• 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 5–9 odd, 13–25 odd, 33, 37, 39, 45 57, 61–63, 79–81

Additional Resources

Separation of Variables

We looked at this for the first time in 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.

\[\begin{align} M(x) + N(y) \frac{dy}{dx} = 0 \end{align}\]

This form might not be immediately obvious, so some working out might be required.

Example 1

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

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

At this point, there are two things that need addressing. The first is the absolute value, which we can deal with by rewriting with $\pm$.

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

Next is the fact that $e^C>0$, meaning we can’t rewrite this is as $C$ … at least not yet. The $\pm$ allows us to have negatives, so we are left with a $C$ that can be anything but zero.

Something you can check easily—and should check for each differential—is if $y=0$ is a solution. Remember that you need $y$ and $y’$ to check if a function is a solution The derivative of $y=0$ is $y’=0$, so this is an easy solution to check. And it turns out that yes, $y=0$ is a solution in this case, so since we have everything covered, we can now safely say the general solution is

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

Example 2: Particular Solutions

Given the initial solution of $y(0)=1$, find the particular solution of the equation \(\begin{align} xy\, dx + e^{-x^2}\left(y^2-1\right)\, dy = 0 \end{align}\)

This looks intimidating, but $dy$ and $dx$ are already split into two terms, so we can move them to opposite sides and start moving things as needed.

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

There’s no sense in solving for $y$ here. You have a mix of a logarithmic and non-logarithmic functions, which has no algebraic solutions, meaning transcendental numbers and this is a rabbit hole we are not going down.

Anyway, we can use the initial condition now to find the $C$ needed for our particular solution.

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

Implicit functions, where both variables are on the same side, are preferred when it can’t be defined explicitly.

\[\begin{align} \frac{y^2}{2} - \ln|y| + \frac{1}{2}e^{x^2} &= 1 \\ y^2 - \ln y^2 + e^{x^2} &= 2 \end{align}\]

Make sure you look through the remaining examples in the book. We’ll go through a few more in class, too.