu   ORCID: orcid.org/0000-0002-6058-5472 2  

In this work, a variable coefficient bilinear residual network method is proposed to solve nonlinear partial differential equations with variable coefficient, including two types of neural network models: 2–2 and 3–3. Various soliton solutions of a (2+1)-dimensional coupled nonlinear equation with variable coefficients are presented based on the variable coefficient bilinear residual network method. The interaction between lump wave and solitons is discussed through a mixed function of rational and exponential functions. Finally, the interaction between lump wave and periodic wave is analyzed through a mixed function of rational and trigonometric functions. Meanwhile, some three-dimensional and density maps are used to describe the dynamic properties of the obtained results.

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Project supported by research on the cultivation path of information-based teaching ability of general education teachers in higher vocational colleges (Grant No. 23JS0401), Doctoral Research Foundation of Jiangxi University of Chinese Medicine (Grant No. 2021WBZR007) and Jiangxi University of Chinese Medicine Science and Technology Innovation Team Development Program (Grant No. CXTD22015).

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Wu, XS., Liu, JG. Solving the variable coefficient nonlinear partial differential equations based on the bilinear residual network method. Nonlinear Dyn (2024). https://doi.org/10.1007/s11071-024-09472-4

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  11. 2. Separation of Variables

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  18. 8.3.3 Separation of Variables

    STEP 1: Separate all y terms on one side and all x terms on the other side. STEP 2: Integrate both sides. STEP 3: Include one "overall" constant of integration. STEP 4: Use the initial or boundary condition to find the particular solution. STEP 5: Write the particular solution in sensible, or required, format.

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  20. The Fractional Separation of Variables Method For Solving System ...

    However, the existing methods, such as the method of separation of variables, the homotopy perturbation method, the variational iteration method, and the Laplace variational iteration method, encounter significant challenges when searching for solutions to systems of nonlinear fractional partial differential equations. Thus, the method of ...

  21. Ordinary Differential Equations (ODE) Calculator

    Free ordinary differential equations (ODE) calculator - solve ordinary differential equations (ODE) step-by-step We've updated our ... use methods such as separation of variables, linear equations, exact equations, homogeneous equations, or numerical methods. ... is a mathematical equation involving a single independent variable and one or more ...

  22. Solving the variable coefficient nonlinear partial differential

    In this work, a variable coefficient bilinear residual network method is proposed to solve nonlinear partial differential equations with variable coefficient, including two types of neural network models: 2-2 and 3-3. Various soliton solutions of a (2+1)-dimensional coupled nonlinear equation with variable coefficients are presented based on the variable coefficient bilinear residual ...