The Wronskian & Variation of Parameters

Study Sheet

The Wronskian & Variation of Parameters

Independence certificates, and the method that handles any forcing

The General Machinery

Tip
The Wronskian certifies independence

For solutions y1,y2y_1, y_2 of the same second-order linear homogeneous equation, the Wronskian W=y1y2′−y1′y2W = y_1y_2' - y_1'y_2 is either NEVER zero or identically zero (Abel's theorem: W=Ce−∫p dxW = Ce^{-\int p\,dx}).

Nonzero WW means {y1,y2}\{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 WW is a 2×22\times2 determinant: for cos⁡βx,sin⁡βx\cos\beta x, \sin\beta x it is the constant β\beta.

Tip
Variation of parameters: the universal particular solution

For y′′+py′+qy=fy'' + p y' + q y = f with fundamental pair y1,y2y_1, y_2: seek yp=u1y1+u2y2y_p = u_1y_1 + u_2y_2 with FUNCTIONS uiu_i. Imposing the standard side condition u1′y1+u2′y2=0u_1'y_1 + u_2'y_2 = 0 leads to u1′=−y2fWu_1' = -\dfrac{y_2 f}{W}, u2′=y1fWu_2' = \dfrac{y_1 f}{W} — two integrations and done.

Unlike undetermined coefficients this needs NO guess and works for any continuous forcing: sec⁡x\sec x, exx\tfrac{e^x}{x}, ln⁡x\ln x. The cost is two possibly-ugly integrals; the gain is universality.

Example
Worked: y″ + y = sec x, which no guess can touch

Step 1 — fundamental pair: y1=cos⁡xy_1 = \cos x, y2=sin⁡xy_2 = \sin x, W=1W = 1.

Step 2 — the formulas: u1′=−sin⁡xsec⁡x=−tan⁡xu_1' = -\sin x\sec x = -\tan x and u2′=cos⁡xsec⁡x=1u_2' = \cos x\sec x = 1.

Step 3 — integrate: u1=ln⁡(cos⁡x)u_1 = \ln(\cos x), u2=xu_2 = x.

Step 4 — assemble: yp=cos⁡x ln⁡(cos⁡x)+xsin⁡xy_p = \cos x\,\ln(\cos x) + x\sin x.

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

Side note
Significance: the Wronskian outlives the course

Wronskians appear wherever independence of functions matters: quantum mechanics uses them to normalize scattering states, and numerical ODE solvers monitor them to detect degenerating solution bases. Variation of parameters, meanwhile, is the finite-dimensional ancestor of the "Green's function" method that solves PDEs by superposing impulse responses.

Try it
Try it: a Wronskian in one line

Compute W(e2x,xe2x)W(e^{2x}, xe^{2x}). Work: W=e2x(e2x+2xe2x)−2e2x⋅xe2x=e4x(1+2x−2x)=e4xW = e^{2x}(e^{2x} + 2xe^{2x}) - 2e^{2x}\cdot xe^{2x} = e^{4x}(1 + 2x - 2x) = e^{4x} — never zero, so the repeated-root pair really is independent, justifying the (A+Bx)e2x(A + Bx)e^{2x} general solution.

Proofs & Why It Matters

Tip
Proof: the variation-of-parameters formulas

With yp=u1y1+u2y2y_p = u_1y_1 + u_2y_2 and side condition u1′y1+u2′y2=0u_1'y_1 + u_2'y_2 = 0: then yp′=u1y1′+u2y2′y_p' = u_1y_1' + u_2y_2', and yp′′=u1′y1′+u2′y2′+u1y1′′+u2y2′′y_p'' = u_1'y_1' + u_2'y_2' + u_1y_1'' + u_2y_2''. Substitute into the ODE; the uiu_i-times-homogeneous-solution terms vanish (each yiy_i solves the homogeneous equation), leaving u1′y1′+u2′y2′=fu_1'y_1' + u_2'y_2' = f.

Together with the side condition this is a 2×22\times2 linear system in u1′,u2′u_1', u_2' whose determinant is exactly WW; Cramer's rule gives u1′=−y2f/Wu_1' = -y_2f/W, u2′=y1f/Wu_2' = y_1f/W. ■\blacksquare

Tip
Proof: Abel's theorem

Differentiate W=y1y2′−y1′y2W = y_1y_2' - y_1'y_2: W′=y1y2′′−y1′′y2W' = y_1y_2'' - y_1''y_2 (the cross terms cancel). Substitute yi′′=−pyi′−qyiy_i'' = -py_i' - qy_i: W′=y1(−py2′−qy2)−(−py1′−qy1)y2=−p(y1y2′−y1′y2)=−pWW' = y_1(-py_2' - qy_2) - (-py_1' - qy_1)y_2 = -p(y_1y_2' - y_1'y_2) = -pW. So W=Ce−∫pW = Ce^{-\int p} — never zero unless C=0C = 0, in which case it vanishes identically. Independence is an all-or-nothing affair. ■\blacksquare

Going Deeper: Worked Problems

Example
Worked: variation of parameters with a non-standard forcing

Solve y′′−2y′+y=exxy'' - 2y' + y = \dfrac{e^x}{x} for x>0x > 0.

Step 1 — homogeneous: (r−1)2=0(r-1)^2 = 0, double root: y1=exy_1 = e^x, y2=xexy_2 = xe^x.

Step 2 — Wronskian: W=ex(ex+xex)−ex⋅xex=e2xW = e^x(e^x + xe^x) - e^x\cdot xe^x = e^{2x}.

Step 3 — the formulas with f=exxf = \tfrac{e^x}{x}: u1′=−y2fW=−xex⋅ex/xe2x=−1u_1' = -\dfrac{y_2f}{W} = -\dfrac{xe^x\cdot e^x/x}{e^{2x}} = -1, and u2′=y1fW=ex⋅ex/xe2x=1xu_2' = \dfrac{y_1f}{W} = \dfrac{e^x\cdot e^x/x}{e^{2x}} = \dfrac1x.

Step 4 — integrate: u1=−xu_1 = -x, u2=ln⁡xu_2 = \ln x.

Step 5 — yp=−xex+xexln⁡xy_p = -xe^x + xe^x\ln x; the −xex-xe^x is a homogeneous solution and can be dropped: yp=xexln⁡xy_p = xe^x\ln x. General solution: y=(c1+c2x)ex+xexln⁡xy = (c_1 + c_2x)e^x + xe^x\ln x. Undetermined coefficients has no guess for exx\tfrac{e^x}{x} — this method never needed one.

Example
Worked: using the Wronskian to test a claimed fundamental set

Are y1=xy_1 = x and y2=xln⁡xy_2 = x\ln x a fundamental set of solutions of x2y′′−xy′+y=0x^2y'' - xy' + y = 0 on x>0x > 0?

Step 1 — check each solves the equation: for y=xy = x: 0−x+x=00 - x + x = 0 ✓; for y=xln⁡xy = x\ln x: y′=ln⁡x+1y' = \ln x + 1, y′′=1xy'' = \tfrac1x, so x2⋅1x−x(ln⁡x+1)+xln⁡x=x−xln⁡x−x+xln⁡x=0x^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⋅xln⁡x=xW = x(\ln x + 1) - 1\cdot x\ln x = x, nonzero for x>0x > 0.

Step 3 — conclusion: yes — independent solutions, so every solution is c1x+c2xln⁡xc_1x + c_2x\ln x.

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