Worked examples — Non-homogeneous — variation of parameters

4.6.15 · D3 · Maths › Ordinary Differential Equations › Non-homogeneous — variation of parameters

Yeh page ek shooting range hai. Parent note ne machine banayi hai; yahan hum use har tarah ke target pe fire karte hain taaki tum kisi bhi case se anjaane na raho.

Shuru karne se pehle, poori recipe ka ek reminder — taaki neeche ke har example ko same andaaz se padha ja sake:

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Scenario matrix

Socho "problem mujhe kya throw kar sakti hai?" ek grid ki tarah. Har row ek case class hai; last column mein us example ka naam hai jo use handle karta hai.

#	Case class	Tricky kyun hai	Handle karta hai
A	Distinct real roots, exponential $g$	$W\ne 1$, RHS homogeneous solution se overlap karta hai (resonance)	Ex. 1
B	Complex roots, RHS undetermined coeff. se undoable ($\tan x$)	$g$ mein genuine singularity hai; domain dekho	Ex. 2
C	Repeated root, polynomial-ish $g$	$y_2=x{e}^{rx}$, "$x\cdot$" factor $W$ ko change karta hai	Ex. 3
D	Cauchy–Euler (non-constant coefficients)	Pehle $x^2$ se divide karna zaroori; $g$ raw RHS nahi hai	Ex. 4
E	Degenerate: $g\equiv 0$	Sanity check — method ko $y_p=0$ return karna chahiye	Ex. 5
F	Sign / quadrant care: $g=1/x$, log aata hai	$\ln$ ko $	x
G	Definite-integral form (koi elementary antiderivative nahi)	Answer integral ke roop mein rehta hai — yeh allowed hai	Ex. 7
H	Real-world word problem (forced spring, no damping, off-resonance)	Physics → ODE → VoP translate karo	Ex. 8

Ab hum har cell ko hit karte hain.

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Example 1 — Cell A: distinct roots, resonant exponential

Forecast (pehle andaza lagao): RHS $e^{2x}$ khud ek homogeneous solution hai. Undetermined coefficients ko extra $x$ ki zaroorat hoti. Kya tumhe lagta hai VoP automatically woh $x$ conjure karega? (Haan — dekho.)

1. Homogeneous solutions. $r^2-3r+2=(r-1)(r-2)=0\Rightarrow y_1=e^x,y_2=e^{2x}$. Yeh step kyun? $y_p$ sirf inhi do blocks se banta hai.
2. Standard form. $y^{\prime \prime }$ ka coefficient pehle se $1$ hai, isliye $g=e^{2x}$. Kyun? Formula ka $g$ normalized equation ke RHS se aana chahiye.
3. Wronskian. $W=e^x\cdot 2e^{2x}-e^{2x}\cdot e^x=2e^{3x}-e^{3x}=e^{3x}$. Kyun? $W$ har denominator mein hota hai.
4. Derivatives. $u_1^{\prime }=-\frac{e^{2x}\cdot e^{2x}}{e^{3x}}=-e^x,u_2^{\prime }=\frac{e^x\cdot e^{2x}}{e^{3x}}=1.$ Kyun? Seedha $u_1^{\prime }=-y_{2}g/W,u_2^{\prime }=+y_{1}g/W$ se aaya.
5. Integrate karo. $u_1=-e^x,u_2=x$. (Constants chhhod do.)
6. Assemble karo. $y_p=e^x(-e^x)+e^{2x}(x)=-e^{2x}+x{e}^{2x}.$ $-e^{2x}$ ek homogeneous piece hai ($c_2$ mein absorb ho jaata hai). Toh $y_p=x{e}^{2x}$ — VoP ne resonance factor $x$ khud banaya.

General solution: $y=c_1{e}^x+c_2{e}^{2x}+x{e}^{2x}$.

Recall Verify

Plug karo $y_p=x{e}^{2x}$: $y_p^{\prime }=e^{2x}+2x{e}^{2x}$, $y_p^{\prime \prime }=4e^{2x}+4x{e}^{2x}$. $y_p^{\prime \prime }-3y_p^{\prime }+2y_p=(4e^{2x}+4x{e}^{2x})-3(e^{2x}+2x{e}^{2x})+2x{e}^{2x}=e^{2x}$. ✓ RHS match karta hai.

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Example 2 — Cell B: complex roots, $g=\tan x$

Forecast: $\tan x$ $\pm \pi /2$ par blow up karta hai — toh solution sirf inhi do deewaaron ke beech valid hai. Guess: answer mein $\ln |\sec x+\tan x|$ hoga (standard $\int \sec$).

1. Homogeneous. $r^2+1=0\Rightarrow r=\pm i\Rightarrow y_1=\cos x,y_2=\sin x$. Kyun? Complex roots $\alpha \pm i\beta$ dete hain $e^{\alpha x}(\cos \beta x,\sin \beta x)$; yahan $\alpha =0,\beta =1$.
2. Standard form. Pehle se normalized hai, $g=\tan x$.
3. Wronskian. $W=\cos x\cos x-\sin x(-\sin x)=1$.
4. Derivatives. $u_1^{\prime }=-\frac{\sin x\tan x}{1}=-\frac{{\sin }^{2}x}{\cos x},u_2^{\prime }=\frac{\cos x\tan x}{1}=\sin x.$
5. Integrate karo. $u_2=-\cos x$ (aasaan). $u_1$ ke liye, $\frac{{\sin }^{2}x}{\cos x}=\frac{1-{\cos }^{2}x}{\cos x}=\sec x-\cos x$ likhte hain. Toh $u_1=-\int (\sec x-\cos x)dx=-\ln |\sec x+\tan x|+\sin x.$ Rewrite kyun? ${\sin }^2/\cos$ standard integral nahi hai; ise split karne par $\sec x$ (known) minus $\cos x$ (trivial) milta hai.
6. Assemble karo. $y_p=\cos x(\sin x-\ln |\sec x+\tan x|)+\sin x(-\cos x)=-\cos x\ln |\sec x+\tan x|.$ $\cos x\sin x$ terms cancel ho jaate hain.

General: $y=c_1\cos x+c_2\sin x-\cos x\ln |\sec x+\tan x|$.

Recall Verify

Maano $y_p=-\cos xL$ jahan $L=\ln |\sec x+\tan x|$, $L^{\prime }=\sec x$. $y_p^{\prime }=\sin xL-\cos x\sec x=\sin xL-1$. $y_p^{\prime \prime }=\cos xL+\sin x\sec x=\cos xL+\tan x$. $y_p^{\prime \prime }+y_p=(\cos xL+\tan x)+(-\cos xL)=\tan x$. ✓

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Example 3 — Cell C: repeated root

Forecast: repeated root $r=2$, toh $y_2=x{e}^{2x}$. $\ln x$ undetermined coefficients se impossible hai — VoP ek hi tool hai. Guess: ek $\ln x$ aur ek $x^2$ bachenge.

1. Homogeneous. $r^2-4r+4=(r-2{)}^2=0$, double root $r=2$. Reduction of Order se second solution: $y_1=e^{2x},y_2=x{e}^{2x}$.
2. Standard form. Pehle se normalized hai, $g=e^{2x}\ln x$.
3. Wronskian. $y_1^{\prime }=2e^{2x}$, $y_2^{\prime }=e^{2x}+2x{e}^{2x}$. $W=e^{2x}(e^{2x}+2x{e}^{2x})-x{e}^{2x}(2e^{2x})=e^{4x}+2x{e}^{4x}-2x{e}^{4x}=e^{4x}.$
4. Derivatives. $u_1^{\prime }=-\frac{x{e}^{2x}\cdot e^{2x}\ln x}{e^{4x}}=-x\ln x,u_2^{\prime }=\frac{e^{2x}\cdot e^{2x}\ln x}{e^{4x}}=\ln x.$
5. Integrate karo (integration by parts): $u_2=\int \ln xdx=x\ln x-x,$ $u_1=-\int x\ln xdx=-\left(\frac{x^2}{2}\ln x-\frac{x^2}{4}\right)=-\frac{x^2}{2}\ln x+\frac{x^2}{4}.$
6. Assemble karo. $y_p=e^{2x}\left(-\frac{x^2}{2}\ln x+\frac{x^2}{4}\right)+x{e}^{2x}\left(x\ln x-x\right).$ Collect karo: $\ln x$ terms $=e^{2x}{x}^2\ln x\left(-\frac{1}{2}+1\right)=\frac{1}{2}{x}^2{e}^{2x}\ln x$; non-log terms $=e^{2x}(\frac{x^2}{4}-x^2)=-\frac{3}{4}{x}^2{e}^{2x}$. $y_p=\frac{1}{2}{x}^2{e}^{2x}\ln x-\frac{3}{4}{x}^2{e}^{2x}.$ $-\frac{3}{4}{x}^2{e}^{2x}$ term homogeneous nahi hai ($x^2{e}^{2x}$ span mein nahi hai), isliye ise rakhte hain.

General: $y=c_1{e}^{2x}+c_{2}x{e}^{2x}+\frac{1}{2}{x}^2{e}^{2x}\ln x-\frac{3}{4}{x}^2{e}^{2x}$.

Recall Verify

$x=1$ par numerically check karo: $x=1$ par, $\ln 1=0$, toh $y_p(1)=-\frac{3}{4}{e}^2$. VERIFY block symbolically confirm karta hai ki residual $y_p^{\prime \prime }-4y_p^{\prime }+4y_p=e^{2x}\ln x$ sahi hai.

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Example 4 — Cell D: Cauchy–Euler, standard form trap

Forecast: #1 galti $g=x^2$ raw use karna hai. $x^2$ se divide karne par real $g=1$ milta hai. Guess: $x^2/3$ aayega.

1. Homogeneous diya gaya: $y_1=x$, $y_2=x^{-1}$.
2. Standard form — skip mat karo. $x^2$ se divide karo: $y^{\prime \prime }+\frac{1}{x}{y}^{\prime }-\frac{1}{x^2}y=1\Rightarrow g=1.$ Kyun? Formula ko leading coefficient $1$ chahiye; RHS $x^2$ ko $x^2$ se divide karne par $g=1$ milta hai.
3. Wronskian. $y_1^{\prime }=1,y_2^{\prime }=-x^{-2}$. $W=x(-x^{-2})-x^{-1}(1)=-x^{-1}-x^{-1}=-\frac{2}{x}.$
4. Derivatives. $u_1^{\prime }=-\frac{x^{-1}\cdot 1}{-2/x}=-\frac{x^{-1}}{-2/x}=\frac{1}{2},u_2^{\prime }=\frac{x\cdot 1}{-2/x}=-\frac{x^2}{2}.$
5. Integrate karo. $u_1=\frac{x}{2}$, $u_2=-\frac{x^3}{6}$.
6. Assemble karo. $y_p=x\cdot \frac{x}{2}+x^{-1}\left(-\frac{x^3}{6}\right)=\frac{x^2}{2}-\frac{x^2}{6}=\frac{x^2}{3}.$

General: $y=c_{1}x+c_2{x}^{-1}+\frac{x^2}{3}$.

Recall Verify

Plug karo $y_p=\frac{x^2}{3}$: $y_p^{\prime }=\frac{2x}{3},y_p^{\prime \prime }=\frac{2}{3}$. $x^2\cdot \frac{2}{3}+x\cdot \frac{2x}{3}-\frac{x^2}{3}=\frac{2x^2}{3}+\frac{2x^2}{3}-\frac{x^2}{3}=x^2$. ✓

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Example 5 — Cell E: degenerate check, $g\equiv 0$

Forecast: jab system ko force karne wala kuch nahi, absorb karne ke liye bhi kuch nahi. Guess: $y_p=0$.

1. Homogeneous. $y_1=\cos x,y_2=\sin x$, $W=1$.
2. Standard form. $g=0$.
3. Derivatives. $u_1^{\prime }=-\frac{\sin x\cdot 0}{1}=0$, $u_2^{\prime }=\frac{\cos x\cdot 0}{1}=0$.
4. Integrate karo. $u_1=$ const, $u_2=$ const — jo hum drop karte hain.
5. Assemble karo. $y_p=0$. Yeh kyun matter karta hai: yeh prove karta hai ki VoP consistent hai — jab koi forcing nahi hoti toh yeh koi spurious particular solution invent nahi karta. Yeh "sanity anchor" hai jo har method ko pass karna chahiye.

Recall Verify

$y_p=0$ trivially $y_p^{\prime \prime }+y_p=0$ satisfy karta hai. ✓ (neeche check kiya)

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Example 6 — Cell F: sign care, $g=1/x$ aur absolute value

Forecast: $1/x$ ka antiderivative $\ln |x|$ hai — absolute value matter karta hai, aur yahi reason hai ki solution $x$ ke sign se split hoti hai. Saath hi $\int e^{\pm x}/xdx$ ek non-elementary integral hai (exponential integral), toh honest answer mein integral sign rakhna padega.

1. Homogeneous. $r^2-1=0\Rightarrow r=\pm 1$, toh $y_1=e^x,y_2=e^{-x}$.
2. Standard form. $g=\frac{1}{x}$.
3. Wronskian. $y_1^{\prime }=e^x,y_2^{\prime }=-e^{-x}$. $W=e^x(-e^{-x})-e^{-x}(e^x)=-1-1=-2.$
4. Derivatives. $u_1^{\prime }=-\frac{e^{-x}\cdot \frac{1}{x}}{-2}=\frac{e^{-x}}{2x},u_2^{\prime }=\frac{e^x\cdot \frac{1}{x}}{-2}=-\frac{e^x}{2x}.$
5. Integrate karo — elementary form mein ho nahi sakta. Integrals ke roop mein rakhte hain: $u_1=\frac{1}{2}\int \frac{e^{-x}}{x}dx,u_2=-\frac{1}{2}\int \frac{e^x}{x}dx.$ Yahan kyun ruka? $\int e^{-x}/xdx$ ka koi elementary antiderivative nahi hai (yeh $-\mathrm{Ei}(-x)$ hai). Sahi answer yahan ruk jaata hai — clean definite-integral packaging ke liye Cell G dekho.
6. $x$ ka sign. Integrals sirf wahan defined hain jahan $1/x$ continuous hai: alag-alag $x>0$ aur $x<0$ par. $x=0$ cross karne wala koi solution nahi hai kyunki $g$ wahan infinite discontinuity rakhta hai. Isliye hi interval batana zaroori hai.

General (kisi bhi side par): $y=c_1{e}^x+c_2{e}^{-x}+\frac{e^x}{2}\int \frac{e^{-x}}{x}dx-\frac{e^{-x}}{2}\int \frac{e^x}{x}dx$.

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Example 7 — Cell G: definite-integral (Green's-function) packaging

Forecast: do alag indefinite integrals ki jagah, hum inhe ek integral mein fuse kar sakte hain jisme $g(t)$ aur ek kernel ho. Guess: kernel $\sin (x-t)$ hoga.

1. Ingredients. $y_1=\cos x,y_2=\sin x$, $W=1$.
2. VoP as definite integrals. $u_1(x)=-{\int }_a^x\frac{\sin tg(t)}{1}dt,u_2(x)={\int }_a^x\frac{\cos tg(t)}{1}dt.$ Definite kyun? $a$ ko lower limit rakhne se constants fix ho jaate hain taaki $y_p(a)=y_p^{\prime }(a)=0$ — ek clean, unambiguous particular solution.
3. Assemble karo & fuse karo. $y_p=\cos x\left(-{\int }_a^x\sin tgdt\right)+\sin x\left({\int }_a^x\cos tgdt\right)={\int }_a^x(\sin x\cos t-\cos x\sin t)g(t)dt.$ Sine-subtraction identity se $\sin x\cos t-\cos x\sin t=\sin (x-t)$: $\boxed{y_p(x)={\int }_a^x\sin (x-t)g(t)dt}$ Kernel $\sin (x-t)$ is operator ka Green's function hai.

Yeh itna powerful kyun hai: ek formula har $g$ ke liye ek saath equation solve karta hai — bas apna $g$ integrand mein daal do. Yeh Cell A–F sab packaged hai.

Recall Verify

Maano $g=\sec x$, $a=0$: $y_p={\int }_0^x\sin (x-t)\sec tdt$. Do baar differentiate karne par (Leibniz rule) milta hai $y_p^{\prime \prime }+y_p=\sec x$; aur yeh parent note ke example ko homogeneous pieces tak reproduce karta hai. $x=1$ par numeric spot-check VERIFY mein hai.

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Example 8 — Cell H: real-world forced oscillator

Forecast: kyunki drive frequency ($1$) $\ne$ natural frequency ($2$), koi resonance nahi — response ek bounded combination honi chahiye $\cos t$ aur $\cos 2t,\sin 2t$ ki. $\cos t$ part ka amplitude $\approx 1$ guess karo.

1. Homogeneous. $r^2+4=0\Rightarrow r=\pm 2i$, toh $y_1=\cos 2t,y_2=\sin 2t$.
2. Standard form. $g=3\cos t$.
3. Wronskian. $W=\cos 2t(2\cos 2t)-\sin 2t(-2\sin 2t)=2({\cos }^{2}2t+{\sin }^{2}2t)=2$.
4. Derivatives. $u_1^{\prime }=-\frac{\sin 2t\cdot 3\cos t}{2},u_2^{\prime }=\frac{\cos 2t\cdot 3\cos t}{2}.$
5. Integrate karo product-to-sum use karke ($\sin 2t\cos t=\frac{1}{2}(\sin 3t+\sin t)$, $\cos 2t\cos t=\frac{1}{2}(\cos 3t+\cos t)$): $u_1=-\frac{3}{4}\int (\sin 3t+\sin t)dt=-\frac{3}{4}\left(-\frac{\cos 3t}{3}-\cos t\right)=\frac{1}{4}\cos 3t+\frac{3}{4}\cos t,$ $u_2=\frac{3}{4}\int (\cos 3t+\cos t)dt=\frac{3}{4}\left(\frac{\sin 3t}{3}+\sin t\right)=\frac{1}{4}\sin 3t+\frac{3}{4}\sin t.$
6. Assemble karo. $y_p=\cos 2t\left(\frac{1}{4}\cos 3t+\frac{3}{4}\cos t\right)+\sin 2t\left(\frac{1}{4}\sin 3t+\frac{3}{4}\sin t\right).$ Sum identities se group karo: $\frac{1}{4}(\cos 2t\cos 3t+\sin 2t\sin 3t)=\frac{1}{4}\cos t$; $\frac{3}{4}(\cos 2t\cos t+\sin 2t\sin t)=\frac{3}{4}\cos t$. Total $y_p=\cos t$.
7. Initial conditions apply karo. General $y=c_1\cos 2t+c_2\sin 2t+\cos t$. $y(0)=c_1+1=0\Rightarrow c_1=-1$. $y^{\prime }(t)=-2c_1\sin 2t+2c_2\cos 2t-\sin t$; $y^{\prime }(0)=2c_2=0\Rightarrow c_2=0$. $\boxed{y(t)=\cos t-\cos 2t}$

Units sanity: har term pattern mein dimensionless hai lekin metres represent karta hai; dono cosines $1$ se bounded hain, toh $|y|\le 2$ m — bounded, jaise off-resonance ke liye forecast kiya tha. Figure beat-like superposition dikhata hai.

Recall Verify

$y=\cos t-\cos 2t$: $y^{\prime \prime }=-\cos t+4\cos 2t$. $y^{\prime \prime }+4y=(-\cos t+4\cos 2t)+4(\cos t-\cos 2t)=3\cos t$. ✓ Aur $y(0)=1-1=0$, $y^{\prime }(0)=0$. ✓

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Matrix, revisited

Recall Kis example ne kaun sa cell fill kiya?

A resonant-exponential ::: Ex. 1 B complex roots + $\tan x$ ::: Ex. 2 C repeated root + $\ln x$ ::: Ex. 3 D Cauchy–Euler standard-form trap ::: Ex. 4 E degenerate $g\equiv 0$ ::: Ex. 5 F sign/quadrant $1/x$ aur $\ln |x|$ ::: Ex. 6 G definite-integral / Green's kernel ::: Ex. 7 H real-world forced spring with initial conditions ::: Ex. 8

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Connections

• Wronskian — $W$ jo har denominator mein hai; non-zero hona guarantee karta hai ki linear system Cramer's Rule se solvable hai.
• Cramer's Rule — isse $u_1^{\prime },u_2^{\prime }$ $2\times 2$ system se extract kiye gaye.
• Method of Undetermined Coefficients — woh shortcut jo Ex. 2, 3, 6, 7 par fail hota hai.
• Second-order linear homogeneous ODE — $y_1,y_2$ ka source.
• Reduction of Order — Ex. 3 mein repeated-root $y_2=x{e}^{2x}$ aise milta hai.
• Green's function — Ex. 7 ka fused kernel $\sin (x-t)$.