Non-homogeneous — variation of parameters — Hinglish notes

4.6.15 · Maths › Ordinary Differential Equations

Intuition Badi idea kya hai

Maan lo tumhe homogeneous equation $y^{''} + p y^{'} + q y = 0$ ke do independent solutions $y_{1}, y_{2}$ pehle se pata hain. General homogeneous solution hai $c_{1} y_{1} + c_{2} y_{2}$ jisme $c_{1}, c_{2}$ constant hain.

KYA karte hain: non-homogeneous equation $y^{''} + p y^{'} + q y = g (x)$ solve karne ke liye hum constants ko vary karte hain — $c_{1}, c_{2}$ ki jagah unknown functions $u_{1} (x), u_{2} (x)$ rakhte hain aur demand karte hain ki $u_{1} y_{1} + u_{2} y_{2}$ poori equation satisfy kare.

KYUN kaam karta hai: homogeneous solutions already left side ko "kill" kar dete hain. Unhe functions se tweak karke hum bacha hua $g (x)$ absorb kar sakte hain. Yeh undetermined coefficients se zyada powerful hai kyunki yeh kisi bhi $g (x)$ ke liye kaam karta hai (jaise $sec x$ , $ln x$ , $1/ x$ ) — sirf polynomials/exponentials/sines ke liye nahi.

Derivation setup (scratch se)

KAISE do unknown functions ko pin down karte hain: hamare paas ek equation hai (ODE) lekin do unknowns hain. Toh hum apni marzi se ek extra condition impose kar sakte hain. Hum woh choose karte hain jo algebra ko simple bana de.

Ek baar differentiate karo: $y_{p}^{'} = u_{1} y_{1}^{'} + u_{2} y_{2}^{'} + set = 0 (u_{1}^{'} y_{1} + u_{2}^{'} y_{2})$

Toh Condition 1: $u_{1}^{'} y_{1} + u_{2}^{'} y_{2} = 0$ .

Ab $y_{p}^{'} = u_{1} y_{1}^{'} + u_{2} y_{2}^{'}$ . Dobara differentiate karo: $y_{p}^{''} = u_{1} y_{1}^{''} + u_{2} y_{2}^{''} + u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'}$

$y_{p}, y_{p}^{'}, y_{p}^{''}$ ko $y^{''} + p y^{'} + q y = g$ mein plug karo: $(u_{1} y_{1}^{''} + u_{2} y_{2}^{''} + u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'}) + p (u_{1} y_{1}^{'} + u_{2} y_{2}^{'}) + q (u_{1} y_{1} + u_{2} y_{2}) = g$

$u_{1}$ aur $u_{2}$ ke hisaab se group karo: $u_{1} = 0 (y_{1}^{''} + p y_{1}^{'} + q y_{1}) + u_{2} = 0 (y_{2}^{''} + p y_{2}^{'} + q y_{2}) + (u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'}) = g$

Do bracketed terms vanish hote hain kyunki $y_{1}, y_{2}$ homogeneous equation solve karte hain — yahi toh pura point hai. Bachi Condition 2: $u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} = g$ .

Cramer's rule se solve karo. Matrix ka determinant Wronskian hai: $W = y_{1} y_{1}^{'} y_{2} y_{2}^{'} = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}$

Figure — Non-homogeneous — variation of parameters

Worked example 1 — $sec x$ (undetermined coefficients ke liye impossible)

Solve karo $y^{''} + y = sec x$ .

Step 1 — homogeneous solutions. $r^{2} + 1 = 0 \Rightarrow r = \pm i$ , toh $y_{1} = cos x, y_{2} = sin x$ . Yeh step kyun? Hume hamesha pehle homogeneous solutions chahiye — yahi raw material hai jisse $y_{p}$ banta hai.

Step 2 — Wronskian. $W = cos x \cdot cos x - sin x \cdot (- sin x) = cos^{2} x + sin^{2} x = 1$ . Kyun? $W$ denominator mein aata hai; clean $W = 1$ integrals simple kar deta hai.

Step 3 — $u_{1}^{'}, u_{2}^{'}$ . $g = sec x$ ke saath: $u_{1}^{'} = - \frac{s i n x s e c x}{1} = - tan x, u_{2}^{'} = \frac{c o s x s e c x}{1} = 1$

Step 4 — integrate. $u_{1} = - \int tan x d x = ln ∣ cos x ∣, u_{2} = \int 1 d x = x$ $+ C$ kyun drop karte hain? Constants sirf homogeneous part regenerate karte hain; hume sirf ek particular solution chahiye.

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, $W \neq = 1$

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

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

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

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

Step 4. $u_{1} = - x, u_{2} = - e^{- x}$ .

Step 5. $y_{p} = e^{x} (- x) + e^{2 x} (- e^{- x}) = - x e^{x} - e^{x}$ $- e^{x}$ ek homogeneous piece hai ( $c_{1}$ mein absorb ho jaata hai), toh essentially $y_{p} = - x e^{x}$ . General: $y = c_{1} e^{x} + c_{2} e^{2 x} - x e^{x}$ . (Undetermined coefficients se match karta hai — achha sanity check.)

Worked example 3 — non-constant coefficients

Solve karo $x^{2} y^{''} - 2 x y^{'} + 2 y = x^{3}$ , given ki $y_{1} = x, y_{2} = x^{2}$ homogeneous part solve karte hain.

Step 0 — standard form! $x^{2}$ se divide karo: $y^{''} - \frac{2}{x} y^{'} + \frac{2}{x ^{2}} y = x$ . Toh $g = x$ hai (NOT $x^{3}$ ). Yeh step kyun? Formula assume karta hai leading coefficient $1$ . Yeh bhoolna #1 error hai.

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

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

Step 4. $u_{1} = - \frac{x ^{2}}{2}, u_{2} = x$ .

Step 5. $y_{p} = x (- \frac{x ^{2}}{2}) + x^{2} (x) = - \frac{x ^{3}}{2} + x^{3} = \frac{x ^{3}}{2} .$ General: $y = c_{1} x + c_{2} x^{2} + \frac{x ^{3}}{2}$ .

Common mistake Classic errors ko steel-man karna

(1) Standard form ke bina raw RHS use karna. Sahi lagta hai kyunki " $g$ toh bas right-hand side hai." Fix: pehle leading coefficient se divide karo; $g$ normalized equation ka RHS hai. (2) $u_{1}^{'}$ mein minus sign bhool jaana. "Symmetric lagta hai" dono ko $+$ likhna. Fix: Cramer's rule $y_{2}$ (second column source) ko sign ke saath rakhta hai — $u_{1}^{'} = - y_{2} g / W$ , $u_{2}^{'} = + y_{1} g / W$ . Yaad rakho "minus us wale ke saath jaata hai jo index se match nahi karta." (3) Wronskian ka sign. $W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}$ (first row times cross). $y_{1}, y_{2}$ swap karne se $W$ ka sign aur $g$ ki placement ka sign flip hota hai, toh final answer unchanged rehta hai — lekin conventions mix karne se sign error aata hai. (4) $u_{1}, u_{2}$ mein integration constants rakhna. Yeh sirf complementary function reproduce karte hain; harmless hai par redundant. Inhe drop karo.

Recall Feynman: ek 12-saal ke bachhe ko explain karo

Tumhare paas already do "building-block" wiggles hain jo ek seesaw ko perfectly balance karte hain jab koi push nahi karta. Ab koi seesaw ko push kar raha hai (woh hai $g (x)$ ). Brand-new wiggles invent karne ki jagah, tum apne do building blocks lete ho aur unki strengths ko time ke saath change karte ho — kabhi block 1 zyada use karo, kabhi block 2 — bilkul utna jitna pusher ko cancel karne ke liye chahiye. Do choti equations tumhe, moment by moment, batati hain ki kitna mix karna hai. Wronskian bas ek "fairness scale" hai jo ensure karta hai ki do blocks genuinely alag hain aur secretly same nahi hain.

Flashcards

Variation of parameters mein konsa ansatz define karta hai?

y_{p} = u_{1} (x) y_{1} (x) + u_{2} (x) y_{2} (x)

, jahan

y_{1}, y_{2}

homogeneous equation solve karte hain aur

u_{1}, u_{2}

find karne wale functions hain.

Konsi convenient extra condition impose ki jaati hai, aur kyun?

u_{1}^{'} y_{1} + u_{2}^{'} y_{2} = 0

; yeh unknowns ke second derivatives ko appear hone se rokti hai, system ko first-order rakhti hai.

u_{1}^{'}, u_{2}^{'}

ke liye do simultaneous equations batao.

u_{1}^{'} y_{1} + u_{2}^{'} y_{2} = 0

aur

u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} = g (x)

y_{1}, y_{2}

ka Wronskian

W

define karo.

W = y_{1} y_{2}^{'} - y_{2} y_{1}^{'}

Cramer's rule se

u_{1}^{'}

aur

u_{2}^{'}

do.

u_{1}^{'} = - y_{2} g / W

u_{2}^{'} = + y_{1} g / W

g (x)

read off karne se pehle kya karna chahiye?

ODE ko standard form mein rakho (leading coefficient 1) divide karke;

g

tab RHS hai.

Variation of parameters undetermined coefficients se kyun better hai?

Yeh KISI BHI continuous

g

ke liye kaam karta hai (jaise

sec x

ln x

1/ x

), sirf polynomial/exp/sin/cos forms ke liye nahi.

y^{''} + y = sec x

ke liye

W

aur

y_{p}

kya hai?

W = 1

;

y_{p} = cos x ln ∣ cos x ∣ + x sin x

u_{1}, u_{2}

mein integration constants kyun drop kar sakte hain?

Woh sirf complementary function

c_{1} y_{1} + c_{2} y_{2}

reproduce karte hain; hume bas ek particular solution chahiye.

Connections

Wronskian — linear independence measure karta hai; non-zero $W$ guarantee karta hai ki method kaam karega.
Method of Undetermined Coefficients — faster hai par special $g (x)$ tak limited; VoP general tool hai.
Second-order linear homogeneous ODE — $y_{1}, y_{2}$ supply karta hai, jo prerequisite hain.
Cramer's Rule — $u_{1}^{'}, u_{2}^{'}$ ke liye $2 \times 2$ system solve karta hai.
Green's function — VoP integral exactly Green's-function representation of the solution hai.
Reduction of Order — alternative jab sirf ek $y_{1}$ pata ho.

Derivation setup (scratch se)

Worked example 1 — secx (undetermined coefficients ke liye impossible)

Worked example 2 — exponential RHS, W=1

Worked example 3 — non-constant coefficients

Flashcards

Connections

Concept Map

Worked example 1 — $sec x$ (undetermined coefficients ke liye impossible)

Worked example 2 — exponential RHS, $W \neq = 1$