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Solutions to Differential Equations

- Calculus
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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**.

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?

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.

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.

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.

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.

- 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 \(y'(x)\equiv 0\).

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.

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.

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.

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