1.
(a) Pick the Method.
\begin{equation*}
y'' + 4y = u_{3}(t), \qquad y(0) = 0, \quad y'(0) = 0
\end{equation*}
-
The Laplace transform method—step-function forcing is exactly what it’s built for.
- Correct! The equation is linear with constant coefficients, the initial conditions sit at \(t = 0\text{,}\) and no earlier method can absorb a forcing function that switches on at \(t = 3\text{.}\)
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Undetermined coefficients, guessing \(y_p = A\,u_3(t)\text{.}\)
- Undetermined coefficients only handles forcing built from polynomials, exponentials, sines, and cosines—a step function is none of these, and its jump breaks the guess-and-match machinery.
-
The characteristic equation alone.
- The characteristic equation handles the homogeneous part, but the step forcing still needs to enter the solution—that’s the Laplace method’s job.
-
Separation of variables.
- Separation of variables is a first-order tool; this equation is second order.
(b) Pick the Method.
\begin{equation*}
\frac{dy}{dt} = y\left(1 - y\right)
\end{equation*}
-
Separation of variables—the Laplace method doesn’t apply.
- Correct! The \(y^2\) hiding in \(y(1-y)\) makes this logistic equation nonlinear, and our Laplace method requires linear constant-coefficient equations. But the right side factors as a function of \(y\) alone, so it separates (and qualitative tools describe its behavior beautifully).
-
The Laplace transform method.
- Transforming the nonlinear term \(y^2\) is beyond the Laplace machinery we’ve built—there’s no table entry for the transform of a product of unknown functions. The equation is nonlinear, so look to first-order tools instead.
-
Undetermined coefficients.
- Undetermined coefficients is for linear constant-coefficient equations with forcing; this equation is nonlinear.
-
The characteristic equation.
- Substituting \(y = e^{rt}\) only works for linear equations—the \(y^2\) term breaks it.
(c) More Than One Way.
\begin{equation*}
y'' + 3y' + 2y = 12e^{t}, \qquad y(0) = 1, \quad y'(0) = 0
\end{equation*}
-
Either undetermined coefficients or the Laplace method—both apply; Laplace folds the initial conditions in automatically.
- Correct! Linear, constant coefficients, forcing \(12e^t\) in the undetermined-coefficients family (\(y_p = 2e^t\)), and an IVP at \(t = 0\)—so both roads are open. Undetermined coefficients needs a follow-up step to fit the constants; Laplace handles them during the forward transform.
-
Only the Laplace method—undetermined coefficients can’t handle initial conditions.
- Undetermined coefficients handles this fine: find the general solution \(y = y_h + y_p\text{,}\) then use the initial conditions to fix the constants. Laplace just merges those steps.
-
Only undetermined coefficients—Laplace can’t handle exponential forcing.
- Exponentials are one of the easiest things to transform: \(\lap{12e^{t}} = \frac{12}{s-1}\text{.}\) Both methods apply here.
-
Neither—the forcing function rules both out.
- \(12e^t\) is in the undetermined-coefficients family and has a known Laplace transform, so both methods apply.
(d) Check the Fine Print.
\begin{equation*}
y' + t\,y = 0
\end{equation*}
-
Separation of variables—the coefficient \(t\) puts this outside our Laplace method.
- Correct! The variable coefficient \(t\) means this isn’t a constant-coefficient equation, so our Laplace playbook doesn’t cover it. But \(y' = -t\,y\) separates immediately: \(\frac{dy}{y} = -t\,dt\) gives \(y = Ce^{-t^2/2}\text{.}\)
-
The Laplace transform method.
- The variable coefficient \(t\) multiplying \(y\) takes this outside the constant-coefficient equations our Laplace method covers. A first-order tool is the way in—notice the equation is separable.
-
The characteristic equation.
- Characteristic equations require constant coefficients; the coefficient here is \(t\text{.}\)
-
No method applies.
- Isolate the derivative: \(y' = -t\,y\text{.}\) Can the right side be written as a function of \(t\) times a function of \(y\text{?}\)
(e) Match the Tool to the Question.
Find the general solution:
\begin{equation*}
y'' + 9y = 0
\end{equation*}
-
The characteristic equation—Laplace is the wrong shape for a general-solution request.
- Correct! \(r^2 + 9 = 0\) gives \(r = \pm 3i\) and \(y = C_1\cos 3t + C_2\sin 3t\) in two lines. The Laplace method wants specific initial conditions at \(t=0\text{;}\) with none given, the characteristic equation is the natural tool.
-
The Laplace transform method.
- With no initial conditions given, the forward transform would carry the unknowns \(y(0)\) and \(y'(0)\) through every step. It can be done, but the characteristic equation gives the general solution far more directly.
-
Undetermined coefficients.
- The forcing function is \(0\text{,}\) so there is no particular solution to build—the characteristic equation alone finishes the problem.
-
The integrating factor method.
- The integrating factor method is a first-order tool; this equation is second order.
(f) Piecewise Forcing Protocol.
What is the right first step for this IVP?
\begin{equation*}
y' + 4y = f(t), \qquad y(0) = 0, \qquad \text{where } f(t) = \begin{cases} 5, & 0 \le t \lt 2 \\ 0, & t \ge 2 \end{cases}
\end{equation*}
-
Rewrite \(f(t)\) with unit step functions as \(5\left(1 - u_2(t)\right)\text{,}\) then run the Laplace method.
- Correct! Piecewise forcing enters the Laplace machinery through unit step notation—rewrite first, transform second. Here \(f(t)\) is ON at value \(5\) until it switches OFF at \(t = 2\text{.}\)
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Solve on \(0 \le t \lt 2\) and \(t \ge 2\) separately with an integrating factor, then glue the pieces.
- This can be made to work, but matching the pieces at \(t = 2\) by hand is exactly the bookkeeping the unit-step approach eliminates. Rewrite the forcing with step functions and transform once.
-
Apply undetermined coefficients with guess \(y_p = A\text{.}\)
- A constant guess matches constant forcing—but this forcing isn’t constant; it shuts off at \(t = 2\text{.}\) Undetermined coefficients can’t represent the switch.
-
Take the Laplace transform immediately, integrating the piecewise definition term by term from the definition.
- Retreating to the integral definition works but is slow. The whole point of the unit-step rules is to skip it: rewrite \(f(t) = 5\left(1 - u_2(t)\right)\) and use the table.
