The Wronskian & Variation of Parameters

Why this determinant: to match arbitrary data $y(x_0) = a$, $y'(x_0) = b$ with $c_1y_1 + c_2y_2$, you must solve $c_1y_1(x_0) + c_2y_2(x_0) = a$ and $c_1y_1'(x_0) + c_2y_2'(x_0) = b$ — a linear system whose determinant is $W(x_0)$. Nonzero $W$ means $\{y_1, y_2\}$ is a fundamental set — every solution is a combination of them, and every initial-value problem is solvable within their span. Computing $W$ is a $2\times2$ determinant: for $\cos\beta x, \sin\beta x$ it is the constant $\beta$.

Unlike undetermined coefficients this needs no guess and works for any continuous forcing: $\sec x$, $\tfrac{e^x}{x}$, $\ln x$. The cost is two possibly-ugly integrals; the gain is universality. The side condition is a choice, not a law: two unknown functions leave room for two conditions, and this one keeps $u_1', u_2'$ out of $y_p'$, so that $y_p''$ contains only first derivatives of the $u_i$ and the equation becomes a linear system for $u_1'$ and $u_2'$.

Step 5 — sanity: the $x\sin x$ piece echoes resonance ($\sec x$ contains the natural frequency), and differentiating twice confirms $y_p'' + y_p = \sec x$ ✓.

Step 6 — where it lives: $\sec x$ is continuous only on $|x| < \tfrac\pi2$ (there $\cos x > 0$, so $\ln(\cos x)$ makes sense), and on that interval $y_p(0) = 0$ and $y_p'(0) = -\sin 0\,\ln(\cos 0) + 0 = 0$: this $y_p$ is the response of the system started from rest.

The formula is Cramer's rule for the two conditions on $u_1', u_2'$: the side condition $u_1'y_1 + u_2'y_2 = 0$ and the equation's requirement $u_1'y_1' + u_2'y_2' = f$. The coefficient matrix has determinant $W$, never zero for a fundamental pair, so the system always has a solution — that is why the method cannot fail when $f$ is continuous. The $f$ must come from the equation with leading coefficient $1$, because the derivation substitutes into $y'' + py' + qy = f$.

Step 3 — integrate: $u_1 = 2\cos 2x$ and $u_2 = 2\sin 2x$.

Step 4 — $y_p = 2\cos^2 2x + 2\sin^2 2x = 2$.

Step 5 — compare: guessing a constant $y_p = A$ gives $4A = 8$, so $A = 2$, the same answer. The machinery is heavier than needed here, but it produces exactly what guessing produces whenever guessing works.

Step 6 — from rest, $y(0) = y'(0) = 0$ forces $c_1 = -2$, $c_2 = 0$: $y = 2 - 2\cos 2x$, which oscillates between $0$ and $4$ about $y_p = 2$.

Step 3 — integrate: $u_1 = \sin x - \ln(\sec x + \tan x)$ and $u_2 = -\cos x$.

Step 4 — assemble: $y_p = \cos x\sin x - \cos x\ln(\sec x + \tan x) - \sin x\cos x = -\cos x\ln(\sec x + \tan x)$.

Step 5 — the general solution is $y = c_1\cos x + c_2\sin x - \cos x\ln(\sec x + \tan x)$. Undetermined coefficients could not have found this: $\tan x$ has infinitely many independent derivatives, so there is no finite family to guess from.

Step 3 — $u_1' = -\dfrac{x^2\cdot x}{x^2} = -x$ and $u_2' = \dfrac{x\cdot x}{x^2} = 1$.

Step 4 — $u_1 = -\tfrac{x^2}{2}$ and $u_2 = x$, so $y_p = -\tfrac{x^3}{2} + x^3 = \tfrac{x^3}{2}$.

Step 5 — verify in the original equation: $x^2\cdot 3x - 2x\cdot\tfrac{3x^2}{2} + 2\cdot\tfrac{x^3}{2} = 3x^3 - 3x^3 + x^3 = x^3$ ✓. The general solution is $y = c_1x + c_2x^2 + \tfrac{x^3}{2}$.

By the quotient rule, $\left(\dfrac{y_2}{y_1}\right)' = \dfrac{y_1y_2' - y_1'y_2}{y_1^2} = \dfrac{W}{y_1^2}$. Abel's theorem says $W = Ce^{-\int p}$ for any pair of solutions, so choosing $C = 1$ and integrating once recovers the ratio $\tfrac{y_2}{y_1}$, hence $y_2$. Because $W \ne 0$, the ratio is not constant, so $y_2$ really is independent of $y_1$.

Step 2 — direct computation: $W = x^2\cdot(-2x^{-3}) - 2x\cdot x^{-2} = -\dfrac2x - \dfrac2x = -\dfrac4x$.

Step 3 — this has Abel's shape with $C = -4$, and it is never zero for $x > 0$, so $x^2$ and $x^{-2}$ form a fundamental set.

Step 3 — $y_2 = x^2\displaystyle\int\frac{x^3}{x^4}\,dx = x^2\ln x$.

Step 4 — verify: $y_2' = 2x\ln x + x$ and $y_2'' = 2\ln x + 3$, so $x^2(2\ln x + 3) - 3x(2x\ln x + x) + 4x^2\ln x = 0$ ✓. The indicial equation $r(r - 1) - 3r + 4 = (r - 2)^2$ has a double root, and $\ln x$ plays the role that the extra factor $x$ plays for constant coefficients.

Together with the side condition this is a $2\times2$ linear system in $u_1', u_2'$ whose determinant is exactly $W$; Cramer's rule gives $u_1' = -y_2f/W$, $u_2' = y_1f/W$. $\blacksquare$

Step 1 — homogeneous: $(r-1)^2 = 0$, double root: $y_1 = e^x$, $y_2 = xe^x$.

Step 2 — Wronskian: $W = e^x(e^x + xe^x) - e^x\cdot xe^x = e^{2x}$.

Step 3 — the formulas with $f = \tfrac{e^x}{x}$: $u_1' = -\dfrac{y_2f}{W} = -\dfrac{xe^x\cdot e^x/x}{e^{2x}} = -1$, and $u_2' = \dfrac{y_1f}{W} = \dfrac{e^x\cdot e^x/x}{e^{2x}} = \dfrac1x$.

Step 4 — integrate: $u_1 = -x$, $u_2 = \ln x$.

Step 5 — $y_p = -xe^x + xe^x\ln x$; the $-xe^x$ is a homogeneous solution and can be dropped: $y_p = xe^x\ln x$. General solution: $y = (c_1 + c_2x)e^x + xe^x\ln x$. Undetermined coefficients has no guess for $\tfrac{e^x}{x}$ — this method never needed one.

Step 1 — check each solves the equation: for $y = x$: $0 - x + x = 0$ ✓; for $y = x\ln x$: $y' = \ln x + 1$, $y'' = \tfrac1x$, so $x^2\cdot\tfrac1x - x(\ln x + 1) + x\ln x = x - x\ln x - x + x\ln x = 0$ ✓.

Step 2 — Wronskian: $W = x(\ln x + 1) - 1\cdot x\ln x = x$, nonzero for $x > 0$.

Step 3 — conclusion: yes — independent solutions, so every solution is $c_1x + c_2x\ln x$.

Step 4 — connect: this is a Cauchy–Euler equation with double indicial root $r = 1$, and the $\ln x$ companion is the $x = e^t$ shadow of the $te^{rt}$ rule. The Wronskian being exactly $x$ matches Abel's formula $W = Ce^{-\int p}$ with $p = -\tfrac1x$.