1.
(a) Pick the Method.
\begin{equation*}
\frac{dy}{dx} = x^2 e^{-x}
\end{equation*}
-
Direct integration.
- Correct! The right-hand side involves only \(x\text{,}\) so integrating both sides gives the general solution.
-
Separation of variables (and direct integration won’t work).
- Separation technically applies with \(g(y) = 1\text{,}\) but there’s nothing to separate—the equation is already set up to integrate directly.
-
Integrating factor, since the equation is linear.
- The equation is linear, but with no \(y\) term at all there’s nothing for an integrating factor to fix—just integrate both sides.
-
None of our methods apply.
- Look again: the right-hand side is a function of \(x\) alone.
(b) Pick the Method.
\begin{equation*}
\frac{dy}{dx} = x y^2
\end{equation*}
-
Separation of variables.
- Correct! The right side is the product \(f(x)\cdot g(y)\) with \(f(x) = x\) and \(g(y) = y^2\text{.}\)
-
Integrating factor.
- The \(y^2\) term makes this equation nonlinear, so the integrating factor method does not apply.
-
Direct integration.
- The right side involves \(y\text{,}\) so we cannot simply integrate both sides with respect to \(x\text{.}\)
-
None of our methods apply.
- Check the separability test: can the right side be written as (a function of \(x\)) \(\times\) (a function of \(y\))?
(c) Pick the Method.
\begin{equation*}
y' + 3y = e^{2x}
\end{equation*}
-
Integrating factor.
- Correct! The equation is first-order linear, already in standard form with \(P(x) = 3\text{,}\) and it is not separable.
-
Separation of variables.
- Isolating the derivative gives \(y' = e^{2x} - 3y\text{,}\) a difference, not a product of an \(x\)-part and a \(y\)-part. Not separable.
-
Direct integration.
- The left side is not the derivative of a single expression as written—that’s exactly what multiplying by an integrating factor will fix.
-
None of our methods apply.
- Check the linearity test: \(y\) and \(y'\) each appear once, to the first power.
(d) Pick the Method.
\begin{equation*}
x\, y' = y + x^2 \sin x
\end{equation*}
-
Integrating factor.
- Correct! Dividing by \(x\) gives the standard form \(y' - \frac{1}{x}y = x\sin x\text{,}\) which is linear but not separable.
-
Separation of variables.
- Isolating the derivative gives \(y' = \frac{y}{x} + x\sin x\text{,}\) a sum that cannot be factored into \(f(x)\cdot g(y)\text{.}\)
-
Direct integration.
- The right side involves \(y\text{,}\) so both sides cannot be integrated as-is.
-
None of our methods apply.
- Try dividing both sides by \(x\) and rearranging into standard linear form.
(e) Pick the Method.
\begin{equation*}
\frac{dy}{dx} = \left(1 + y^2\right) e^{x}
\end{equation*}
-
Separation of variables.
- Correct! The right side factors as \(f(x)\cdot g(y)\) with \(f(x) = e^x\) and \(g(y) = 1 + y^2\text{.}\)
-
Integrating factor.
- The \(y^2\) term makes the equation nonlinear, so the integrating factor method is out.
-
Direct integration.
- The right side involves \(y\text{,}\) so we cannot simply integrate both sides.
-
None of our methods apply.
- Check the separability test—the right side is already factored for you.
(f) More Than One Way.
\begin{equation*}
\frac{dy}{dx} = 2y
\end{equation*}
-
Either separation of variables or an integrating factor—both apply.
- Correct! It separates as \(\frac{dy}{y} = 2\,dx\text{,}\) and rewriting it as \(y' - 2y = 0\) shows it’s also first-order linear. Both routes give \(y = Ce^{2x}\text{.}\)
-
Separation of variables only.
- Separation works, but so does an integrating factor: \(y' - 2y = 0\) is first-order linear with \(Q(x) = 0\text{.}\)
-
Integrating factor only.
- An integrating factor works, but so does separation: \(\frac{dy}{y} = 2\,dx\text{.}\)
-
Direct integration only.
- The right side involves \(y\text{,}\) so direct integration is the one method that doesn’t apply here.
(g) The Fastest Route.
This equation can be solved by more than one of our methods. Which observation gives the fastest route to the solution?
\begin{equation*}
x^2 y' + 2x y = 4x
\end{equation*}
-
The left side is already the derivative \(\frac{d}{dx}\left[x^2 y\right]\text{,}\) so integrate both sides directly.
- Correct! The left side is a completed product rule, so direct integration gives \(x^2 y = 2x^2 + C\) in one step. Always scan for this shortcut first.
-
Divide by \(x^2\) to reach standard form, then compute the integrating factor \(\mu(x)\text{.}\)
- This works—but the integrating factor you’d build is \(x^2\text{,}\) which the equation already has in place! Recognizing the completed product rule skips the whole computation.
-
Isolate \(y'\) and separate the variables.
- Surprisingly, this works too: \(y' = \frac{2(2 - y)}{x}\) is separable. But it’s the slowest route—look at the left side of the original equation again.
-
No method applies without more information.
- Look at the left side: does it match the expanded form of a product rule?
(h) Know When to Fold.
\begin{equation*}
y' = y^2 + x
\end{equation*}
-
None of our formula methods apply.
- Correct! The \(y^2\) term rules out linearity, and \(y^2 + x\) is a sum that won’t factor into \(f(x)\cdot g(y)\text{.}\) Equations like this are why the next two chapters develop qualitative and numerical tools.
-
Separation of variables.
- The right side \(y^2 + x\) is a sum, not a product of an \(x\)-part and a \(y\)-part—it cannot be separated.
-
Integrating factor.
- The \(y^2\) term makes the equation nonlinear, so the integrating factor method does not apply.
-
Direct integration.
- The right side involves \(y\text{,}\) so both sides cannot simply be integrated with respect to \(x\text{.}\)
