Non-homogeneous — variation of parameters: Visual walkthrough

4.6.15 · D2 · Maths › Ordinary Differential Equations › Non-homogeneous — variation of parameters

Step 1 — Balanced seesaw se shuru karo

KYA. Kisi bhi push se pehle, homogeneous equation dekho — wahi equation jisme right side zero kar diya gaya ho: $y^{''} + p y^{'} + q y = 0.$ Iske do independent solutions hum $y_{1} (x)$ aur $y_{2} (x)$ kehte hain. "Independent" ka matlab hai ki ek doosre ki sirf scaled copy nahi hai — ye genuinely alag shapes hain. Koi bhi combination $c_{1} y_{1} + c_{2} y_{2}$ constant numbers $c_{1}, c_{2}$ ke saath bhi ise solve karta hai.

KYUN. Ye do shapes humare raw building material hain. Ye pehle se hi left side ko akele zero kar dete hain — yahi woh property hai jo hum exploit karenge. ( $y_{1}, y_{2}$ kahan se aate hain? Constant $p, q$ ke liye inhe characteristic equation se milate ho — dekho Second-order linear homogeneous ODE.)

PICTURE. Do smooth curves, hamare "blocks." Unke neeche, wahi do blocks fixed strengths $c_{1}, c_{2}$ ke saath ek combined curve mein mix kiye gaye. Constant strengths = ek curve jo zero equation solve karta hai par push handle nahi kar sakta.

Figure — Non-homogeneous — variation of parameters

Step 2 — Constants ko saans lene do

KYA. Frozen numbers $c_{1}, c_{2}$ ki jagah chalte functions $u_{1} (x), u_{2} (x)$ rakh do. Hum ek particular solution ka guess karte hain is shape mein: $y_{p} = block 1 ki strength u_{1} (x) block 1 y_{1} (x) + block 2 ki strength u_{2} (x) block 2 y_{2} (x) .$ Is educated guess ko ansatz kehte hain (German mein "attempt/setup" ka matlab).

KYUN. Constant strengths ek chalti push $g (x)$ absorb nahi kar sakti. Agar strengths ko moment-to-moment change karne dein, to hume ek live dial milta hai jise hum exactly utna ghuma sakte hain jitna push cancel karne ke liye chahiye. Yahi se naam aata hai: hum parameters ko vary karte hain.

PICTURE. Step 1 ke do fixed dials knobs ban jaate hain jinki settings $x$ ke saath slide karti hain. Blended curve ab ek wiggly target follow karne ke liye flex karta hai.

Step 3 — Ek equation hai, do unknowns hain — toh hum ek free choice lete hain

KYA. Ansatz ko ek baar differentiate karo. Product rule (kisi product ki rate = pehla $\times$ doosre-ki-rate + doosra $\times$ pehle-ki-rate) har term pe use karo: $y_{p}^{'} = blocks badal rahe hain u_{1} y_{1}^{'} + u_{2} y_{2}^{'} + strengths badal rahi hain (u_{1}^{'} y_{1} + u_{2}^{'} y_{2}) .$

Yahan ek key observation hai. Hum do unknown functions $u_{1}, u_{2}$ pin down karne ki koshish kar rahe hain, lekin ODE sirf ek equation hai. Do unknowns, ek constraint — hum underdetermined hain, isliye hume apni marzi se ek extra equation banane ki ijazat hai. Hum choose karte hain: $u_{1}^{'} y_{1} + u_{2}^{'} y_{2} = 0 (Condition 1) .$

KYUN. Upar wala doosra bracket dekho. Agar hum ise survive karne dein, to dobara differentiate karne par $u_{1}^{''}$ aur $u_{2}^{''}$ aayenge — unknowns ke second derivatives, jo us problem se zyada mushkil hai jisse humne shuru kiya tha. Us bracket ko abhi kill karne se puri system strengths mein first-order rehti hai. Yeh ek deal hai jo hum extra freedom ki wajah se karne ke haqdar hain.

PICTURE. Ek balance beam: do "strength-change" terms $u_{1}^{'} y_{1}$ aur $u_{2}^{'} y_{2}$ equal aur opposite hain, isliye cancel ho jaate hain — hum deliberately inhe ek doosre ke against set karte hain.

Step 4 — Dobara differentiate karo aur ODE mein daalo

KYA. Condition 1 ke baad, first derivative simplify hokar $y_{p}^{'} = u_{1} y_{1}^{'} + u_{2} y_{2}^{'}$ ban jaata hai. Ise differentiate karo (phir se product rule): $y_{p}^{''} = blocks ki curvature u_{1} y_{1}^{''} + u_{2} y_{2}^{''} + naya leftover u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} .$ Ab $y_{p}, y_{p}^{'}, y_{p}^{''}$ ko $y^{''} + p y^{'} + q y = g$ mein substitute karo aur $u_{1}$ aur $u_{2}$ ke hisaab se terms gather karo: $u_{1} = 0 (y_{1}^{''} + p y_{1}^{'} + q y_{1}) + u_{2} = 0 (y_{2}^{''} + p y_{2}^{'} + q y_{2}) + (u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'}) = g .$

KYUN. Har bada bracket bilkul wahi homogeneous equation hai jo $y_{1}$ ya $y_{2}$ pe apply ki gayi hai — aur woh Step 1 se zero hain. Yahi poora magic hai: hamare building blocks exactly isliye choose kiye gaye the taaki ye terms vanish ho jaayein. Jo bachta hai woh hai: $u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} = g (x) (Condition 2) .$

PICTURE. Terms ke do columns " $= 0$ " marked (cross out kiye kyunki $y_{1}, y_{2}$ homogeneous ODE solve karte hain), sirf woh chhota green survivor bachta hai jo push $g$ ke equal hona chahiye.

Step 5 — Do blocks do saaf equations ban gaye

KYA. Conditions 1 aur 2 collect karo: ${u_{1}^{'} y_{1} + u_{2}^{'} y_{2} = 0 u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} = g (haari free choice se) (ODE se)$ Unknowns ko $u_{1}^{'}$ aur $u_{2}^{'}$ padhо — woh rates jinse haari strengths change hoti hain. Ise ek matrix ke roop mein likho jo un rates pe act kare: $(y_{1} y_{1}^{'} y_{2} y_{2}^{'}) (u_{1}^{'} u_{2}^{'}) = (0 g) .$

KYUN. Ek two-by-two linear system kuch aisa hai jise hum Cramer's Rule se khol sakte hain — koi cleverness nahi chahiye. Sirf ek cheez galat ho sakti hai agar matrix ka koi inverse na ho, jo next step ki warning hai.

PICTURE. System ek machine ki tarah: jaana-pehchana matrix (blocks aur unki slopes se bana) times unknown rate-vector equals demand vector $(0, g)$ .

Step 6 — Cramer's rule, aur Wronskian appear hota hai

KYA. Cramer's Rule kehta hai: har unknown pane ke liye, uska column right-hand side se replace karo aur matrix determinant se divide karo. Hamare matrix ka determinant hai: $W = y_{1} y_{1}^{'} y_{2} y_{2}^{'} = y_{1} y_{2}^{'} - y_{2} y_{1}^{'},$ jise Wronskian kehte hain. Rule apply karne par: $u_{1}^{'} = \frac{0 g y _{2} y _{2}^{'}}{W} = \frac{- y _{2} g}{W}, u_{2}^{'} = \frac{y _{1} y _{1}^{'} 0 g}{W} = \frac{y _{1} g}{W} .$

Term by term: $u_{1}^{'}$ mein numerator determinant hai $0 \cdot y_{2}^{'} - y_{2} \cdot g = - y_{2} g$ — woh stray minus galti nahi hai jo fix karni ho, yeh Cramer's rule apna kaam kar raha hai. $u_{2}^{'}$ mein yeh hai $y_{1} \cdot g - 0 = + y_{1} g$ .

KYUN. $W$ denominator mein hai, isliye hume $W \neq = 0$ chahiye. Wronskian exactly woh honesty-check hai ki $y_{1}, y_{2}$ truly independent hain: $W = 0$ ka matlab hoga ek block doosre ki copy hai, matrix singular hai, aur koi unique solution nahi hai. Independence guarantee karta hai $W \neq = 0$ , isliye hum divide kar sakte hain.

PICTURE. Wronskian vectors $(y_{1}, y_{1}^{'})$ aur $(y_{2}, y_{2}^{'})$ se bane parallelogram ka signed area — nonzero area = genuinely alag directions = method chalega.

Step 7 — Rates ko integrate karke strengths banao, phir assemble karo

KYA. Ab hume rates $u_{1}^{'}, u_{2}^{'}$ pata hain. Strengths $u_{1}, u_{2}$ recover karne ke liye hum derivative undo karte hain — integrate karte hain: $u_{1} = - \int \frac{y _{2} g}{W} d x, u_{2} = \int \frac{y _{1} g}{W} d x .$ Inhe ansatz mein daalo aur finished particular solution pao: $y_{p} = - y_{1} \int \frac{y _{2} g}{W} d x + y_{2} \int \frac{y _{1} g}{W} d x .$

KYUN. Derivative ne hume har strength ka slope bataya tha; integration strength khud recover karta hai. Hum integration ke $+ C$ constants drop kar dete hain: ek constant times $y_{1}$ ya $y_{2}$ sirf ek homogeneous solution hai, jo $c_{1} y_{1} + c_{2} y_{2}$ wala part pehle se cover karta hai. Hume sirf ek particular solution chahiye.

PICTURE. Left panel: do rate curves $u_{1}^{'}, u_{2}^{'}$ . Middle: integration area accumulate karke strength curves $u_{1}, u_{2}$ banata hai. Right: blocks $\times$ strengths milke particular solution $y_{p}$ banate hain jo push track karta hai.

Step 8 — Degenerate aur edge cases (kabhi chhupaaye nahi jaate)

KYA & KYUN — teen tarike jisme yeh sideways ja sakta hai, aur kya hota hai:

$W = 0$ (blocks secretly identical). Tab $y_{2} = k y_{1}$ kisi constant $k$ ke liye; Step 6 ka parallelogram zero area ki ek line mein flat ho jaata hai. Cramer's rule zero se divide karta hai — method ruk jaata hai. Fix: pehle ek genuinely independent second solution dhundho (jaise Reduction of Order se).
$g (x) = 0$ (koi push nahi). Tab $u_{1}^{'} = u_{2}^{'} = 0$ , isliye $u_{1}, u_{2}$ constants hain aur $y_{p}$ wapas $c_{1} y_{1} + c_{2} y_{2}$ mein collapse ho jaata hai. Machinery sahi se report karti hai "kuch extra karne ki zaroorat nahi" — ek reassuring consistency check.
$g$ kisi homogeneous solution se overlap kare (jaise $g = e^{x}$ jab $y_{1} = e^{x}$ ). Undetermined coefficients ko yahan ek awkward $x$ -multiplier rule ki zaroorat hoti; variation of parameters sirf integrate karta hai aur zaroorat wala $x$ ka factor automatically appear hota hai (jaise parent ke Example 2 mein, jahan $y_{p} = - x e^{x}$ ). Koi special-casing nahi.

PICTURE. Teen mini-panels: (1) do collapsed parallel blocks $W = 0$ ke saath crossed out; (2) ek flat "no push" line jahan $y_{p}$ sirf homogeneous curve hai; (3) ek resonant case jahan answer naturally ek $x$ factor grow karta hai.

Ek-picture summary

Sab ek saath: do balanced blocks se shuru karo → unki strengths vary karne do → free choice (Condition 1) aur ODE (Condition 2) do equations dete hain → Cramer's rule with Wronskian neeche → rates integrate karo → $y_{p}$ assemble karo.

Recall Poore walkthrough ki Feynman retelling

Tumhare paas do "wiggle blocks" hain jo ek still seesaw ko perfectly balance karte hain — inhe kisi bhi fixed amount mein mix karo aur seesaw solved rehta hai (Step 1). Ab koi seesaw ko push karta hai; fixed amounts keep up nahi kar sakti, toh tum amounts ko time ke saath change karne dete ho (Step 2). Do unknown dials, lekin satisfy karne ke liye sirf ek rule (ODE) — isliye tumhe apna ek convenient rule invent karne ki ijazat hai, aur tum woh choose karte ho jo tumhari algebra ko second derivatives mein explode hone se bachata hai (Step 3). Dobara differentiate karo, equation mein daalo, aur block terms vanish ho jaate hain kyunki blocks pehle se quiet seesaw solve karte hain — ek clean equation bachti hai jo kehti hai "changing dials exactly push ke equal hone chahiye" (Step 4). Do rules, do unknown dial-rates: yeh ek baby 2×2 system hai (Step 5). Cramer's rule ise crack karta hai, aur neeche baitha number — Wronskian — tumhara fairness scale hai jo check karta hai ki do blocks sach mein alag hain (Step 6). Rules tumhe dials ki speeds dete hain; dials khud pane ke liye integrate karo, phir blocks ko dials se mix karo aur apna answer pao (Step 7). Aur agar blocks secret twins hain toh scale zero padhta hai aur tumhe ek naya block laana padega; agar koi push nahi karta, dials freeze ho jaate hain aur kuch extra nahi milta — exactly jaisa hona chahiye (Step 8).

Recall

$W$ denominator mein kyun hai, aur $W = 0$ ka matlab kya hai? ::: $W$ woh determinant hai jisse Cramer's rule divide karta hai; $W = 0$ ka matlab hai $y_{1}, y_{2}$ linearly dependent hain (zero area ka parallelogram), system singular hai, aur method fail ho jaata hai. Kaun si condition free choice hai aur kaun si ODE se aati hai? ::: $u_{1}^{'} y_{1} + u_{2}^{'} y_{2} = 0$ haari free choice hai; $u_{1}^{'} y_{1}^{'} + u_{2}^{'} y_{2}^{'} = g$ ODE mein substitute karne se forced hai. VoP mein koi special "resonance" rule kyun nahi aata? ::: Integration automatically koi bhi zaroorati $x$ ka factor produce kar deta hai; correct karne ke liye koi guess nahi hai.

Connections

Wronskian — denominator $W$ ; nonzero = independent blocks = method chalega.
Cramer's Rule — 2×2 system crack karta hai aur $u_{1}^{'}$ mein minus sign supply karta hai.
Second-order linear homogeneous ODE — jahan se $y_{1}, y_{2}$ aate hain.
Reduction of Order — jab $W = 0$ ho toh doosra independent block kaise banao.
Method of Undetermined Coefficients — tez, narrow alternative.
Green's function — is poore integral ko ek single kernel mein package karna.