Exercises — Non-homogeneous — variation of parameters

WHAT we do: plug into $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$. $y_1=\cos x\Rightarrow y_1^{\prime }=-\sin x$. $y_2=\sin x\Rightarrow y_2^{\prime }=\cos x$. $W=\cos x\cdot \cos x-\sin x\cdot (-\sin x)={\cos }^{2}x+{\sin }^{2}x=1.$ WHY it matters: $W$ sits in the denominator of both fractions. A clean $W=1$ means no extra division later. $\boxed{W=1}$.

WHAT we do: differentiate each solution, then cross-multiply per $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$. $y_1^{\prime }=e^x,y_2^{\prime }=2e^{2x}$. $W=e^x\cdot 2e^{2x}-e^{2x}\cdot e^x=2e^{3x}-e^{3x}=e^{3x}.$ WHY the sum of exponents: $e^x\cdot e^{2x}=e^{x+2x}=e^{3x}$ every time, so both products land on $e^{3x}$ and simply subtract as coefficients $2-1=1$. $\boxed{W=e^{3x}}$. It is never zero — the two exponentials are genuinely different building blocks, so the method is guaranteed to work (this is exactly what a non-vanishing Wronskian certifies).

WHAT we do: divide every term by the leading coefficient $2x^2$. $y^{\prime \prime }+\frac{x}{2x^2}{y}^{\prime }-\frac{1}{2x^2}y=\frac{6x}{2x^2}$ $y^{\prime \prime }+\frac{1}{2x}{y}^{\prime }-\frac{1}{2x^2}y=\frac{3}{x}.$ WHY: the formula's $g$ is the RHS of the normalized equation, not the raw $6x$. So $\boxed{g(x)=3/x}$, not $6x$.

Step 1 — homogeneous. $r^2+1=0\Rightarrow r=\pm i$, so $y_1=\cos x,y_2=\sin x$ (see Second-order linear homogeneous ODE). Step 2 — Wronskian. From Ex 1.1, $W=1$. Step 3 — the two derivatives. With $g=\tan x$: $u_1^{\prime }=-\frac{y_{2}g}{W}=-\sin x\tan x=-\frac{{\sin }^{2}x}{\cos x},u_2^{\prime }=\frac{y_{1}g}{W}=\cos x\tan x=\sin x.$ WHY rewrite $\sin x\tan x$ as ${\sin }^{2}x/\cos x$: because a bare $\sin x\tan x$ has no standard antiderivative, but ${\sin }^{2}x/\cos x$ can be turned into secants and cosines, which do. Step 4 — integrate. The second is easy: $u_2=-\cos x$. For $u_1$, write ${\sin }^{2}x=1-{\cos }^{2}x$ so the fraction splits: $u_1=-\int \frac{1-{\cos }^{2}x}{\cos x}dx=-\int (\sec x-\cos x)dx=-\ln |\sec x+\tan x|+\sin x.$ WHY this split: $\frac{1}{\cos x}=\sec x$ (a known integral $\ln |\sec x+\tan x|$) and $\frac{{\cos }^{2}x}{\cos x}=\cos x$ (trivial). Turning one hard fraction into two known integrals is the whole trick. Step 5 — assemble. $y_p=\cos x(\sin x-\ln |\sec x+\tan x|)+\sin x(-\cos x).$ The $\cos x\sin x$ terms cancel: $\boxed{y_p=-\cos x\ln |\sec x+\tan x|}.$ General: $y=c_1\cos x+c_2\sin x-\cos x\ln |\sec x+\tan x|$.

Domain caveat. $\tan x$ blows up at $x=\pm \frac{\pi }{2}$, and $\sec x+\tan x>0$ throughout $(-\frac{\pi }{2},\frac{\pi }{2})$, so the logarithm's argument is positive and the absolute value is unnecessary on this interval — but keep it if you extend to other branches, where $\sec x+\tan x$ changes sign. The solution is only valid on one such open interval between consecutive singularities of $\tan x$; you cannot cross $x=\frac{\pi }{2}$.

$y_1=e^x,y_2=e^{2x}$, $W=e^{3x}$ (Ex 1.2). With $g=e^{3x}$: $u_1^{\prime }=-\frac{e^{2x}\cdot e^{3x}}{e^{3x}}=-e^{2x},u_2^{\prime }=\frac{e^x\cdot e^{3x}}{e^{3x}}=e^x.$ WHY the fractions collapse so cleanly: numerator $e^{2x}{e}^{3x}=e^{5x}$ divided by $W=e^{3x}$ leaves $e^{2x}$ — exponentials always subtract exponents, so no messy integral survives. Integrate: $u_1=-\frac{1}{2}{e}^{2x}$ (WHY the $\frac{1}{2}$: $\int e^{2x}dx=\frac{1}{2}{e}^{2x}$, the inner-derivative factor of $2$), and $u_2=e^x$. $y_p=e^x\left(-\frac{1}{2}{e}^{2x}\right)+e^{2x}(e^x)=-\frac{1}{2}{e}^{3x}+e^{3x}=\frac{1}{2}{e}^{3x}.$ $\boxed{y_p=\frac{1}{2}{e}^{3x}}$. General: $y=c_1{e}^x+c_2{e}^{2x}+\frac{1}{2}{e}^{3x}$. (Since $e^{3x}$ is not a homogeneous solution, Method of Undetermined Coefficients would give the same answer with a guess $A{e}^{3x}$ — a good sanity check.)

Step 0 — standard form. Divide by $x^2$: $y^{\prime \prime }-\frac{1}{x}{y}^{\prime }+\frac{1}{x^2}y=\frac{1}{x}\Rightarrow g=\frac{1}{x}.$ Step 2 — Wronskian. $y_1=x\Rightarrow y_1^{\prime }=1$. $y_2=x\ln x\Rightarrow y_2^{\prime }=\ln x+1$. WHY $y_2^{\prime }=\ln x+1$: product rule on $x\cdot \ln x$ gives $1\cdot \ln x+x\cdot \frac{1}{x}=\ln x+1$. $W=x(\ln x+1)-x\ln x\cdot 1=x\ln x+x-x\ln x=x.$ Step 3. $u_1^{\prime }=-\frac{y_{2}g}{W}=-\frac{x\ln x\cdot \frac{1}{x}}{x}=-\frac{\ln x}{x},u_2^{\prime }=\frac{y_{1}g}{W}=\frac{x\cdot \frac{1}{x}}{x}=\frac{1}{x}.$ Step 4 — integrate. $u_1=-\int \frac{\ln x}{x}dx=-\frac{(\ln x{)}^2}{2}$. WHY substitute $t=\ln x$: then $dt=\frac{1}{x}dx$, so the integral becomes $\int tdt=\frac{t^2}{2}$ — the $\frac{1}{x}$ was exactly the missing piece that makes $\ln x$'s own derivative appear. $u_2=\int \frac{1}{x}dx=\ln x$. Step 5. $y_p=x\left(-\frac{(\ln x{)}^2}{2}\right)+x\ln x(\ln x)=-\frac{x(\ln x{)}^2}{2}+x(\ln x{)}^2=\frac{x(\ln x{)}^2}{2}.$ $\boxed{y_p=\frac{1}{2}x(\ln x{)}^2}$. General: $y=c_{1}x+c_{2}x\ln x+\frac{1}{2}x(\ln x{)}^2$. Domain caveat. $\ln x$ needs $x>0$; $y_2=x\ln x$ and the whole solution live only on $x>0$. The point $x=0$ is a singular point of the original Euler equation, so no solution crosses it.

Step 0 — standard form. Divide by $x^2$: $y^{\prime \prime }-\frac{2}{x^2}y=1$, so $g=1$ (not $x^2$!). Step 2 — Wronskian. $y_1^{\prime }=2x$, $y_2^{\prime }=-x^{-2}$. $W=x^2(-x^{-2})-x^{-1}(2x)=-1-2=-3.$ A negative constant $W$ — perfectly fine. Step 3. $u_1^{\prime }=-\frac{x^{-1}\cdot 1}{-3}=\frac{1}{3x},u_2^{\prime }=\frac{x^2\cdot 1}{-3}=-\frac{x^2}{3}.$ Step 4. $u_1=\frac{1}{3}\ln |x|=\frac{1}{3}\ln x$ (we are on $x>0$), $u_2=-\frac{x^3}{9}$. WHY $\int \frac{1}{x}dx=\ln x$ and $\int x^{2}dx=\frac{x^3}{3}$ — the standard power rule, then times $-\frac{1}{3}$. Step 5. $y_p=x^2\left(\frac{1}{3}\ln x\right)+x^{-1}\left(-\frac{x^3}{9}\right)=\frac{1}{3}{x}^2\ln x-\frac{1}{9}{x}^2.$ WHY we may drop $-\frac{1}{9}{x}^2$: it is a constant multiple of $y_1=x^2$, i.e. already inside the complementary function $c_1{x}^2$, so it changes only $c_1$ and adds nothing new. $\boxed{y_p=\frac{1}{3}{x}^2\ln x}$. General: $y=c_1{x}^2+c_2{x}^{-1}+\frac{1}{3}{x}^2\ln x$. Domain caveat. $y_2=x^{-1}$ is singular at $x=0$; we work on $x>0$ (or symmetrically $x<0$), never across it.

Step 1. $r^2-2r+1=(r-1{)}^2=0$, a repeated root. So $y_1=e^x$ and $y_2=x{e}^x$ (the second from Reduction of Order). Step 2 — Wronskian. $y_1^{\prime }=e^x$, $y_2^{\prime }=e^x+x{e}^x$. $W=e^x(e^x+x{e}^x)-x{e}^x\cdot e^x=e^{2x}+x{e}^{2x}-x{e}^{2x}=e^{2x}.$ WHY the $x{e}^{2x}$ cancels: the cross-terms are equal and opposite — this cancellation is exactly why the Wronskian of $e^x,x{e}^x$ stays a clean $e^{2x}$. Step 3. $g=e^x$: $u_1^{\prime }=-\frac{x{e}^x\cdot e^x}{e^{2x}}=-x,u_2^{\prime }=\frac{e^x\cdot e^x}{e^{2x}}=1.$ WHY a polynomial $-x$ appears: dividing $e^{2x}$ by $e^{2x}$ kills the exponential entirely, leaving the bare factor $x$ from $y_2=x{e}^x$. That surviving $x$ is what will integrate into the $x^2$ resonance term. Step 4. $u_1=-\frac{x^2}{2}$, $u_2=x$. Step 5. $y_p=e^x\left(-\frac{x^2}{2}\right)+x{e}^x(x)=-\frac{x^2}{2}{e}^x+x^2{e}^x=\frac{x^2}{2}{e}^x.$ $\boxed{y_p=\frac{1}{2}{x}^2{e}^x}$ — exactly the $x^2{e}^x$ shape resonance forces, and VoP produced it with no guessing. General: $y=(c_1+c_{2}x)e^x+\frac{1}{2}{x}^2{e}^x$.

Step 1. $r^2-1=0\Rightarrow r=\pm 1$, so $y_1=e^x,y_2=e^{-x}$. Step 2 — Wronskian. $y_1^{\prime }=e^x$, $y_2^{\prime }=-e^{-x}$. $W=e^x(-e^{-x})-e^{-x}{e}^x=-1-1=-2.$ Step 3. $g=\frac{2}{1+e^x}$: $u_1^{\prime }=-\frac{e^{-x}\cdot \frac{2}{1+e^x}}{-2}=\frac{e^{-x}}{1+e^x},u_2^{\prime }=\frac{e^x\cdot \frac{2}{1+e^x}}{-2}=-\frac{e^x}{1+e^x}.$ Step 4 — integrate. For $u_2$: $\int \frac{e^x}{1+e^x}dx=\ln (1+e^x)$, WHY — the numerator $e^x$ is precisely the derivative of the denominator $1+e^x$, so it is a direct $\int \frac{f^{\prime }}{f}=\ln f$ pattern (and $1+e^x>0$ always, so no absolute value is ever needed here). Thus $u_2=-\ln (1+e^x)$. For $u_1$: let $t=e^x$, so $dt=e^{x}dx=tdx$, i.e. $dx=dt/t$. WHY this substitution — it converts the mixed $e^{-x}/(1+e^x)$ into a pure rational function of $t$, which partial fractions can crack: $u_1=\int \frac{e^{-x}}{1+e^x}dx=\int \frac{1/t}{1+t}\cdot \frac{dt}{t}=\int \frac{dt}{t^2(1+t)}.$ Partial fractions — why this shape. A denominator $t^2(1+t)$ demands one term per distinct power of each factor: a $\frac{A}{t}$ (simple), a $\frac{B}{t^2}$ (repeated factor $t$ needs both powers), and a $\frac{C}{1+t}$ (the linear factor). Solving $1=At(1+t)+B(1+t)+C{t}^2$ gives $B=1$ (set $t=0$), $C=1$ (set $t=-1$), $A=-1$ (match $t^2$). Hence $\frac{1}{t^2(1+t)}=\frac{-1}{t}+\frac{1}{t^2}+\frac{1}{1+t}.$ Integrate term by term: $u_1=-\ln t-\frac{1}{t}+\ln (1+t)=-x-e^{-x}+\ln (1+e^x)(\text{since }t=e^x>0).$ Step 5. $y_p=e^x(-x-e^{-x}+\ln (1+e^x))+e^{-x}(-\ln (1+e^x)).$ $=-x{e}^x-1+e^x\ln (1+e^x)-e^{-x}\ln (1+e^x).$ The $-1$ is a constant, but $1$ is not a homogeneous solution here (those are $e^{\pm x}$), so keep it. Cleanly: $\boxed{y_p=-x{e}^x-1+(e^x-e^{-x})\ln (1+e^x)}.$ Domain caveat. $1+e^x>0$ for every real $x$, so this solution is valid on the whole real line — no branch cuts, no excluded points. Contrast Ex 2.1, where $\tan x$ forced us onto a single interval.

$y_1=\cos x,y_2=\sin x,W=1$. Then $u_1^{\prime }=-\sin xg(x),u_2^{\prime }=\cos xg(x).$ Integrate from a base point $x_0$ to $x$ (use dummy variable $t$): $u_1(x)=-{\int }_{x_0}^x\sin tg(t)dt,u_2(x)={\int }_{x_0}^x\cos tg(t)dt.$ WHY definite integrals with a base point $x_0$? An indefinite integral hides an arbitrary $+C$; pinning the lower limit to a chosen $x_0$ fixes that constant. How to choose $x_0$: pick it to match your initial data — if the problem specifies $y(x_0)=y^{\prime }(x_0)=0$, then this very formula already satisfies those conditions (both $u_1,u_2$ vanish at $x_0$, so $y_p(x_0)=0$, and one checks $y_p^{\prime }(x_0)=0$ too). Any other $x_0$ merely shifts $y_p$ by a homogeneous piece $c_1\cos x+c_2\sin x$, absorbed into the general solution. So $x_0$ is free for the general solution and set by the initial condition for an initial-value problem. Assemble: $y_p=\cos x\left(-{\int }_{x_0}^x\sin tg(t)dt\right)+\sin x{\int }_{x_0}^x\cos tg(t)dt.$ Pull both under one integral (both are integrals in $t$; $\cos x,\sin x$ are constants w.r.t. $t$, so they slide inside): $y_p={\int }_{x_0}^x(\sin x\cos t-\cos x\sin t)g(t)dt={\int }_{x_0}^x\sin (x-t)g(t)dt.$ WHY the bracket became $\sin (x-t)$: it is the sine subtraction identity $\sin (x-t)=\sin x\cos t-\cos x\sin t$ — the algebra was designed to reveal it. $\boxed{y_p(x)={\int }_{x_0}^x\sin (x-t)g(t)dt}$ The kernel $K(x,t)=\sin (x-t)$ is precisely the Green's function for $y^{\prime \prime }+y$: VoP is the constructive route to it. Notice the kernel depends only on the gap $x-t$ (a "how long ago did the push at time $t$ act" factor) — see the figure below. Check with $g=\sec x$ (parent Example 1) recovers $\cos x\ln |\cos x|+x\sin x$ up to homogeneous terms.

Step 1. $r^2+4=0\Rightarrow r=\pm 2i$, so $y_1=\cos 2x,y_2=\sin 2x$. Step 2 — Wronskian. $y_1^{\prime }=-2\sin 2x$, $y_2^{\prime }=2\cos 2x$. $W=\cos 2x(2\cos 2x)-\sin 2x(-2\sin 2x)=2({\cos }^{2}2x+{\sin }^{2}2x)=2.$ Step 3. $g=\csc 2x=1/\sin 2x$: $u_1^{\prime }=-\frac{\sin 2x\cdot \frac{1}{\sin 2x}}{2}=-\frac{1}{2},u_2^{\prime }=\frac{\cos 2x\cdot \frac{1}{\sin 2x}}{2}=\frac{\cos 2x}{2\sin 2x}=\frac{\cot 2x}{2}.$ WHY $u_1^{\prime }$ collapses to a constant: $\sin 2x$ in the numerator cancels the $\csc 2x=1/\sin 2x$, leaving just $-\frac{1}{2}$ — the cleanest possible integrand. Step 4. $u_1=-\frac{1}{2}x$. $u_2=\frac{1}{2}\int \cot 2xdx=\frac{1}{2}\cdot \frac{1}{2}\ln |\sin 2x|=\frac{1}{4}\ln |\sin 2x|$. WHY the extra $\frac{1}{2}$: $\frac{d}{dx}\ln |\sin 2x|=\frac{2\cos 2x}{\sin 2x}=2\cot 2x$, so $\int \cot 2xdx=\frac{1}{2}\ln |\sin 2x|$ — the inner derivative of $2x$ demands the $\frac{1}{2}$. Step 5. $y_p=\cos 2x\left(-\frac{1}{2}x\right)+\sin 2x\left(\frac{1}{4}\ln |\sin 2x|\right).$ $\boxed{y_p=-\frac{x}{2}\cos 2x+\frac{1}{4}\sin 2x\ln |\sin 2x|}.$ General: $y=c_1\cos 2x+c_2\sin 2x-\frac{x}{2}\cos 2x+\frac{1}{4}\sin 2x\ln |\sin 2x|$. Domain caveat. $\csc 2x$ is singular wherever $\sin 2x=0$, i.e. $x=0,\frac{\pi }{2},\pi ,\dots$. On $0<x<\frac{\pi }{2}$ we have $\sin 2x>0$, so $\ln |\sin 2x|=\ln (\sin 2x)$ and the absolute value is decorative there; keep it if you move to an adjacent interval where $\sin 2x<0$. The solution never crosses these singularities.

1. Normalize to leading coefficient $1$ — get the true $g$.
2. Find $y_1,y_2$ (characteristic roots, or given).
3. $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$ — confirm $W\ne 0$.
4. $u_1^{\prime }=-y_{2}g/W$, $u_2^{\prime }=+y_{1}g/W$ (minus on the first).
5. Integrate, drop constants — mind chain-rule factors and the valid interval.
6. $y_p=u_1{y}_1+u_2{y}_2$; absorb any homogeneous leftovers into $c_1,c_2$.