Worked examples — Non-homogeneous — variation of parameters

This page is a shooting range. The parent note built the machine; here we fire it at every kind of target so you never meet a case you haven't already solved.

Before we start, one reminder of the whole recipe, so every example below reads the same way:

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The scenario matrix

Think of "what could the problem throw at me?" as a grid. Each row is a case class; the last column names the example that handles it.

#	Case class	What makes it tricky	Handled by
A	Distinct real roots, exponential $g$	$W\ne 1$, RHS overlaps a homogeneous solution (resonance)	Ex. 1
B	Complex roots, RHS undoable by undetermined coeff. ($\tan x$)	$g$ has a genuine singularity; watch the domain	Ex. 2
C	Repeated root, polynomial-ish $g$	$y_2=x{e}^{rx}$, the "$x\cdot$" factor changes $W$	Ex. 3
D	Cauchy–Euler (non-constant coefficients)	Must divide by $x^2$ first; $g$ is not the raw RHS	Ex. 4
E	Degenerate: $g\equiv 0$	The sanity check — method must return $y_p=0$	Ex. 5
F	Sign / quadrant care: $g=1/x$, log appears	$\ln$ needs $	x
G	Definite-integral form (no elementary antiderivative)	Answer stays as an integral — that's allowed	Ex. 7
H	Real-world word problem (forced spring, no damping, off-resonance)	Translate physics → ODE → VoP	Ex. 8

We now hit each cell.

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Example 1 — Cell A: distinct roots, resonant exponential

Forecast (guess first): the RHS $e^{2x}$ is itself a homogeneous solution. Undetermined coefficients would need an extra $x$. Do you think VoP will conjure that $x$ automatically? (Yes — watch.)

1. Homogeneous solutions. $r^2-3r+2=(r-1)(r-2)=0\Rightarrow y_1=e^x,y_2=e^{2x}$. Why this step? $y_p$ is built only from these two blocks.
2. Standard form. Coefficient of $y^{\prime \prime }$ is already $1$, so $g=e^{2x}$. Why? The formula's $g$ must be the RHS of the normalized equation.
3. Wronskian. $W=e^x\cdot 2e^{2x}-e^{2x}\cdot e^x=2e^{3x}-e^{3x}=e^{3x}$. Why? $W$ sits in every denominator.
4. The derivatives. $u_1^{\prime }=-\frac{e^{2x}\cdot e^{2x}}{e^{3x}}=-e^x,u_2^{\prime }=\frac{e^x\cdot e^{2x}}{e^{3x}}=1.$ Why? Straight from $u_1^{\prime }=-y_{2}g/W,u_2^{\prime }=+y_{1}g/W$.
5. Integrate. $u_1=-e^x,u_2=x$. (Drop constants.)
6. Assemble. $y_p=e^x(-e^x)+e^{2x}(x)=-e^{2x}+x{e}^{2x}.$ The $-e^{2x}$ is a homogeneous piece (absorb into $c_2$). So $y_p=x{e}^{2x}$ — VoP produced the resonance factor $x$ by itself.

General solution: $y=c_1{e}^x+c_2{e}^{2x}+x{e}^{2x}$.

Recall Verify

Plug $y_p=x{e}^{2x}$: $y_p^{\prime }=e^{2x}+2x{e}^{2x}$, $y_p^{\prime \prime }=4e^{2x}+4x{e}^{2x}$. $y_p^{\prime \prime }-3y_p^{\prime }+2y_p=(4e^{2x}+4x{e}^{2x})-3(e^{2x}+2x{e}^{2x})+2x{e}^{2x}=e^{2x}$. ✓ RHS matches.

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Example 2 — Cell B: complex roots, $g=\tan x$

Forecast: $\tan x$ blows up at $\pm \pi /2$ — so the solution is only valid between those walls. Guess: the answer will contain a $\ln |\sec x+\tan x|$ (the standard $\int \sec$).

1. Homogeneous. $r^2+1=0\Rightarrow r=\pm i\Rightarrow y_1=\cos x,y_2=\sin x$. Why? Complex roots $\alpha \pm i\beta$ give $e^{\alpha x}(\cos \beta x,\sin \beta x)$; here $\alpha =0,\beta =1$.
2. Standard form. Already normalized, $g=\tan x$.
3. Wronskian. $W=\cos x\cos x-\sin x(-\sin x)=1$.
4. Derivatives. $u_1^{\prime }=-\frac{\sin x\tan x}{1}=-\frac{{\sin }^{2}x}{\cos x},u_2^{\prime }=\frac{\cos x\tan x}{1}=\sin x.$
5. Integrate. $u_2=-\cos x$ (easy). For $u_1$, rewrite $\frac{{\sin }^{2}x}{\cos x}=\frac{1-{\cos }^{2}x}{\cos x}=\sec x-\cos x$. So $u_1=-\int (\sec x-\cos x)dx=-\ln |\sec x+\tan x|+\sin x.$ Why the rewrite? ${\sin }^2/\cos$ isn't a standard integral; splitting it turns it into $\sec x$ (known) minus $\cos x$ (trivial).
6. Assemble. $y_p=\cos x(\sin x-\ln |\sec x+\tan x|)+\sin x(-\cos x)=-\cos x\ln |\sec x+\tan x|.$ The $\cos x\sin x$ terms cancel.

General: $y=c_1\cos x+c_2\sin x-\cos x\ln |\sec x+\tan x|$.

Recall Verify

Let $y_p=-\cos xL$ with $L=\ln |\sec x+\tan x|$, $L^{\prime }=\sec x$. $y_p^{\prime }=\sin xL-\cos x\sec x=\sin xL-1$. $y_p^{\prime \prime }=\cos xL+\sin x\sec x=\cos xL+\tan x$. $y_p^{\prime \prime }+y_p=(\cos xL+\tan x)+(-\cos xL)=\tan x$. ✓

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Example 3 — Cell C: repeated root

Forecast: repeated root $r=2$, so $y_2=x{e}^{2x}$. The $\ln x$ is impossible for undetermined coefficients — VoP is the only tool. Guess: a $\ln x$ and an $x^2$ survive.

1. Homogeneous. $r^2-4r+4=(r-2{)}^2=0$, double root $r=2$. Second solution via Reduction of Order: $y_1=e^{2x},y_2=x{e}^{2x}$.
2. Standard form. Normalized already, $g=e^{2x}\ln x$.
3. Wronskian. $y_1^{\prime }=2e^{2x}$, $y_2^{\prime }=e^{2x}+2x{e}^{2x}$. $W=e^{2x}(e^{2x}+2x{e}^{2x})-x{e}^{2x}(2e^{2x})=e^{4x}+2x{e}^{4x}-2x{e}^{4x}=e^{4x}.$
4. Derivatives. $u_1^{\prime }=-\frac{x{e}^{2x}\cdot e^{2x}\ln x}{e^{4x}}=-x\ln x,u_2^{\prime }=\frac{e^{2x}\cdot e^{2x}\ln x}{e^{4x}}=\ln x.$
5. Integrate (integration by parts): $u_2=\int \ln xdx=x\ln x-x,$ $u_1=-\int x\ln xdx=-\left(\frac{x^2}{2}\ln x-\frac{x^2}{4}\right)=-\frac{x^2}{2}\ln x+\frac{x^2}{4}.$
6. Assemble. $y_p=e^{2x}\left(-\frac{x^2}{2}\ln x+\frac{x^2}{4}\right)+x{e}^{2x}\left(x\ln x-x\right).$ Collect: $\ln x$ terms $=e^{2x}{x}^2\ln x\left(-\frac{1}{2}+1\right)=\frac{1}{2}{x}^2{e}^{2x}\ln x$; non-log terms $=e^{2x}(\frac{x^2}{4}-x^2)=-\frac{3}{4}{x}^2{e}^{2x}$. $y_p=\frac{1}{2}{x}^2{e}^{2x}\ln x-\frac{3}{4}{x}^2{e}^{2x}.$ The $-\frac{3}{4}{x}^2{e}^{2x}$ term is not homogeneous ($x^2{e}^{2x}$ isn't in the span), so keep it.

General: $y=c_1{e}^{2x}+c_{2}x{e}^{2x}+\frac{1}{2}{x}^2{e}^{2x}\ln x-\frac{3}{4}{x}^2{e}^{2x}$.

Recall Verify

Numerically check at $x=1$: at $x=1$, $\ln 1=0$, so $y_p(1)=-\frac{3}{4}{e}^2$. The VERIFY block confirms the residual $y_p^{\prime \prime }-4y_p^{\prime }+4y_p=e^{2x}\ln x$ symbolically.

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Example 4 — Cell D: Cauchy–Euler, standard form trap

Forecast: the #1 error is using $g=x^2$ raw. After dividing by $x^2$ the real $g$ is $1$. Guess: an $x^2/3$ shows up.

1. Homogeneous given: $y_1=x$, $y_2=x^{-1}$.
2. Standard form — do NOT skip. Divide by $x^2$: $y^{\prime \prime }+\frac{1}{x}{y}^{\prime }-\frac{1}{x^2}y=1\Rightarrow g=1.$ Why? The formula needs leading coefficient $1$; the RHS $x^2$ divided by $x^2$ gives $g=1$.
3. Wronskian. $y_1^{\prime }=1,y_2^{\prime }=-x^{-2}$. $W=x(-x^{-2})-x^{-1}(1)=-x^{-1}-x^{-1}=-\frac{2}{x}.$
4. Derivatives. $u_1^{\prime }=-\frac{x^{-1}\cdot 1}{-2/x}=-\frac{x^{-1}}{-2/x}=\frac{1}{2},u_2^{\prime }=\frac{x\cdot 1}{-2/x}=-\frac{x^2}{2}.$
5. Integrate. $u_1=\frac{x}{2}$, $u_2=-\frac{x^3}{6}$.
6. Assemble. $y_p=x\cdot \frac{x}{2}+x^{-1}\left(-\frac{x^3}{6}\right)=\frac{x^2}{2}-\frac{x^2}{6}=\frac{x^2}{3}.$

General: $y=c_{1}x+c_2{x}^{-1}+\frac{x^2}{3}$.

Recall Verify

Plug $y_p=\frac{x^2}{3}$: $y_p^{\prime }=\frac{2x}{3},y_p^{\prime \prime }=\frac{2}{3}$. $x^2\cdot \frac{2}{3}+x\cdot \frac{2x}{3}-\frac{x^2}{3}=\frac{2x^2}{3}+\frac{2x^2}{3}-\frac{x^2}{3}=x^2$. ✓

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Example 5 — Cell E: the degenerate check, $g\equiv 0$

Forecast: with nothing forcing the system, there is nothing to absorb. Guess: $y_p=0$.

1. Homogeneous. $y_1=\cos x,y_2=\sin x$, $W=1$.
2. Standard form. $g=0$.
3. Derivatives. $u_1^{\prime }=-\frac{\sin x\cdot 0}{1}=0$, $u_2^{\prime }=\frac{\cos x\cdot 0}{1}=0$.
4. Integrate. $u_1=$ const, $u_2=$ const — which we drop.
5. Assemble. $y_p=0$. Why this matters: it proves VoP is consistent — it never invents a spurious particular solution when there's no forcing. This is the "sanity anchor" every method must pass.

Recall Verify

$y_p=0$ trivially satisfies $y_p^{\prime \prime }+y_p=0$. ✓ (checked below)

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Example 6 — Cell F: sign care, $g=1/x$ and the absolute value

Forecast: the antiderivative of $1/x$ is $\ln |x|$ — the absolute value matters, and it's the reason the solution splits by sign of $x$. Also $\int e^{\pm x}/xdx$ is a non-elementary integral (the exponential integral), so the honest answer keeps an integral sign.

1. Homogeneous. $r^2-1=0\Rightarrow r=\pm 1$, so $y_1=e^x,y_2=e^{-x}$.
2. Standard form. $g=\frac{1}{x}$.
3. Wronskian. $y_1^{\prime }=e^x,y_2^{\prime }=-e^{-x}$. $W=e^x(-e^{-x})-e^{-x}(e^x)=-1-1=-2.$
4. Derivatives. $u_1^{\prime }=-\frac{e^{-x}\cdot \frac{1}{x}}{-2}=\frac{e^{-x}}{2x},u_2^{\prime }=\frac{e^x\cdot \frac{1}{x}}{-2}=-\frac{e^x}{2x}.$
5. Integrate — cannot be done in elementary form. Keep as integrals: $u_1=\frac{1}{2}\int \frac{e^{-x}}{x}dx,u_2=-\frac{1}{2}\int \frac{e^x}{x}dx.$ Why leave it? $\int e^{-x}/xdx$ has no elementary antiderivative (it's $-\mathrm{Ei}(-x)$). A correct answer stops here — see Cell G for the clean definite-integral packaging.
6. Sign of $x$. The integrals are defined only where $1/x$ is continuous: separately on $x>0$ and $x<0$. There is no solution crossing $x=0$ because $g$ has an infinite discontinuity there. This is exactly why the interval must be stated.

General (on either side): $y=c_1{e}^x+c_2{e}^{-x}+\frac{e^x}{2}\int \frac{e^{-x}}{x}dx-\frac{e^{-x}}{2}\int \frac{e^x}{x}dx$.

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Example 7 — Cell G: definite-integral (Green's-function) packaging

Forecast: instead of two separate indefinite integrals, we can fuse them into one integral of $g(t)$ times a kernel. Guess: the kernel is $\sin (x-t)$.

1. Ingredients. $y_1=\cos x,y_2=\sin x$, $W=1$.
2. VoP as definite integrals. $u_1(x)=-{\int }_a^x\frac{\sin tg(t)}{1}dt,u_2(x)={\int }_a^x\frac{\cos tg(t)}{1}dt.$ Why definite? Using $a$ as the lower limit fixes the constants so that $y_p(a)=y_p^{\prime }(a)=0$ — a clean, unambiguous particular solution.
3. Assemble & fuse. $y_p=\cos x\left(-{\int }_a^x\sin tgdt\right)+\sin x\left({\int }_a^x\cos tgdt\right)={\int }_a^x(\sin x\cos t-\cos x\sin t)g(t)dt.$ By the sine-subtraction identity $\sin x\cos t-\cos x\sin t=\sin (x-t)$: $\boxed{y_p(x)={\int }_a^x\sin (x-t)g(t)dt}$ The kernel $\sin (x-t)$ is the Green's function for this operator.

Why this is powerful: one formula solves the equation for every $g$ at once — you just drop your $g$ into the integrand. This is Cell A–F all packaged.

Recall Verify

Take $g=\sec x$, $a=0$: $y_p={\int }_0^x\sin (x-t)\sec tdt$. Differentiating twice (Leibniz rule) gives $y_p^{\prime \prime }+y_p=\sec x$; and it reproduces Ex. from the parent up to homogeneous pieces. The numeric spot-check at $x=1$ is in VERIFY.

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Example 8 — Cell H: real-world forced oscillator

Forecast: since drive frequency ($1$) $\ne$ natural frequency ($2$), no resonance — the response should be a bounded combination of $\cos t$ and $\cos 2t,\sin 2t$. Guess amplitude of the $\cos t$ part $\approx 1$.

1. Homogeneous. $r^2+4=0\Rightarrow r=\pm 2i$, so $y_1=\cos 2t,y_2=\sin 2t$.
2. Standard form. $g=3\cos t$.
3. Wronskian. $W=\cos 2t(2\cos 2t)-\sin 2t(-2\sin 2t)=2({\cos }^{2}2t+{\sin }^{2}2t)=2$.
4. Derivatives. $u_1^{\prime }=-\frac{\sin 2t\cdot 3\cos t}{2},u_2^{\prime }=\frac{\cos 2t\cdot 3\cos t}{2}.$
5. Integrate using product-to-sum ($\sin 2t\cos t=\frac{1}{2}(\sin 3t+\sin t)$, $\cos 2t\cos t=\frac{1}{2}(\cos 3t+\cos t)$): $u_1=-\frac{3}{4}\int (\sin 3t+\sin t)dt=-\frac{3}{4}\left(-\frac{\cos 3t}{3}-\cos t\right)=\frac{1}{4}\cos 3t+\frac{3}{4}\cos t,$ $u_2=\frac{3}{4}\int (\cos 3t+\cos t)dt=\frac{3}{4}\left(\frac{\sin 3t}{3}+\sin t\right)=\frac{1}{4}\sin 3t+\frac{3}{4}\sin t.$
6. Assemble. $y_p=\cos 2t\left(\frac{1}{4}\cos 3t+\frac{3}{4}\cos t\right)+\sin 2t\left(\frac{1}{4}\sin 3t+\frac{3}{4}\sin t\right).$ Group with sum identities: $\frac{1}{4}(\cos 2t\cos 3t+\sin 2t\sin 3t)=\frac{1}{4}\cos t$; $\frac{3}{4}(\cos 2t\cos t+\sin 2t\sin t)=\frac{3}{4}\cos t$. Total $y_p=\cos t$.
7. Apply initial conditions. General $y=c_1\cos 2t+c_2\sin 2t+\cos t$. $y(0)=c_1+1=0\Rightarrow c_1=-1$. $y^{\prime }(t)=-2c_1\sin 2t+2c_2\cos 2t-\sin t$; $y^{\prime }(0)=2c_2=0\Rightarrow c_2=0$. $\boxed{y(t)=\cos t-\cos 2t}$

Units sanity: each term is dimensionless in the pattern but represents metres; both cosines are bounded by $1$, so $|y|\le 2$ m — bounded, as forecast for off-resonance. The figure shows the beat-like superposition.

Recall Verify

$y=\cos t-\cos 2t$: $y^{\prime \prime }=-\cos t+4\cos 2t$. $y^{\prime \prime }+4y=(-\cos t+4\cos 2t)+4(\cos t-\cos 2t)=3\cos t$. ✓ And $y(0)=1-1=0$, $y^{\prime }(0)=0$. ✓

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The matrix, revisited

Recall Which example filled which cell?

A resonant-exponential ::: Ex. 1 B complex roots + $\tan x$ ::: Ex. 2 C repeated root + $\ln x$ ::: Ex. 3 D Cauchy–Euler standard-form trap ::: Ex. 4 E degenerate $g\equiv 0$ ::: Ex. 5 F sign/quadrant $1/x$ and $\ln |x|$ ::: Ex. 6 G definite-integral / Green's kernel ::: Ex. 7 H real-world forced spring with initial conditions ::: Ex. 8

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Connections

• Wronskian — the $W$ in every denominator; non-zero guarantees the linear system is solvable via Cramer's Rule.
• Cramer's Rule — how $u_1^{\prime },u_2^{\prime }$ were extracted from the $2\times 2$ system.
• Method of Undetermined Coefficients — the shortcut that fails on Ex. 2, 3, 6, 7.
• Second-order linear homogeneous ODE — source of $y_1,y_2$.
• Reduction of Order — how the repeated-root $y_2=x{e}^{2x}$ in Ex. 3 is found.
• Green's function — the fused kernel $\sin (x-t)$ of Ex. 7.