1.
(a) Pick the Method.
\begin{equation*}
y'' - 5y' + 6y = 0
\end{equation*}
-
Characteristic equation—this is an LHCC equation.
- Correct! Linear, constant coefficients, zero forcing. The characteristic equation \(r^2 - 5r + 6 = 0\) hands you the general solution.
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Undetermined coefficients, to find \(y_p\text{.}\)
- The forcing function is \(0\text{,}\) so there is no \(y_p\) to find—the homogeneous solution is the whole story.
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Integrating factor.
- The integrating factor method is a first-order tool; this equation is second order.
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Separation of variables.
- Separation of variables is a first-order tool; this equation is second order.
(b) Pick the Method.
\begin{equation*}
y'' - 5y' + 6y = e^{x}
\end{equation*}
-
Characteristic equation for \(y_h\text{,}\) then undetermined coefficients with guess \(y_p = Ae^{x}\text{.}\)
- Correct! This is an LNCC equation whose forcing function \(e^x\) is in the undetermined-coefficients family, and \(e^x\) does not overlap \(y_h = C_1 e^{2x} + C_2 e^{3x}\text{.}\)
-
Characteristic equation alone—the general solution comes entirely from its roots.
- The roots only produce \(y_h\text{.}\) With a nonzero forcing function you also need a particular solution: \(y = y_h + y_p\text{.}\)
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Integrating factor with \(\mu = e^{\int -5\, dx}\text{.}\)
- The integrating factor method is a first-order tool; this equation is second order.
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None of our methods apply.
- Check the table: linear, constant coefficients, and \(e^x\) is squarely in the undetermined-coefficients family.
(c) Two Toolkits, One Equation.
Which statement is true about the equation below?
\begin{equation*}
y' + 3y = 6
\end{equation*}
-
Both the integrating factor method and \(y_h + y_p\) reasoning (undetermined coefficients) apply.
- Correct! It’s first-order linear, so \(\mu = e^{3x}\) works—and it’s also an LNCC equation with constant forcing, so \(y_h = Ce^{-3x}\) plus the constant guess \(y_p = A\) works too. Both give \(y = 2 + Ce^{-3x}\text{.}\)
-
Only the integrating factor method applies, because the equation is first order.
- First order doesn’t disqualify the new machinery: this is a linear constant-coefficient equation with constant forcing, so \(y_h + y_p\) reasoning applies just as well.
-
Only undetermined coefficients applies, because the forcing function is a constant.
- Undetermined coefficients works here, but so does the integrating factor method—the equation is first-order linear.
-
Neither method applies.
- Quite the opposite—this equation belongs to both toolkits. Check the linearity and coefficient tests again.
(d) Know the Limits of the Guess.
\begin{equation*}
y'' + 4y = \tan x
\end{equation*}
-
The characteristic equation gives \(y_h\text{,}\) but none of our methods produce \(y_p\text{.}\)
- Correct! The forcing function \(\tan x\) is not built from polynomials, exponentials, sines, and cosines, so undetermined coefficients has no form to guess. Equations like this need a method beyond our current toolkit (its name, for the curious: variation of parameters).
-
Undetermined coefficients with guess \(y_p = A\tan x + B\sec x\text{.}\)
- Tempting, but derivatives of \(\tan x\) generate an endless stream of new function types—the guess never closes. Undetermined coefficients only works for forcing functions whose derivative families are finite: polynomials, exponentials, sines, and cosines.
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Separation of variables.
- Separation of variables is a first-order tool; this equation is second order.
-
No method applies, not even for \(y_h\text{.}\)
- The homogeneous side is a perfectly good LHCC equation—\(y_h = C_1\cos 2x + C_2 \sin 2x\text{.}\) It’s only the particular solution that’s out of reach.
(e) Check the Fine Print.
\begin{equation*}
y'' + y^2 = 0
\end{equation*}
-
None of our methods apply—the equation is nonlinear.
- Correct! The \(y^2\) term makes this nonlinear, and the characteristic equation shortcut is built on linearity. Substituting \(y = e^{rx}\) here would give \(r^2 e^{rx} + e^{2rx} = 0\text{,}\) which no constant \(r\) can satisfy.
-
Characteristic equation: \(r^2 + r^2 = 0\text{.}\)
- Careful—\(y^2\) is the square of the function, not a second derivative. The equation is nonlinear, and the characteristic equation only applies to linear constant-coefficient equations.
-
Undetermined coefficients.
- Undetermined coefficients builds particular solutions for linear equations with forcing; here the equation is nonlinear with no forcing at all.
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Direct integration, twice.
- Integrating twice would need the right side free of \(y\text{;}\) the \(y^2\) term blocks that.
(f) Check the Fine Print.
\begin{equation*}
x\, y'' + y' = 0
\end{equation*}
-
The characteristic-equation method does not apply—the coefficients are not constant.
- Correct! The leading coefficient is \(x\text{,}\) not a constant, so the exponential-guess machinery breaks down. (If you’re curious: the substitution \(v = y'\) turns this into the separable first-order equation \(x v' + v = 0\text{,}\) leading to \(y = C\ln x + D\) for \(x \gt 0\text{.}\))
-
Characteristic equation: \(xr^2 + r = 0\text{.}\)
- A characteristic equation must have constant coefficients—an \(x\) inside means the method was never valid to begin with.
-
Undetermined coefficients.
- Undetermined coefficients assumes constant coefficients and a nonzero forcing function; this equation has neither.
-
Integrating factor, directly on \(y''\text{.}\)
- The integrating factor method as we know it handles first-order linear equations. (There is a first-order equation hiding here, though—what happens if you rename \(y'\) as a new unknown?)
(g) Any Order Welcome.
\begin{equation*}
y''' - y' = 0
\end{equation*}
-
Characteristic equation—it works for LHCC equations of any order.
- Correct! \(r^3 - r = r(r-1)(r+1) = 0\) gives roots \(0, 1, -1\text{,}\) so \(y = C_1 + C_2 e^{x} + C_3 e^{-x}\text{.}\)
-
No method applies—the characteristic equation only handles second-order equations.
- The characteristic equation scales to any order: an order-\(n\) LHCC equation gives a degree-\(n\) polynomial with \(n\) roots and \(n\) terms in the general solution.
-
Direct integration, three times.
- After one integration you’d have \(y'' - y = C\text{,}\) which still involves \(y\)—direct integration can’t finish the job.
-
Undetermined coefficients.
- The forcing function is \(0\text{,}\) so there is no particular solution to build—the characteristic equation alone gives the general solution.
