differential equations by separation of variables homework

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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

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Video transcript

Introduction

A differential equation is an equation involving derivatives.

Solving them is an art, like integrating.

Integration can be used directly to solve some differential equations.

15.1 Definition

15.2 Separation of Variables

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.

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.

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?

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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Differential Equations by Separation of Variables -Classwork

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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:

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

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Separation of Variables (Worksheet)

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Name: ______________________________

Section: _____________________________

Student ID#:__________________________

Work in groups on these problems. You should try to answer the questions without referring to your textbook. If you get stuck, try asking another group for help.

Learning Objectives

• Identify separable simple differential equations.
• Solve separable differential equations and initial value problems.
• Determine the interval(s) (with respect to the independent variable) on which a solution to a separable differential equation is defined.

The "separation of variables" is any of several methods for solving ordinary and partial differential equations, in which algebra allows one to rewrite an equation so that each of two variables occurs on a different side of an equation. The method of separation of variables is also used to solve a wide range of linear partial differential equations with boundary and initial conditions.

Separable Differential Equations

A separable differential equation is a differential equation whose algebraic structure permits the variables present to be separated in a particular way. For instance, consider the equation

\[\dfrac{dy}{dt} = t y.\]

We would like to separate the variables \(t\) and \(y\) so that all occurrences of \(t\) appear on the right-hand side, and all occurrences of \(y\) appears on the left and multiply \(dy/dt\). We may do this in the preceding differential equation by dividing both sides by \(y\):

\[\dfrac{1}{y} \dfrac{dy}{dt} = t. \]

Note particularly that when we attempt to separate the variables in a differential equation, we require that the left-hand side be a product in which the derivative \(dy/dt\) is one term. Not every differential equation is separable. For example, if we consider the equation

\[\dfrac{dy}{ dt} = t − y,\]

it may seem natural to separate it by writing

\[y + \dfrac{dy}{dt} = t.\]

As we will see, this will not be helpful since the left-hand side is not a product of a function of \(y\) with \(\frac{dy}{dt}\).

Which of the following differential equations are separable? If the equation is separable, write the equation in the revised form \(g(y) \dfrac{dy}{dt} = h(t).\)

• \( \dfrac{dy}{dt} = −3y\)
• \( \dfrac{dy}{dt}t = t y − y. \)
• \( \dfrac{dy}{dt} = t + 1. \)
• \( \dfrac{dy}{dt} = t^2 − y^2 .\)

Why do we include the term “\(+C\)” in the expression \[\int x\, dx = \dfrac{x^2}{2} + C?\]

Suppose we know that a certain function \(f\) satisfies the equation \[\int f'(x) \,dx = \int x\, dx.\] What can you conclude about \(f\)?

Answer the following question for this function of \(t\) and \(x\):

\[f(x, t) = e^{−3t} \cos(2x)\]

Which part(s) of \(f (x,t)\) vary with \(x\)?

Which part(s) of \(f(x,t)\) are constant when \(x\) is varied?

What is \( \dfrac{\partial f}{\partial x}\)?

If \(f (x,t) = a(x)b(t)\), using the definition of \(f(x,t)\) above, what is \(a(x)\)?

What is \(b(t)\)?

For any function defined as \(u(x, t) = X(x)T (t )\), write the general expression for \(\dfrac{\partial u}{ \partial x}\) in terms of \(u(x, t ) = X(x)T (t )\).

Similarly, how would you write \( \dfrac{\partial u}{\partial t}\) in terms of \(u(x, t ) = X(x)T(t)\)?

The equation for the vibration of a string is (\(x\) is the distance along the string and \(t\) is time)

\[\dfrac{\partial ^2u(x,t)}{\partial t^2}=\dfrac{1}{v^2}\dfrac{\partial ^2u(x,t)}{\partial x^2}\]

Given that \(u(x, t ) = X(x)T(t)\), the following questions will guide you through how to rewrite this equation in terms of only \(X(x)\) and \(T(t )\).

Rewrite the equation for the vibration of a string so that one side depends only on \(x\) and the other depends only on \(t\):

Since \(x\) and \(t\) vary independently, each side of the equation given above must be equal to a constant, the same constant. Set both sides equal to \(K\) (we call this the separation constant ) and rewrite it as two equations, one with only \(x\) and one with only \(t\):

By separating variables, we’ve managed to turn a partial differential equation in two variables into two differential equations, each with in one variable. This should be much easier to solve. What about \(u(x,t)\) makes separation of variables possible?

Using the answer to the above question, is \(u(x,t) = x e^{-3t} \cos (2x)\) separable? What about \(u(x,t) = e^{-ix} \cos \left(\dfrac{2x}{\pi t}\right)\)? Why or why not?

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5: Separation of Variables on Rectangular Domains

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• Page ID 8314

• Niels Walet
• University of Manchester

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In this section we shall investigate two dimensional equations defined on rectangular domains. We shall either look at finite rectangles, when we have two space variables, or at semi-infinite rectangles when one of the variables is time. We shall study all three different types of partial differental equations: parabolic, hyperbolic and elliptical.

• 5.1: Cookbook Let me start with a recipe that describes the approach to separation of variables, as exemplified in the following sections, and in later chapters.
• 5.2: Parabolic Equation Parabolic PDEs are used to describe a wide variety of time-dependent phenomena, including heat conduction, and particle diffusion.
• 5.3: Hyperbolic Equation Many of the equations of mechanics are hyperbolic and the model hyperbolic equation is the wave equation.  The solutions of hyperbolic equations are "wave-like". If a disturbance is made in the initial data of a hyperbolic differential equation, then not every point of space feels the disturbance at once.
• 5.4: Laplace’s Equation Laplace's equation are the simplest examples of elliptic partial differential equations. The solutions of Laplace's equation are the harmonic functions, which are important in many fields of science, notably the fields of electromagnetism, astronomy, and fluid dynamics, because they can be used to accurately describe the behavior of electric, gravitational, and fluid potentials. In the study of heat conduction, the Laplace equation is the steady-state heat equation.
• 5.5: More Complex Initial/Boundary Conditions It is not always possible on separation of variables to separate initial or boundary conditions in a condition on one of the two functions. We can either map the problem into simpler ones by using superposition of boundary conditions, a way discussed below, or we can carry around additional integration constants.
• 5.6: Inhomogeneous Equations Inhomogeneous equations can often be solved (for constant coefficient PDEs, always be solved) by finding the fundamental solution (the solution for a point source), then taking the convolution with the boundary conditions to get the solution.

Thumbnail: A visualization of a solution to the two-dimensional heat equation with temperature represented by the third dimension. Imaged used wth permission (Public Domain; Oleg Alexandrov ). The heat equation is a parabolic partial differential equation that describes the distribution of heat (or variation in temperature) in a given region over time.