• Recognize and solve differential equations that can be solved by separation of variables.
  • Use differential equations to model and solve applied problems.

Assignment

  • Vocabulary and teal boxes
  • p393 (21 problems) 5–9 odd, 13–25 odd, 33, 37, 39, 45, 57, 61–63, 79–81

Additional Resources

Separation of Variables

We saw this for the first time in the last section, but we’ll look at more varied examples here. In general, you can separate variables if it can be written in the form below.

\[\begin{align} \frac{dy}{dx} = f(x)g(y) \end{align}\]

If you can get it in this form (and I do suggest it), you should check when $g(y)=0$. This will give you your constant solutions (e.g., $y=0$ or $y=2$). This works because if $g(y)=0$ then $\frac{dy}{dx}=0$.

We don’t do this for $x$ since we are only concerned with functions, and horizontal lines are never functions, no matter how you restrict the domain or range.

Example: Find a General Solution

Find the general solution of $(x^2+4)\frac{dy}{dx} = xy$.

Solving for $\frac{dy}{dx}$ gives us $\frac{dy}{dx}=\frac{xy}{x^2+4}$. So, $f(x)=\frac{x}{x^2+4}$ and $g(y)=y$. From there, if $g(y)=0$ then $y=0$, so we have one constant solution. Now we can move on to our general solution.

\[\begin{align} \frac{dy}{dx}&=\frac{xy}{x^2+4} \\ \frac{1}{y}\,dy &= \frac{x}{x^2+4}\,dx \\ \int \frac{1}{y}\,dy &= \int \frac{x}{x^2+4}\,dx \\ \ln |y| &= \frac{1}{2}\ln(x^2 + 4) + C \\ \ln |y| &= e^{\frac{1}{2}\ln(x^2 + 4) + C} \\ \ln |y| &= e^{\frac{1}{2}\ln(x^2 + 4)}e^C \\ \ln |y| &= e^{\ln\sqrt{x^2 + 4}}e^C \\ |y| &= e^C\sqrt{x^2+4} \\ \end{align}\]

There’s a lot of exponent and log rules happening in those last few steps. Make sure you review them if you’re not sure how each step one about.

At this point, there are some things that need addressing. First is the absolute value. Ideally we want $y=$, so to get that we can rewrite $e^C$ with a $\pm$.

\[\begin{align} y &= \pm e^C\sqrt{x^2+4} \\ \end{align}\]

Now, we could put $C$ instead of $\pm e^C$, but we have to verify that every constant is covered first. In this case, ${\pm e^C \neq 0}$, but we already found that $y=0$ is a solution, so we’re in the clear.

\[\begin{align} y &= C\sqrt{x^2+4} \\ \end{align}\]

$\blacksquare$

Example 2: Particular Solutions

Given the initial solution of $y(0)=1$, find the particular solution of the differential equation.

\[\begin{align} xy\, dx + e^{-x^2}\left(y^2-1\right)\, dy = 0 \end{align}\]

Finding a particular solution just means an extra step after getting the general one. So, we’ll work this one out like before. Rearrange so we can find constant solutions, then separate and integrate.

\[\begin{align} xy\, dx + e^{-x^2}\left(y^2-1\right)\, dy = 0 \\ e^{-x^2}\left(y^2-1\right)\, dy = -xy\, dx \\ \frac{dy}{dx} = -\frac{xy}{e^{-x^2}\left(y^2-1\right)} \\ \end{align}\]

$g(x)=\frac{y}{y^2-1}$, so $y=0$ is a solution again. Also, we can quickly see that $y\neq\pm1$.

\[\begin{align} \frac{y^2-1}{y}\, dy &= -xe^{x^2}\, dx \\ \int \frac{y^2-1}{y}\, dy &= \int -xe^{x^2}\, dx \\ \int y - \frac{1}{y}\, dy &= \int -xe^{x^2}\, dx &\text{\color{grey}Let $u=x^2$}\\ \frac{1}{2}y^2 - \ln{|y|} &= -\frac{1}{2}e^{x^2}+C \\ y^2 - 2\ln{|y|} + e^{x^2} &= C \end{align}\]

There’s no sense in solving for $y$ here since we have a mix of a logarithmic and non-logarithmic functions. Traditional algebra won’t get you anywhere, so usually the cleanest approach is to get all the $x$ and $y$ terms on one side.

Anyway, we can use the initial condition now to find the $C$ needed for our particular solution. $y(0)=1$, so we plug in $x=0$ and $y=1$.

\[\begin{align} (1)^2 - 2\ln{1} + e^{0} &= C \\ 1 - 0 + 1 &= C \\ 2 &= C \\[1em] y^2 - 2\ln{|y|} + e^{x^2} &= 2 \end{align}\]

Here’s the slope field and general solution in GeoGebra. Desmos struggles with graphing this one a bit, which isn’t uncommon with the more complicated implicit functions.

$\blacksquare$