Exercises — Non-homogeneous — variation of parameters

KYA karte hain: $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$ mein plug karo। $y_1=\cos x\Rightarrow y_1^{\prime }=-\sin x$। $y_2=\sin x\Rightarrow y_2^{\prime }=\cos x$। $W=\cos x\cdot \cos x-\sin x\cdot (-\sin x)={\cos }^{2}x+{\sin }^{2}x=1.$ KYUN matter karta hai: $W$ dono fractions ke denominator mein baithta hai। Ek clean $W=1$ matlab baad mein koi extra division nahi। $\boxed{W=1}$।

KYA karte hain: har solution differentiate karo, phir $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$ ke according cross-multiply karo। $y_1^{\prime }=e^x,y_2^{\prime }=2e^{2x}$। $W=e^x\cdot 2e^{2x}-e^{2x}\cdot e^x=2e^{3x}-e^{3x}=e^{3x}.$ KYUN exponents ka sum: $e^x\cdot e^{2x}=e^{x+2x}=e^{3x}$ har baar, isliye dono products $e^{3x}$ par land karte hain aur simply coefficients ke roop mein subtract hote hain $2-1=1$। $\boxed{W=e^{3x}}$। Ye kabhi zero nahi — do exponentials genuinely alag building blocks hain, isliye method kaam karne ki guarantee hai (exactly yahi ek non-vanishing Wronskian certify karta hai)।

KYA karte hain: har term ko leading coefficient $2x^2$ se divide karo। $y^{\prime \prime }+\frac{x}{2x^2}{y}^{\prime }-\frac{1}{2x^2}y=\frac{6x}{2x^2}$ $y^{\prime \prime }+\frac{1}{2x}{y}^{\prime }-\frac{1}{2x^2}y=\frac{3}{x}.$ KYUN: formula ka $g$ normalized equation ka RHS hai, raw $6x$ nahi। Isliye $\boxed{g(x)=3/x}$, na ki $6x$।

Step 1 — homogeneous. $r^2+1=0\Rightarrow r=\pm i$, isliye $y_1=\cos x,y_2=\sin x$ (dekho Second-order linear homogeneous ODE)। Step 2 — Wronskian. Ex 1.1 se, $W=1$। Step 3 — do derivatives. $g=\tan x$ ke saath: $u_1^{\prime }=-\frac{y_{2}g}{W}=-\sin x\tan x=-\frac{{\sin }^{2}x}{\cos x},u_2^{\prime }=\frac{y_{1}g}{W}=\cos x\tan x=\sin x.$ KYUN $\sin x\tan x$ ko ${\sin }^{2}x/\cos x$ ke roop mein rewrite karo: kyunki bare $\sin x\tan x$ ka koi standard antiderivative nahi hai, lekin ${\sin }^{2}x/\cos x$ ko secants aur cosines mein convert kiya ja sakta hai, jinke integrals hote hain। Step 4 — integrate. Doosra easy hai: $u_2=-\cos x$। $u_1$ ke liye, ${\sin }^{2}x=1-{\cos }^{2}x$ likho taaki fraction split ho: $u_1=-\int \frac{1-{\cos }^{2}x}{\cos x}dx=-\int (\sec x-\cos x)dx=-\ln |\sec x+\tan x|+\sin x.$ KYUN ye split: $\frac{1}{\cos x}=\sec x$ (ek known integral $\ln |\sec x+\tan x|$) aur $\frac{{\cos }^{2}x}{\cos x}=\cos x$ (trivial)। Ek hard fraction ko do known integrals mein todna hi saari trick hai। Step 5 — assemble. $y_p=\cos x(\sin x-\ln |\sec x+\tan x|)+\sin x(-\cos x).$ $\cos x\sin x$ terms cancel ho jaate hain: $\boxed{y_p=-\cos x\ln |\sec x+\tan x|}.$ General: $y=c_1\cos x+c_2\sin x-\cos x\ln |\sec x+\tan x|$।

Domain caveat. $\tan x$ $x=\pm \frac{\pi }{2}$ par blow up hota hai, aur $\sec x+\tan x>0$ poore $(-\frac{\pi }{2},\frac{\pi }{2})$ mein, isliye logarithm ka argument positive hai aur absolute value is interval par unnecessary hai — lekin agar tum doosri branches tak extend karo to rakhna, jahan $\sec x+\tan x$ ka sign badalta hai। Solution sirf $\tan x$ ki consecutive singularities ke beech ek open interval par valid hai; tum $x=\frac{\pi }{2}$ cross nahi kar sakte।

$y_1=e^x,y_2=e^{2x}$, $W=e^{3x}$ (Ex 1.2)। $g=e^{3x}$ ke saath: $u_1^{\prime }=-\frac{e^{2x}\cdot e^{3x}}{e^{3x}}=-e^{2x},u_2^{\prime }=\frac{e^x\cdot e^{3x}}{e^{3x}}=e^x.$ KYUN fractions itne cleanly collapse hote hain: numerator $e^{2x}{e}^{3x}=e^{5x}$ ko $W=e^{3x}$ se divide karne par $e^{2x}$ bachta hai — exponentials hamesha exponents subtract karte hain, isliye koi messy integral survive nahi karta। Integrate: $u_1=-\frac{1}{2}{e}^{2x}$ (KYUN $\frac{1}{2}$: $\int e^{2x}dx=\frac{1}{2}{e}^{2x}$, inner-derivative factor of $2$), aur $u_2=e^x$। $y_p=e^x\left(-\frac{1}{2}{e}^{2x}\right)+e^{2x}(e^x)=-\frac{1}{2}{e}^{3x}+e^{3x}=\frac{1}{2}{e}^{3x}.$ $\boxed{y_p=\frac{1}{2}{e}^{3x}}$। General: $y=c_1{e}^x+c_2{e}^{2x}+\frac{1}{2}{e}^{3x}$। (Kyunki $e^{3x}$ koi homogeneous solution nahi hai, Method of Undetermined Coefficients bhi $A{e}^{3x}$ guess se same answer deta — ek accha sanity check।)

Step 0 — standard form. $x^2$ se divide karo: $y^{\prime \prime }-\frac{1}{x}{y}^{\prime }+\frac{1}{x^2}y=\frac{1}{x}\Rightarrow g=\frac{1}{x}.$ Step 2 — Wronskian. $y_1=x\Rightarrow y_1^{\prime }=1$। $y_2=x\ln x\Rightarrow y_2^{\prime }=\ln x+1$। KYUN $y_2^{\prime }=\ln x+1$: $x\cdot \ln x$ par product rule deta hai $1\cdot \ln x+x\cdot \frac{1}{x}=\ln x+1$। $W=x(\ln x+1)-x\ln x\cdot 1=x\ln x+x-x\ln x=x.$ Step 3. $u_1^{\prime }=-\frac{y_{2}g}{W}=-\frac{x\ln x\cdot \frac{1}{x}}{x}=-\frac{\ln x}{x},u_2^{\prime }=\frac{y_{1}g}{W}=\frac{x\cdot \frac{1}{x}}{x}=\frac{1}{x}.$ Step 4 — integrate. $u_1=-\int \frac{\ln x}{x}dx=-\frac{(\ln x{)}^2}{2}$। KYUN substitute $t=\ln x$: tab $dt=\frac{1}{x}dx$, isliye integral ban jaata hai $\int tdt=\frac{t^2}{2}$ — $\frac{1}{x}$ exactly woh missing piece tha jo $\ln x$ ki apni derivative ko appear karaata hai। $u_2=\int \frac{1}{x}dx=\ln x$। Step 5. $y_p=x\left(-\frac{(\ln x{)}^2}{2}\right)+x\ln x(\ln x)=-\frac{x(\ln x{)}^2}{2}+x(\ln x{)}^2=\frac{x(\ln x{)}^2}{2}.$ $\boxed{y_p=\frac{1}{2}x(\ln x{)}^2}$। General: $y=c_{1}x+c_{2}x\ln x+\frac{1}{2}x(\ln x{)}^2$। Domain caveat. $\ln x$ ko $x>0$ chahiye; $y_2=x\ln x$ aur poora solution sirf $x>0$ par live karta hai। Original Euler equation ka $x=0$ ek singular point hai, isliye koi solution use cross nahi karta।

Step 0 — standard form. $x^2$ se divide karo: $y^{\prime \prime }-\frac{2}{x^2}y=1$, isliye $g=1$ (na ki $x^2$!)। Step 2 — Wronskian. $y_1^{\prime }=2x$, $y_2^{\prime }=-x^{-2}$। $W=x^2(-x^{-2})-x^{-1}(2x)=-1-2=-3.$ Ek negative constant $W$ — bilkul theek hai। Step 3. $u_1^{\prime }=-\frac{x^{-1}\cdot 1}{-3}=\frac{1}{3x},u_2^{\prime }=\frac{x^2\cdot 1}{-3}=-\frac{x^2}{3}.$ Step 4. $u_1=\frac{1}{3}\ln |x|=\frac{1}{3}\ln x$ (hum $x>0$ par hain), $u_2=-\frac{x^3}{9}$। KYUN $\int \frac{1}{x}dx=\ln x$ aur $\int x^{2}dx=\frac{x^3}{3}$ — standard power rule, phir times $-\frac{1}{3}$। Step 5. $y_p=x^2\left(\frac{1}{3}\ln x\right)+x^{-1}\left(-\frac{x^3}{9}\right)=\frac{1}{3}{x}^2\ln x-\frac{1}{9}{x}^2.$ KYUN hum $-\frac{1}{9}{x}^2$ drop kar sakte hain: ye $y_1=x^2$ ka ek constant multiple hai, yaani complementary function $c_1{x}^2$ ke andar pehle se hai, isliye sirf $c_1$ change hota hai aur kuch naya add nahi hota। $\boxed{y_p=\frac{1}{3}{x}^2\ln x}$। General: $y=c_1{x}^2+c_2{x}^{-1}+\frac{1}{3}{x}^2\ln x$। Domain caveat. $y_2=x^{-1}$ $x=0$ par singular hai; hum $x>0$ par (ya symmetrically $x<0$ par) kaam karte hain, kabhi cross nahi karte।

Step 1. $r^2-2r+1=(r-1{)}^2=0$, ek repeated root। Isliye $y_1=e^x$ aur $y_2=x{e}^x$ (Reduction of Order se doosra)। Step 2 — Wronskian. $y_1^{\prime }=e^x$, $y_2^{\prime }=e^x+x{e}^x$। $W=e^x(e^x+x{e}^x)-x{e}^x\cdot e^x=e^{2x}+x{e}^{2x}-x{e}^{2x}=e^{2x}.$ KYUN $x{e}^{2x}$ cancel hota hai: cross-terms equal aur opposite hain — yahi cancellation exactly isliye hoti hai ki $e^x,x{e}^x$ ka Wronskian ek clean $e^{2x}$ rehta hai। Step 3. $g=e^x$: $u_1^{\prime }=-\frac{x{e}^x\cdot e^x}{e^{2x}}=-x,u_2^{\prime }=\frac{e^x\cdot e^x}{e^{2x}}=1.$ KYUN ek polynomial $-x$ appear hota hai: $e^{2x}$ ko $e^{2x}$ se divide karne par exponential completely kill ho jaata hai, sirf $y_2=x{e}^x$ se bare factor $x$ bachta hai। Ye surviving $x$ hi hai jo integrate hokar $x^2$ resonance term banega। Step 4. $u_1=-\frac{x^2}{2}$, $u_2=x$। Step 5. $y_p=e^x\left(-\frac{x^2}{2}\right)+x{e}^x(x)=-\frac{x^2}{2}{e}^x+x^2{e}^x=\frac{x^2}{2}{e}^x.$ $\boxed{y_p=\frac{1}{2}{x}^2{e}^x}$ — exactly woh $x^2{e}^x$ shape jo resonance force karti hai, aur VoP ne ise bina guessing ke produce kiya। General: $y=(c_1+c_{2}x)e^x+\frac{1}{2}{x}^2{e}^x$।

Step 1. $r^2-1=0\Rightarrow r=\pm 1$, isliye $y_1=e^x,y_2=e^{-x}$। Step 2 — Wronskian. $y_1^{\prime }=e^x$, $y_2^{\prime }=-e^{-x}$। $W=e^x(-e^{-x})-e^{-x}{e}^x=-1-1=-2.$ Step 3. $g=\frac{2}{1+e^x}$: $u_1^{\prime }=-\frac{e^{-x}\cdot \frac{2}{1+e^x}}{-2}=\frac{e^{-x}}{1+e^x},u_2^{\prime }=\frac{e^x\cdot \frac{2}{1+e^x}}{-2}=-\frac{e^x}{1+e^x}.$ Step 4 — integrate. $u_2$ ke liye: $\int \frac{e^x}{1+e^x}dx=\ln (1+e^x)$, KYUN — numerator $e^x$ exactly denominator $1+e^x$ ki derivative hai, isliye ye ek direct $\int \frac{f^{\prime }}{f}=\ln f$ pattern hai (aur $1+e^x>0$ hamesha, isliye yahan absolute value kabhi nahi chahiye)। Thus $u_2=-\ln (1+e^x)$। $u_1$ ke liye: $t=e^x$ let karo, isliye $dt=e^{x}dx=tdx$, yaani $dx=dt/t$। KYUN ye substitution — ye mixed $e^{-x}/(1+e^x)$ ko $t$ ki ek pure rational function mein convert karti hai, jise partial fractions crack kar sakti hain: $u_1=\int \frac{e^{-x}}{1+e^x}dx=\int \frac{1/t}{1+t}\cdot \frac{dt}{t}=\int \frac{dt}{t^2(1+t)}.$ Partial fractions — kyun ye shape. Denominator $t^2(1+t)$ mein har factor ki har distinct power ke liye ek term chahiye: ek $\frac{A}{t}$ (simple), ek $\frac{B}{t^2}$ (repeated factor $t$ ko dono powers chahiye), aur ek $\frac{C}{1+t}$ (linear factor)। $1=At(1+t)+B(1+t)+C{t}^2$ solve karne par $B=1$ ($t=0$ set karo), $C=1$ ($t=-1$ set karo), $A=-1$ ($t^2$ match karo)। Isliye $\frac{1}{t^2(1+t)}=\frac{-1}{t}+\frac{1}{t^2}+\frac{1}{1+t}.$ Term by term integrate karo: $u_1=-\ln t-\frac{1}{t}+\ln (1+t)=-x-e^{-x}+\ln (1+e^x)(\text{since }t=e^x>0).$ Step 5. $y_p=e^x(-x-e^{-x}+\ln (1+e^x))+e^{-x}(-\ln (1+e^x)).$ $=-x{e}^x-1+e^x\ln (1+e^x)-e^{-x}\ln (1+e^x).$ $-1$ ek constant hai, lekin $1$ yahan koi homogeneous solution nahi hai (woh $e^{\pm x}$ hain), isliye ise rakho। Cleanly: $\boxed{y_p=-x{e}^x-1+(e^x-e^{-x})\ln (1+e^x)}.$ Domain caveat. $1+e^x>0$ har real $x$ ke liye, isliye ye solution poori real line par valid hai — koi branch cuts nahi, koi excluded points nahi। Ex 2.1 se contrast karo, jahan $\tan x$ ne humein ek single interval par force kiya tha।

$y_1=\cos x,y_2=\sin x,W=1$। Tab $u_1^{\prime }=-\sin xg(x),u_2^{\prime }=\cos xg(x).$ Base point $x_0$ se $x$ tak integrate karo (dummy variable $t$ use karo): $u_1(x)=-{\int }_{x_0}^x\sin tg(t)dt,u_2(x)={\int }_{x_0}^x\cos tg(t)dt.$ KYUN $x_0$ ke saath definite integrals? Ek indefinite integral ek arbitrary $+C$ hide karta hai; lower limit ko chosen $x_0$ par pin karne se woh constant fix ho jaata hai। $x_0$ kaise choose karein: ise apne initial data ke saath match karo — agar problem specify karta hai $y(x_0)=y^{\prime }(x_0)=0$, to ye formula in conditions ko already satisfy karta hai (dono $u_1,u_2$ $x_0$ par vanish hote hain, isliye $y_p(x_0)=0$, aur verify kiya ja sakta hai ki $y_p^{\prime }(x_0)=0$ bhi)। Koi bhi doosra $x_0$ sirf $y_p$ ko ek homogeneous piece $c_1\cos x+c_2\sin x$ se shift karta hai, jo general solution mein absorb ho jaata hai। Isliye general solution ke liye $x_0$ free hai aur initial-value problem ke liye initial condition se set hai। Assemble karo: $y_p=\cos x\left(-{\int }_{x_0}^x\sin tg(t)dt\right)+\sin x{\int }_{x_0}^x\cos tg(t)dt.$ Dono ko ek integral ke neeche lo ($\cos x,\sin x$ $t$ ke w.r.t. constants hain, isliye ye andar slide kar sakte hain): $y_p={\int }_{x_0}^x(\sin x\cos t-\cos x\sin t)g(t)dt={\int }_{x_0}^x\sin (x-t)g(t)dt.$ KYUN bracket $\sin (x-t)$ ban gaya: ye sine subtraction identity hai $\sin (x-t)=\sin x\cos t-\cos x\sin t$ — algebra is reveal karne ke liye designed thi। $\boxed{y_p(x)={\int }_{x_0}^x\sin (x-t)g(t)dt}$ Kernel $K(x,t)=\sin (x-t)$ exactly $y^{\prime \prime }+y$ ke liye Green's function hai: VoP iske constructive route hai। Note karo kernel sirf gap $x-t$ par depend karta hai (ek "time $t$ par push kitni der pehle act kiya" factor) — neeche wali figure dekho। Check $g=\sec x$ ke saath (parent Example 1) $\cos x\ln |\cos x|+x\sin x$ recover karta hai homogeneous terms tak।

Step 1. $r^2+4=0\Rightarrow r=\pm 2i$, isliye $y_1=\cos 2x,y_2=\sin 2x$। Step 2 — Wronskian. $y_1^{\prime }=-2\sin 2x$, $y_2^{\prime }=2\cos 2x$। $W=\cos 2x(2\cos 2x)-\sin 2x(-2\sin 2x)=2({\cos }^{2}2x+{\sin }^{2}2x)=2.$ Step 3. $g=\csc 2x=1/\sin 2x$: $u_1^{\prime }=-\frac{\sin 2x\cdot \frac{1}{\sin 2x}}{2}=-\frac{1}{2},u_2^{\prime }=\frac{\cos 2x\cdot \frac{1}{\sin 2x}}{2}=\frac{\cos 2x}{2\sin 2x}=\frac{\cot 2x}{2}.$ KYUN $u_1^{\prime }$ ek constant mein collapse hota hai: numerator mein $\sin 2x$ $\csc 2x=1/\sin 2x$ ko cancel kar deta hai, sirf $-\frac{1}{2}$ bachta hai — possible sabse cleaner integrand। Step 4. $u_1=-\frac{1}{2}x$। $u_2=\frac{1}{2}\int \cot 2xdx=\frac{1}{2}\cdot \frac{1}{2}\ln |\sin 2x|=\frac{1}{4}\ln |\sin 2x|$। KYUN extra $\frac{1}{2}$: $\frac{d}{dx}\ln |\sin 2x|=\frac{2\cos 2x}{\sin 2x}=2\cot 2x$, isliye $\int \cot 2xdx=\frac{1}{2}\ln |\sin 2x|$ — $2x$ ki inner derivative $\frac{1}{2}$ demand karti hai। Step 5. $y_p=\cos 2x\left(-\frac{1}{2}x\right)+\sin 2x\left(\frac{1}{4}\ln |\sin 2x|\right).$ $\boxed{y_p=-\frac{x}{2}\cos 2x+\frac{1}{4}\sin 2x\ln |\sin 2x|}.$ General: $y=c_1\cos 2x+c_2\sin 2x-\frac{x}{2}\cos 2x+\frac{1}{4}\sin 2x\ln |\sin 2x|$। Domain caveat. $\csc 2x$ jahan bhi $\sin 2x=0$ ho singular hai, yaani $x=0,\frac{\pi }{2},\pi ,\dots$। $0<x<\frac{\pi }{2}$ par $\sin 2x>0$ hai, isliye $\ln |\sin 2x|=\ln (\sin 2x)$ aur absolute value wahan decorative hai; ise rakho agar adjacent interval par jao jahan $\sin 2x<0$ ho। Solution kabhi in singularities ko cross nahi karta।

1. Leading coefficient $1$ tak normalize karo — sach wala $g$ nikalo।
2. $y_1,y_2$ find karo (characteristic roots, ya given)।
3. $W=y_1{y}_2^{\prime }-y_2{y}_1^{\prime }$ — confirm karo $W\ne 0$।
4. $u_1^{\prime }=-y_{2}g/W$, $u_2^{\prime }=+y_{1}g/W$ (minus on the first)।
5. Integrate karo, constants drop karo — chain-rule factors aur valid interval ka dhyan rakho।
6. $y_p=u_1{y}_1+u_2{y}_2$; koi bhi homogeneous leftover $c_1,c_2$ mein absorb karo।