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Solutions to Differential Equations

Solutions to Differential Equations

Wouldn't it be nice to have the solution to all of your problems? Or at least your math problems? How about just the differential equations problems? Sadly, you can't even find solutions to all kinds of differential equations. However here you can find at least some kinds of solutions to differential equations.

Verifying Solutions to Differential Equations

Let's start by seeing how to verify if a function is a solution to a differential equation. Suppose you are given a differential equation

\[ y' = f(x,y),\]

and someone tells you that the function \(y(x)\) is a solution to the equation. How do you see if they are right?

To verify that \(y(x)\) is a solution to the differential equation \(y'=f(x,y)\), evaluate \(y'(x) - f(x, y(x))\) and see if you get \(0\). If you do, then \(y(x)\) is a solution.

Let's look at an example.

Verify that

\[y(x) = \ln (x^2 - 4x - 4)\]

is a solution to the differential equation

\[ y' = e^{-y}(2x - 4).\]

Solution

First you will need to find \(y'(x)\), so using the Chain Rule you get

\[ \begin{align} y'(x) &= \frac{\mathrm{d}}{\mathrm{d}x} \ln (x^2 - 4x - 4) \\ &= \frac{1}{x^2 - 4x - 4}\cdot \frac{\mathrm{d}}{\mathrm{d}x} (x^2 - 4x - 4 ) \\ &= \frac{2x - 4}{x^2 - 4x - 4 }.\end{align}\]

Then plugging that into \(y'(x) - f(x, y(x))\) you can see that

\[ \begin{align} y'(x) - f(x, y(x)) &= \frac{2x - 4}{x^2 - 4x - 4 } - (2x-4) \exp\left(-\ln (x^2 - 4x - 4) \right) \\ &= \frac{2x - 4}{x^2 - 4x - 4 } - (2x-4) \exp\left( \ln \left( \frac{1}{x^2 - 4x - 4 }\right)\right) \\ &= \frac{2x - 4}{x^2 - 4x - 4 } - (2x-4) \left( \frac{1}{x^2 - 4x - 4 }\right) \\ &= 0.\end{align}\]

Therefore \(y(x)\) is a solution to the differential equation.

What can you do if you want to get an idea of what a solution looks like without solving the differential equation?

Graphing Solutions to Differential Equations

There are two main methods you can use to get an idea of what the solution to a differential equation looks like and how it behaves without actually solving it.

  • If you want a numerical approximation, you can use Euler's Method.

  • Direction Fields, also called slope fields, use the fact that the derivative is a slope to plot out a "field" of slopes that can allow you to predict how solutions will behave.

The articles on those topics will have lots of examples of how to graph solutions. If you can actually solve the differential equation, you can graph the general solution. Calling it "the general solution" makes it sound like only one solution, but in fact, it is a family of functions. How the solution behaves depends on where the solution starts (also called the initial condition). For more information on this topic see General Solutions to Differential Equations.

Solutions of First Order Linear Differential Equations

A first-order linear differential equation can always be written in the form

\[ \frac{\mathrm{d}y}{\mathrm{d}x} + P(x)y=Q(x),\]

where \(P(x)\) and \(Q(x)\) are functions.

A special case of this is when \(P(x)\) and \(Q(x)\) are constants, so the first order linear equation can be written as

\[ \frac{\mathrm{d}y}{\mathrm{d}x}+ay=b.\]

Then

\[y=Ae^{-ax}+\frac{b}{a},\]

is the solution of the first order linear differential equation with constant coefficients.

Solving first order linear differential equations involves an integrating factor, and there are lots of examples in the articles Linear Differential Equations and Nonhomogeneous Linear Equations.

Exponential Solutions to Differential Equation

Solutions to first-order linear differential equations with constant coefficients are just about the only class of differential equations that are guaranteed to have an exponential solution. However, that doesn't mean other differential equations can't have exponential functions in their solutions. Let's take a look at an example.

Verify that \(y(x) = e^{2x} + e^{4x}\) is a solution to the second order differential equation

\[y'' - 6y' + 8y = 0.\]

Solution

To verify that \(y(x)\) is a solution, you will need both the first and second derivatives. Finding the first derivative gives you

\[ y'(x) = 2e^{2x} + 4e^{4x},\]

and you can use that to find the second derivative is

\[ y''(x) = 4e^{2x} + 16e^{4x}.\]

Then plugging them into the equation you get

\[ \begin{align} y'' - 6y' + 8y &= 4e^{2x} + 16e^{4x} -6(2e^{2x} + 4e^{4x}) + 8(e^{2x} + e^{4x} ) \\ &= 4e^{2x} + 16e^{4x} -12e^{2x}-24e^{4x} + 8e^{2x} + 8e^{4x} \\ &= 0. \end{align}\]

Since you got zero, it verifies that \(y(x)\) is a solution to the second order differential equation.

Notice that in the example the differential equation was

\[y'' - 6y' + 8y = 0.\]

If you think of translating this to a polynomial where the number of derivatives is the power you raise \(r\) to, you get

\[ r^2 - 6r + 8 = 0.\]

This factors into \( (r-2)(r-4) = 0\), which has solutions \(r=2\) and \(r=4\), which turned up in the exponents of the solution! Things like characteristic polynomials and second order linear differential equations are some of the things you will learn about if you take a class in differential equations.

Equilibrium Solutions to Differential Equations

Some differential equations will have an equilibrium solution.

An equilibrium solution \(y(x)\) of a first order differential equation is one that satisfies \(y'(x)\equiv 0\).

In other words, an equilibrium solution to a first-order differential equation is a constant solution! Equilibrium solutions are sometimes called steady state solutions.

One famous differential equation with not just one, but two equilibrium solutions is the logistic one,

\[P' = r\left( 1- \frac{P}{k}\right)P.\]

For more information on the logistic equation, along with the solutions and how they behave, see The Logistic Differential Equation.

One place where equilibrium solutions tend to show up is in separable differential equations. Let's look at an example.

Find the solutions to the separable differential equation

\[ y' = (x+3)(y^2-16).\]

Solution

First, let's look for the constant solutions. Those happen when \(y' = 0\), or in other words when

\[ (x+3)(y^2-16) =0.\]

Remember that you are looking for a solution, or in other words a function \(y(x)\)! So the idea is to solve for \(y\) in the above equation. That means you will get a constant solution when \(y^2 - 16 = 0\), or in other words when \(y = 4\) or \(y = -4\). So there are two constant solutions.

What about other solutions? Since the equation is separable, first write it as

\[ \frac{1}{y^2-16} y' = x+3\]

then integrate both sides with respect to \(x\) to get

\[ \frac{1}{4}\ln\left|\frac{y-2}{y+2}\right| = \frac{1}{2}x^2 + 3x + C.\]

Since you have solved the differential equation, but not solved for \(y\), this is the implicit solution. If at all possible, it is a good idea to write the solution to the differential equation in explicit form, or in other words solve for \(y\).

So doing a bit of algebra you get

\[ \ln\left|\frac{y-2}{y+2}\right| = 2x^2 + 12x + D\]

where \(D\) is just renaming the constant \(4C\). Using properties of logarithms, this means

\[ \left|\frac{y-2}{y+2}\right| = \exp (2x^2 + 12x + D ),\]

so

\[\frac{y-2}{y+2} = \pm \exp (2x^2 + 12x + D ) .\]

Letting \(B = \pm e^D\) (which is really just another constant), and doing a bit more algebra gives you

\[ \begin{align} y-2 &= (y+2)\left(B\exp (2x^2 + 12x) \right) \\ &= By\exp (2x^2 + 12x )+2 B\exp (2x^2 + 12x ), \end{align}\]

so

\[ y - By\exp (2x^2 + 12x ) = By\exp (2x^2 + 12x ) \]

and

\[ y(x) = \frac{ By\exp (2x^2 + 12x ) }{1-B\exp (2x^2 + 12x ) }.\]

If you prefer you can write it as

\[ y(x) = \frac{ Be^{2x^2 + 12x } }{1-Be^{2x^2 + 12x } }.\]

So now you have two equilibrium solutions and a general solution! How do you know which one is the correct one? Well, technically they are all correct. They make up a set of functions that all solve the differential equation. If you had initial values given to you, you would be able to then either pick one of the equilibrium solutions, or solve for \(B\) in the general solution to get a particular solution.

To see an example of a differential equation that can have one, none, or infinitely many solutions depending on the initial value, see our article General Solutions to Differential Equations.

If you are interested in seeing more about solutions to initial value problems, see Particular Solutions to Differential Equations and Differential Equations Initial-Value Problems.

Let's look at another example.

Suppose you have a frozen pizza and you need to bake it. The pizza bakes at \(375^\circ\). The temperature of your kitchen is \(70^\circ\). What is the differential equation that models this, and what is the equilibrium solution?

Solution

First, let's decide what the variables are. Certainly one of them is going to be time, and the other temperature, but you need to figure out which one is the independent variable and which is the dependent variable. Since the temperature of the pizza depends on the time, it means time is the independent variable and temperature is the dependent variable. Giving them each a variable, let

  • \(t\) be the time since coming out of the oven; and
  • \(y(t)\) be the temperature since coming out of the oven.

Now you need to figure out what equation models this situation. Newton's Law of Cooling to the rescue! Remember that for an object cooling (in this case your pizza is cooling to room temperature), the rate of change of the temperature is given by a constant times the difference between the current temperature and room temperature. In other words,

\[y'(t) = k(y(t) - 70),\]

where \(k\) is the cooling constant.

You still need an initial value to complete this as a differential equation.

Now what is the initial value? That is the temperature when it comes out of the oven, so \(y(0) = 375\). So to complete the differential equation as an initial value problem,

\[\begin{align} &y'(t) = k(y(t) - 70) \\ &y(0)=375 \end{align}\]

where \(k\) is the cooling constant.

You also need to find the equilibrium solution. You don't need to find the solution to the initial value problem to find it, just set \( k(y(t) - 70) =0\), which gives you \(y(t) = 70\). So the equilibrium solution is

\[ y = 70^\circ \]

which makes sense because that is the temperature of your kitchen, and you would expect that over time the pizza would cool down to your kitchen temperature.

Now let's look at an associated problem.

Suppose you have a frozen pizza and you need to bake it. The pizza bakes at \(375^\circ\). The temperature of your kitchen is \(70^\circ\), and after \(5\) minutes of sitting on the counter after baking your pizza is \(350^\circ\). Naturally you don't want to burn your mouth eating the pizza, so you want to wait until it is \(300^\circ\) before eating it. How long will you have to wait?

Solution

In the previous example, you saw how to set up this differential equation and find the equilibrium solution, and you found that

\[\begin{align} &y'(t) = k(y(t) - 70) \\ &y(0)=375 \end{align}\]

where \(k\) is the cooling constant. Let's build on that information.

This is a nice separable equation, and writing it in separable form gives you

\[ \frac{1}{y-70}y' = k.\]

Then integrating both sides with respect to \(t\) gives

\[ \ln |y-70| = kt+C.\]

You can either use the information given in the problem to find \(k\) and \(C\) first, or you can find the explicit solution and then find the constants. Either way will give you the same answer.

If you plug in the initial condition \(y(0) = 375\) you get

\[ \ln |375-70| = k\cdot 0 + C,\]

so \( C = \ln 305\).

But how do you find \(k\)? Ask yourself what you information you have that you haven't used yet. Here, you know that after \(5\) minutes of sitting on the counter after baking your pizza is \(350^\circ\), but you haven't used it. Translating that into the variables, \(y(5) = 350\). Plugging it, along with \(C\), into the equation gives you

\[ \ln |350-70| = 5k+\ln 305 .\]

In other words,

\[ \begin{align} 5k &= \ln |350-70| - \ln 305 \\ &= \ln 280 - \ln 305 \\ &= \ln \frac{280}{305}, \end{align}\]

so

\[k= \frac{1}{5} \ln \frac{280}{305} .\]

Then putting it all together, the solution to the initial value problem is

\[\ln |y-70| =\frac{1}{5} \ln \frac{280}{305} t+\ln 305 .\]

Notice that the question didn't ask for an explicit solution, it asked how long you would need to wait for the pizza to be \(300^\circ\). So rather than going to the work of finding an explicit solution, just plug in the temperature and solve for the time. That means

\[ \ln |300-70| = \frac{1}{5}\ln \frac{280}{305} t+\ln 305 \]

so

\[ \ln 230 - \ln 305 = \frac{1}{5}\ln \frac{280}{305} t \]

which means

\[ t = 5\frac{ \ln \frac{230}{305}}{ \ln \frac{280}{305} } \approx 16.5.\]

So you will need to wait about 16.5 minutes before you can eat your pizza without burning your mouth.

Solutions to Differential Equations - Key takeaways

  • To verify that \(y(x)\) is a solution to the differential equation \(y'=f(x,y)\), evaluate \(y'(x) - f(x, y(x))\) and see if you get \(0\). If you do, then \(y(x)\) is a solution.
  • To get a numerical approximation of the solution to a differential equation you can use Euler's Method.
  • Direction Fields, also called slope fields, use the fact that the derivative is a slope to plot out a "field" of slopes that can allow you to predict how solutions will behave.
  • An equilibrium solution (also called a constant solution) \(y(x)\) of a first order differential equation is one that satisfies \(y'(x)\equiv 0\).

Frequently Asked Questions about Solutions to Differential Equations

A homogeneous solution of a differential equation comes from a homogeneous differential equation. In this case, a solution for the differential equation has the form φ(x). But then also any solution cφ(x) where c is any non-zero constant.

If you have a homogeneous differential equation, its solution is a function f(x). Then 2(f(x)) is also a solution of the differential equation.

This isn't generally done since it involves guessing at a solution and then trying to see if you can find coefficients that make it work.

An example is when using this method to solve the equation:


y''Ay'+By=C.


In this case A, B, C are functions of x and then a solution of the type ay1(x)+by2(x) exists. Here you can substitute a and b again for new functions and then use them to find the correct values for a and b.

Exact equations are a very particular kind of differential equation.  Solving those kinds of equations don't involve an integrating factor.  However solving first order linear differential equations often involve integrating factors.

A solution to a differential equation is a function, or family of functions, that satisfies the differential equation.

You find the function f(x) and the possible constants. Then you insert the derivatives of the function and constants into the differential equation, this in turn must satisfy the original equation.

Final Solutions to Differential Equations Quiz

Question

What form does a first order linear initial value problem take?

Show answer

Answer

It will look like 

\[\begin{align} &y' + P(x)y = Q(x) \\ &y(a) = b \end{align}\]

where \(P(x)\) and \(Q(x)\) are functions, and \(a\) and \(b\) are real valued constants.


Show question

Question

What do you call the solution to a first order linear initial value problem?

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Answer

A particular solution.

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Question

What do  you call the solution to a linear first order differential equation without any initial value?

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Answer

A general solution.

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Question

Why do you call something a general solution?

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Answer

Because the solution has a constant in it, making it a general family of functions rather than a single function.

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Question

If you are told that a function is the solution to an initial value problem, what two things does it need to satisfy?

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Answer

It needs to satisfy both the initial value and the differential equation.

Show question

Question

Do all first order linear differential equations have a solution?

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Answer

No.

Show question

Question

Do all first order linear differential equations have a unique solution?

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Answer

No. 

Show question

Question

What needs to be true for a first order linear differential equation 

\[\begin{align} &y' + P(x)y = Q(x) \\ &y(a) = b \end{align}\]

where \(P(x)\) and \(Q(x)\) are functions, and \(a\) and \(b\) are real valued constants to have a unique solution?


Show answer

Answer

If \(a, b \in \mathbb{R}\), and \(P(x)\), \(Q(x)\) are both continuous functions on the interval \((x_1, x_2)\) where \(x_1 < a < x_2 \) then the solution to the initial value problem 

\[\begin{align} &y' + P(x)y = Q(x) \\ &y(a) = b \end{align}\]

exists and is unique.


Show question

Question

If someone tells you that the first order linear differential equation 

\[\begin{align} &y' + P(x)y = Q(x) \\ &y(a) = b \end{align}\]

has a unique solution, what do you know about \(P(x)\) and \(Q(x)\)?

Show answer

Answer

You know that they are both continuous functions in an interval around the initial value.

Show question

Question

What do you call the solution to the initial value problem

\[\begin{align} &y'=f(x)g(y) \\ &y(a)=b? \end{align}\]


Show answer

Answer

A particular solution.

Show question

Question

Is it true that if a differential equation has a unique solution, then that solution exists on the whole real line?

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Answer

No.  Often the solution to a differential equations has an interval of existence which is not the whole real line.

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Question

Can a second order initial value problem have a particular solution?

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Answer

Yes, and if a solution exists which is unique then it has a particular solution.

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Question

How do you verify that a function is a solution to a differential equation?

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Answer

To verify that \(y(x)\) is a solution to the differential equation \(y'=f(x,y)\), evaluate \(y'(x) - f(x, y(x))\) and see if you get \(0\).  If you do, then \(y(x)\) is a solution.

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Question

What are two things you can use if you want an idea of what the solutions to a differential equation look like but you can't solve it?

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Answer

Use Euler's Method or the Direction Field.

Show question

Question

What method do you use to get a numerical approximation to the solution of a differential equation?

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Answer

Euler's Method.

Show question

Question

What is another name for a direction field?

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Answer

A slope field.

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Question

How is a direction field used?

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Answer

Direction Fields, also called slope fields, use the fact that the derivative is a slope to plot out a "field" of slopes that can allow you to predict how solutions will behave.

Show question

Question

What is an equilibrium solution to a first order differential equation?

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Answer

An equilibrium solution \(y(x)\) of a first order differential equation is one that satisfies \(y'(x)\equiv 0\).

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Question

What are two other names for an equilibrium solution?

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Answer

A constant solution or a steady state solution.

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Question

Is it true that a first order differential equation has an equilibrium solution?

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Answer

No, it may not have one at all.

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Question

Is it true that a first order differential equation has at most one equilibrium solution?

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Answer

No.  Equations like the Logistic Equation have two equilibrium solutions.

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Question

How many equilibrium solutions does a first order differential equation have?

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Answer

It can have none, one, or lots, all depending on the differential equation.

Show question

Question

How do you find the equilibrium solution of a first order differential equation?

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Answer

Set \(y' = 0\) and solve for \(y\).  Any constants you get that satisfy \(y' = 0\) are equilibrium solutions.

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Question

Name a first order differential equation that has two equilibrium solutions.

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Answer

The Logistic Equation.

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Question

What is the differential equation that models Newton's Law of Cooling?

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Answer

Let  \(t\) be the time and \(y(t)\) be the temperature.  Then 

\[\begin{align} &y'(t) = k(y(t) - A) \\ &y(0)=I \end{align}\]

where \(k\) is the cooling constant, \(I\) is the initial temperature, and \(A\) is the ambient temperature of the environment.



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Question

In the differential equation for Newton's Law of Cooling, what does the equilibrium solution correspond to?

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Answer

It is the ambient temperature of the environment.  In other words, it is the temperature your object will eventually get really close to.

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Question

Name one type of differential equation that has an exponential function as a solution.

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Answer

A first order linear differential equation with constant coefficients. 

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Question

What is the difference between a general solution and a particular solution?

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Answer

A general solution has a constant of integration in it, meaning that it is actually a family of functions.  A particular solution is one where you have used an initial value to solve for that constant of integration.

Show question

Question

How do you find a particular solution of a differential equation?

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Answer

First you solve the differential equation to find the general solution.  Then you use an initial value to solve for the constant of integration.  The solution that goes through the initial value is the particular solution.

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Question

If you have found the particular solution of a differential equation, will it exist everywhere?

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Answer

It depends on the equation.  There are plenty of equations where the solution to an initial value problem will have a limited interval of existence.

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Question

True or False: The general solution to a differential equation needs to satisfy initial conditions.

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Answer

False.

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Question

True or False:  Nonhomogeneous differential equations have a corresponding homogeneous differential equation associated with them.

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Answer

True.

Show question

Question

What is another name for the general solution to a differential equation?

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Answer

A family of functions.

Show question

Question

Why is the general solution to a differential equation also called a family of functions?

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Answer

Because there is a constant in the general solution that can be replaced by any real number.

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Question

If you look at the general solution \(y_C(x)\) to a differential equation, what sorts of properties will the functions share if you substitute in various values of \(C\)?

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Answer

The same end behavior.

Show question

Question

Which of the following is a general solution for the differential equation 

\[ \begin{align} &y'=3y \\ &y(0)=4? \end{align}\]

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Answer

\(y(t) = Ce^{3t}\).

Show question

Question

For the nonhomogeneous differential equation \(xy' + \sin x = 2y\), which of the following would be the corresponding homogeneous differential equation?

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Answer

\(xy'=2y\).

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Question

Suppose you know that \(y_p(t)\) is a particular solution to a nonhomogeneous differential equation, and that \(y_C(t)\) is the general solution to the corresponding homogeneous differential equation. Which of the following would be the general solution to the nonhomogeneous differential equation?

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Answer

\(y_p(t)+y_C(t)\).

Show question

Question

Suppose you are given a nonhomogeneous differential equation and are told that  \(y(x) = e^x + C\sin x\) is the general solution. Which of the following would be a particular solution to the nonhomogeneous differential equation?

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Answer

\(y(x) = e^x\).

Show question

Question

Suppose you are given a nonhomogeneous differential equation and are told that  \(y(x) = e^x + C\sin x\) is the general solution. Which of the following would be the general solution to the corresponding homogeneous differential equation?

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Answer

\(y(x) = C\sin x\).

Show question

Question

Why would it be an advantage to be able to write the general solution to a nonhomogeneous differential equation as the sum of a particular solution and the general solution to the corresponding homogeneous differential equation? 

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Answer

It is often quite difficult to find the general solution to a nonhomogeneous differential equation.  It can be much easier to find one solution (the particular solution), and a general solution to the corresponding homogeneous differential equation.

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Question

Suppose you know that the corresponding homogeneous differential equation to a nonhomogeneous differential equation is \(xy' + y\sin x = 0\).  Which of the following could be the nonhomogeneous differential equation it came from? 

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Answer

 \(xy' + y\sin x = x^2\).

Show question

Question

Which of the following functions is a general solution to the corresponding homogeneous equation if the nonhomogeneous differential equation is \(y' - y = x\sin x\)?

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Answer

\(y(x)=Ce^x\).

Show question

Question

True or False:  The general solution to a differential equation is a single function.

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Answer

False.

Show question

Question

True or False:  If \(y_C(x) = Ce^{\sin x} +3\) is a general solution to a differential equation, then \(y(x) = 3\) is also a solution to the differential equation.

Show answer

Answer

True.

Show question

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