Non-homogeneous — variation of parameters

Intuition The big idea

Suppose you already know the two independent solutions $y_1, y_2$ of the homogeneous equation $y'' + p y' + q y = 0$ . The general homogeneous solution is $c_1 y_1 + c_2 y_2$ with constant $c_1, c_2$ .

WHAT we do: to solve the non-homogeneous equation $y'' + p y' + q y = g(x)$ , we let the constants vary — replace $c_1, c_2$ by unknown functions $u_1(x), u_2(x)$ and demand that $u_1 y_1 + u_2 y_2$ satisfies the full equation.

WHY it works: the homogeneous solutions already "kill" the left side. By tweaking them with functions, we can absorb the leftover $g(x)$ . This is more powerful than undetermined coefficients because it works for any $g(x)$ (e.g. $\sec x$ , $\ln x$ , $1/x$ ) — not just polynomials/exponentials/sines.

Setting up the derivation (from scratch)

HOW we pin down two unknown functions: we have one equation (the ODE) but two unknowns. So we are free to impose one extra condition of our choosing. We choose it to make the algebra collapse.

Differentiate once: $y_p' = u_1 y_1' + u_2 y_2' + \underbrace{(u_1' y_1 + u_2' y_2)}_{\text{set} = 0}$

So Condition 1: $\;u_1' y_1 + u_2' y_2 = 0$ .

Now $y_p' = u_1 y_1' + u_2 y_2'$ . Differentiate again: $y_p'' = u_1 y_1'' + u_2 y_2'' + u_1' y_1' + u_2' y_2'$

Plug $y_p, y_p', y_p''$ into $y'' + p y' + q y = g$ : $\big(u_1 y_1'' + u_2 y_2''+ u_1'y_1' + u_2'y_2'\big) + p\big(u_1 y_1' + u_2 y_2'\big) + q\big(u_1 y_1 + u_2 y_2\big) = g$

Group by $u_1$ and $u_2$ : $u_1\underbrace{(y_1'' + p y_1' + q y_1)}_{=0} + u_2\underbrace{(y_2'' + p y_2' + q y_2)}_{=0} + (u_1'y_1' + u_2'y_2') = g$

The two bracketed terms vanish because $y_1,y_2$ solve the homogeneous equation — that's the whole point. Left with Condition 2: $\;u_1' y_1' + u_2' y_2' = g$ .

Solve by Cramer's rule. The determinant of the matrix is the Wronskian: $W = \begin{vmatrix} y_1 & y_2 \\ y_1' & y_2'\end{vmatrix} = y_1 y_2' - y_2 y_1'$

Figure — Non-homogeneous — variation of parameters

Worked example 1 — $\sec x$ (impossible for undetermined coeff.)

Solve $y'' + y = \sec x$ .

Step 1 — homogeneous solutions. $r^2+1=0 \Rightarrow r=\pm i$ , so $y_1=\cos x,\ y_2=\sin x$ . Why this step? We always need the homogeneous solutions first — they are the raw material $y_p$ is built from.

Step 2 — Wronskian. $W = \cos x\cdot\cos x - \sin x\cdot(-\sin x) = \cos^2 x+\sin^2 x = 1$ . Why? $W$ appears in the denominator; a clean $W=1$ simplifies the integrals.

Step 3 — $u_1', u_2'$ . With $g=\sec x$ : $u_1' = -\frac{\sin x\sec x}{1} = -\tan x,\qquad u_2' = \frac{\cos x\sec x}{1} = 1$

Step 4 — integrate. $u_1 = -\int\tan x\,dx = \ln|\cos x|, \qquad u_2 = \int 1\,dx = x$ Why drop the $+C$ ? The constants just regenerate the homogeneous part; we only need one particular solution.

Step 5 — assemble. $y_p = \cos x\,\ln|\cos x| + x\sin x$ General solution: $y = c_1\cos x + c_2\sin x + \cos x\ln|\cos x| + x\sin x$ .

Worked example 2 — exponential RHS, $W\neq1$

Solve $y'' - 3y' + 2y = e^{x}$ .

Step 1. $r^2-3r+2=(r-1)(r-2)=0\Rightarrow y_1=e^{x},\ y_2=e^{2x}$ .

Step 2 — Wronskian. $W = \begin{vmatrix} e^x & e^{2x}\\ e^x & 2e^{2x}\end{vmatrix} = e^x\cdot 2e^{2x} - e^{2x}\cdot e^x = e^{3x}$

Step 3. $g=e^x$ : $u_1' = -\frac{e^{2x}e^x}{e^{3x}} = -1,\qquad u_2' = \frac{e^x e^x}{e^{3x}} = e^{-x}$

Step 4. $u_1 = -x,\quad u_2 = -e^{-x}$ .

Step 5. $y_p = e^x(-x) + e^{2x}(-e^{-x}) = -x e^x - e^x$ The $-e^x$ is a homogeneous piece (absorb into $c_1$ ), so essentially $y_p = -xe^x$ . General: $y = c_1 e^x + c_2 e^{2x} - x e^{x}$ . (Matches undetermined coefficients — good sanity check.)

Worked example 3 — non-constant coefficients

Solve $x^2 y'' - 2x y' + 2y = x^3$ , given $y_1=x,\ y_2=x^2$ solve the homogeneous part.

Step 0 — standard form! Divide by $x^2$ : $\;y'' - \tfrac{2}{x}y' + \tfrac{2}{x^2}y = x$ . So $g=x$ (NOT $x^3$ ). Why this step? The formula assumes leading coefficient $1$ . Forgetting this is the #1 error.

Step 2 — Wronskian. $W = x\cdot 2x - x^2\cdot 1 = 2x^2 - x^2 = x^2$ .

Step 3. $u_1' = -\frac{x^2\cdot x}{x^2} = -x,\qquad u_2' = \frac{x\cdot x}{x^2} = 1$

Step 4. $u_1 = -\tfrac{x^2}{2},\quad u_2 = x$ .

Step 5. $y_p = x(-\tfrac{x^2}{2}) + x^2(x) = -\tfrac{x^3}{2} + x^3 = \tfrac{x^3}{2}.$ General: $y = c_1 x + c_2 x^2 + \tfrac{x^3}{2}$ .

Common mistake Steel-manning the classic errors

(1) Using the raw RHS without standard form. Feels right because " $g$ is just the right-hand side." Fix: divide by the leading coefficient first; $g$ is the RHS of the normalized equation. (2) Forgetting the minus sign in $u_1'$ . It "feels symmetric" to write both as $+$ . Fix: Cramer's rule puts $y_2$ (the second column source) with a sign — $u_1' = -y_2 g/W$ , $u_2'=+y_1 g/W$ . Remember "minus goes with the one not matching the index." (3) Sign of the Wronskian. $W = y_1 y_2' - y_2 y_1'$ (first row times cross). Swapping $y_1,y_2$ flips the sign of $W$ and of $g$ 's placement, so the final answer is unchanged — but mixing conventions gives a sign error. (4) Keeping integration constants in $u_1,u_2$ . They merely reproduce the complementary function; harmless but redundant. Drop them.

Recall Feynman: explain it to a 12-year-old

You already have two "building-block" wiggles that perfectly balance a seesaw when nobody pushes it. Now someone is pushing the seesaw (that's $g(x)$ ). Instead of inventing brand-new wiggles, you take your two building blocks and let their strengths change over time — sometimes use more of block 1, sometimes more of block 2 — exactly enough to cancel the pusher. The two little equations tell you, moment by moment, how much of each block to mix. The Wronskian is just a "fairness scale" making sure the two blocks are genuinely different and not secretly the same.

Flashcards

What ansatz defines variation of parameters?

y_p = u_1(x)y_1(x) + u_2(x)y_2(x)

, where

y_1,y_2

solve the homogeneous equation and

u_1,u_2

are functions to find.

What is the convenient extra condition imposed, and why?

u_1'y_1 + u_2'y_2 = 0

; it prevents second derivatives of the unknowns from appearing, keeping the system first-order.

State the two simultaneous equations for

u_1',u_2'

u_1'y_1+u_2'y_2=0

and

u_1'y_1'+u_2'y_2'=g(x)

Define the Wronskian

W

y_1,y_2

W = y_1 y_2' - y_2 y_1'

Give

u_1'

and

u_2'

from Cramer's rule.

u_1' = -y_2 g / W

u_2' = +y_1 g / W

What must you do before reading off

g(x)

Put the ODE in standard form (leading coefficient 1) by dividing through;

g

is then the RHS.

Why does variation of parameters beat undetermined coefficients?

It works for ANY continuous

g

(e.g.

\sec x

\ln x

1/x

), not just polynomial/exp/sin/cos forms.

For

y''+y=\sec x

, what is

W

and

y_p

W=1

;

y_p=\cos x\ln|\cos x| + x\sin x

Why can we drop the integration constants in

u_1,u_2

They only reproduce the complementary function

c_1y_1+c_2y_2

; we need just one particular solution.

Connections

Wronskian — measures linear independence; non-zero $W$ guarantees the method works.
Method of Undetermined Coefficients — faster but limited to special $g(x)$ ; VoP is the general tool.
Second-order linear homogeneous ODE — supplies $y_1,y_2$ , the prerequisite.
Cramer's Rule — solves the $2\times2$ system for $u_1',u_2'$ .
Green's function — VoP integral is exactly the Green's-function representation of the solution.
Reduction of Order — alternative when only one $y_1$ is known.

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, idea bilkul simple hai. Homogeneous equation $y''+py'+qy=0$ ke do independent solutions $y_1$ aur $y_2$ already mil chuke hote hain, aur general solution hota hai $c_1y_1+c_2y_2$ jisme $c_1,c_2$ constants hain. Ab jab right side pe $g(x)$ aa jata hai (non-homogeneous), toh hum ek chhota sa jugaad karte hain — constants ko functions bana dete hain: $c_1\to u_1(x)$ , $c_2\to u_2(x)$ . Isi liye naam hai "variation of parameters" — parameters (constants) ko vary kar rahe hain.

Do unknown functions hain par equation ek hi hai, toh humein freedom milti hai ek extra condition apni marzi se lagane ki. Hum chunte hain $u_1'y_1+u_2'y_2=0$ , taaki second derivatives wala mess na aaye. Phir ODE me daalne par homogeneous wale terms khud-ba-khud zero ho jaate hain (kyunki $y_1,y_2$ unhe satisfy karte hain), aur bachta hai $u_1'y_1'+u_2'y_2'=g$ . Ye do equations Cramer's rule se solve karo: $u_1'=-y_2 g/W$ , $u_2'=+y_1 g/W$ , jahan $W=y_1y_2'-y_2y_1'$ Wronskian hai. Bas integrate karke $u_1,u_2$ nikaalo aur $y_p=u_1y_1+u_2y_2$ likh do.

Important baat: equation ko pehle standard form me lao (leading coefficient 1), tabhi $g$ sahi milega — warna sabse common galti yahi hoti hai. Ye method ki sabse badi taakat ye hai ki ye kisi bhi $g(x)$ ke liye chalti hai — $\sec x$ , $\ln x$ , $1/x$ sab — jabki undetermined coefficients sirf polynomial, exponential, sin/cos pe kaam karti hai. Toh jab guessing fail ho jaye, variation of parameters tumhara pakka hathiyar hai.

Non-homogeneous — variation of parameters

Setting up the derivation (from scratch)

Worked example 1 — $\sec x$ (impossible for undetermined coeff.)

Worked example 2 — exponential RHS, $W\neq1$

Worked example 3 — non-constant coefficients

Flashcards

Connections

Concept Map

Hinglish (regional understanding)

Go deeper — visual, from zero

Test yourself — Ordinary Differential Equations

Connections