Solutions to Differential Equations

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.

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.

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.

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.

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.