Exercises — First-order linear ODEs — integrating factor method (derivation)

(a) Linear. $y$ aur $\frac{dy}{dx}$ sirf first power mein hain, kabhi multiply nahi hue. Already standard hai (coefficient of $\frac{dy}{dx}$ is $1$): $P=3,Q=x^2$. (b) NOT linear. Term $y^2$ mein $y$ second power mein hai — definition ke mutabik $y$ ki koi bhi power first ke alawa allowed nahi hai. (c) Linear. Lekin $\frac{dy}{dx}$ ka coefficient $x$ hai, $1$ nahi. $x$ se divide karo: $\frac{dy}{dx}+\frac{1}{x}y=\frac{\sin x}{x}$, toh $P=\frac{1}{x},Q=\frac{\sin x}{x}$. (d) NOT linear. $\frac{dy}{dx}$ andar $\sin (\cdot )$ ke baithaa hai — yeh derivative ka function hai. Linear matlab $y$ aur $y^{\prime }$ bare aayein, first power mein, kabhi kisi doosre function ke andar nahi.

(a) $\mu =e^{\int 5dx}=e^{5x}$. (b) $\mu =e^{\int \frac{2}{x}dx}=e^{2\ln |x|}=|x{|}^2=x^2$ (apne working domain pe $x^2$ lo). (c) Yahan $P=-\tan x$ hai, toh $\mu =e^{\int -\tan xdx}$. Kyunki $\int \tan xdx=-\ln |\cos x|$ hai, humein milta hai $\int -\tan xdx=\ln |\cos x|$, isliye $\mu =e^{\ln |\cos x|}=|\cos x|=\cos x$ (us interval pe jahan $\cos x>0$ ho).

Standardize: already standard hai, $P=3,Q=6$. Mu: $\mu =e^{\int 3dx}=e^{3x}$. Multiply & collapse: $\frac{d}{dx}(e^{3x}y)=6e^{3x}$. Integrate: $e^{3x}y=\int 6e^{3x}dx=2e^{3x}+C$. Divide: $y=2+C{e}^{-3x}$.

Standardize: already standard hai, $P=\frac{1}{x},Q=3x$. Mu: $\mu =e^{\int \frac{1}{x}dx}=e^{\ln x}=x$. Multiply & collapse: $\frac{d}{dx}(xy)=x\cdot 3x=3x^2$. Integrate: $xy=\int 3x^{2}dx=x^3+C$. Divide: $y=x^2+\frac{C}{x}$.

Standardize: already standard hai, $P=2x,Q=x$. Mu: $\mu =e^{\int 2xdx}=e^{x^2}$. Multiply & collapse: $\frac{d}{dx}(e^{x^2}y)=x{e}^{x^2}$. Integrate: right side hai $\int x{e}^{x^2}dx$. Yahan substitution kyun? Kyunki exponent $x^2$ ka derivative $2x$ hai, aur bahar $x$ pada hai — yeh classic "inner-derivative-present" signal hai. $u=x^2$, $du=2xdx$ lo: $\int x{e}^{x^2}dx=\frac{1}{2}\int e^{u}du=\frac{1}{2}{e}^{x^2}+C.$ Toh $e^{x^2}y=\frac{1}{2}{e}^{x^2}+C$. Divide: $y=\frac{1}{2}+C{e}^{-x^2}$.

Standardize: $x$ se divide karo ($\frac{dy}{dx}$ ka coefficient $1$ hona chahiye): $\frac{dy}{dx}-\frac{2}{x}y=x^3$. Toh $P=-\frac{2}{x},Q=x^3$. Mu: $\mu =e^{\int -\frac{2}{x}dx}=e^{-2\ln x}=x^{-2}=\frac{1}{x^2}$. Multiply & collapse: $\frac{d}{dx}\left(x^{-2}y\right)=x^{-2}\cdot x^3=x$. Integrate: $x^{-2}y=\int xdx=\frac{x^2}{2}+C$. Divide: $x^2$ se multiply karo: $y=\frac{x^4}{2}+C{x}^2$.

Standardize: $P=1,Q=x$. Mu: $\mu =e^{\int 1dx}=e^x$. Multiply & collapse: $\frac{d}{dx}(e^{x}y)=x{e}^x$. Integration by parts: Parts kyun? Humare paas "polynomial $\times$ exponential" ka product hai aur koi chain-rule inner derivative nahi hai, toh substitution kaam nahi karega — parts woh tool hai jo ek mushkil product ko aasaan mein badal deta hai. $u=x,dv=e^{x}dx$ lete hain: $\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C.$ Toh $e^{x}y=x{e}^x-e^x+C$. Divide: $y=x-1+C{e}^{-x}$. IC baad mein lagao: $y(0)=2\Rightarrow 2=0-1+C\Rightarrow C=3$. Answer: $y=x-1+3e^{-x}$.

Standardize: $\cos x$ se divide karo (is interval pe nonzero hai): $\frac{dy}{dx}+\tan xy=\sec x$. Toh $P=\tan x,Q=\sec x$. Mu: $\mu =e^{\int \tan xdx}=e^{-\ln |\cos x|}=\frac{1}{\cos x}=\sec x$ (kyunki yahan $\cos x>0$ hai). Multiply & collapse: $\frac{d}{dx}(\sec xy)=\sec x\cdot \sec x={\sec }^{2}x$. Integrate: $\sec xy=\int {\sec }^{2}xdx=\tan x+C$. Divide: $y=\frac{\tan x+C}{\sec x}=\sin x+C\cos x$ ($\frac{\tan x}{\sec x}=\sin x$ aur $\frac{1}{\sec x}=\cos x$ use karke).

Yeh $n=2$ ke saath ek Bernoulli equation hai — jaisa likha hai woh linear nahi ($x{y}^2$ term powers multiply karta hai). Substitute karo $v=y^{1-n}=y^{-1}$, toh $\frac{dv}{dx}=-y^{-2}\frac{dy}{dx}$. ODE ko $y^2$ se divide karo: $y^{-2}\frac{dy}{dx}+y^{-1}=x$. $v$ ke terms mein: $-\frac{dv}{dx}+v=x$, yaani $\frac{dv}{dx}-v=-x.$ Ab yeh $v$ mein linear hai: $P=-1,Q=-x$. $\mu =e^{\int -1dx}=e^{-x}$. $\frac{d}{dx}(e^{-x}v)=-x{e}^{-x}$. Integration by parts se: $\int -x{e}^{-x}dx=x{e}^{-x}+e^{-x}+C$. Toh $e^{-x}v=x{e}^{-x}+e^{-x}+C\Rightarrow v=x+1+C{e}^x$. Back-substitute karo $v=1/y$: $\frac{1}{y}=x+1+C{e}^x$, toh $\boxed{y=\frac{1}{x+1+C{e}^x}}$.

$\mu =e^x$ poore mein. $0\le x\le 1$ pe ($f=1$): $\frac{d}{dx}(e^{x}y)=e^x\Rightarrow e^{x}y=e^x+C_1\Rightarrow y=1+C_1{e}^{-x}$. IC $y(0)=0$: $0=1+C_1\Rightarrow C_1=-1$, toh $y=1-e^{-x}(0\le x\le 1).$ $x=1$ pe: $y(1)=1-e^{-1}$. $x>1$ pe ($f=0$): $\frac{d}{dx}(e^{x}y)=0\Rightarrow e^{x}y=C_2\Rightarrow y=C_2{e}^{-x}$. $x=1$ pe continuity: $C_2{e}^{-1}=1-e^{-1}\Rightarrow C_2=e-1$, toh $y=(e-1)e^{-x}(x>1).$ Jaise $x\to \infty$, $y\to 0$: driving band hone ke baad, solution decay kar jaata hai.

Rewrite karo: $\frac{dy}{dx}-\frac{1}{x}y=1$, toh $P=-\frac{1}{x},Q=1$. $\mu =e^{\int -\frac{1}{x}dx}=e^{-\ln x}=\frac{1}{x}$. $\frac{d}{dx}\left(\frac{y}{x}\right)=\frac{1}{x}\cdot 1=\frac{1}{x}$. Integrate karo: $\frac{y}{x}=\ln x+C$. Toh $y=x\ln x+Cx$. IC $(1,0)$: $0=1\cdot 0+C\cdot 1\Rightarrow C=0$. Answer: $y=x\ln x$. Shape: $x=1$ pe, $y=0$; red curve $x=1$ ke thoda right mein slightly negative jaati hai (kyunki $x<1$ ke liye $\ln x<0$ ... $x>1$ pe yeh climb karti hai), minimum ko touch karti hai jahan $\frac{dy}{dx}=\ln x+1=0$, yaani $x=e^{-1}$, phir steadily rise karti hai — ek gently curving line jo marked point se guzarti hai.

Case $k=-1$: $\mu =e^x$, $\frac{d}{dx}(e^{x}y)=e^x{e}^{-x}=1$. Integrate karo: $e^{x}y=x+C$, toh $y=(x+C)e^{-x}$. Dhyan do $x$ aage aaya hai — forcing $e^{-x}$ decaying mode $e^{-x}$ se match karta hai, isliye ek constant integrate karne se linear factor aata hai. Case $k\ne -1$: $\frac{d}{dx}(e^{x}y)=e^{(k+1)x}$, integrate karo: $e^{x}y=\frac{1}{k+1}{e}^{(k+1)x}+C$, toh $y=\frac{1}{k+1}{e}^{kx}+C{e}^{-x}$. Koi extra $x$ nahi — forcing aur natural mode alag hain, toh plain exponential particular solution kaam karta hai. Insight: $x$-multiplier exactly tab aata hai jab $Q$ ka exponential rate homogeneous rate $-P$ ke barabar hota hai. (Yeh ODE mein resonance ka analogue hai.)

Maano $\mu =e^{\int P}$, toh ${\mu }^{\prime }=\mu P$ (derivation ki condition se). Likho $\mu y=\int \mu Qdx+C$. Dono sides ko product rule se differentiate karo: $\underset{\frac{d}{dx}(\mu y)}{\underbrace{{\mu }^{\prime }y+\mu y^{\prime }}}=\mu Q.$ ${\mu }^{\prime }=\mu P$ replace karo: $\mu Py+\mu y^{\prime }=\mu Q$. $\mu$ se divide karo (kabhi zero nahi kyunki yeh exponential hai): $y^{\prime }+Py=Q$. ✓ Exact-equation view: $y^{\prime }+Py-Q=0$ ko $\mu$ se multiply karne se left side exact differential $\frac{d}{dx}(\mu y)$ ban jaata hai — integrating factor exactly wahi cheez hai jo ek non-exact form ko exact bana deta hai.

(a) Standardize karo ($x\ne 0$ ke liye): $\frac{dy}{dx}+\frac{1}{x}y=\frac{1}{x}$, $\mu =x$. Phir $\frac{d}{dx}(xy)=x\cdot \frac{1}{x}=1$, toh $xy=x+C\Rightarrow y=1+\frac{C}{x}$. (b) Standard form mein $P=\frac{1}{x}$ aur $Q=\frac{1}{x}$ hain, dono $x=0$ pe undefined — existence-uniqueness theorem ko $P,Q$ continuous chahiye, jo wahan fail hota hai. $y=1+\frac{C}{x}$ dekho: jaise $x\to 0$, term $\frac{C}{x}$ blow up ho jaata hai jab tak $C=0$ na ho. Toh woh akela solution jo $x=0$ pe finite rehta hai $y=1$ hai, jo $y(0)=1$ deta hai. Isliye initial condition $y(0)=y_0$ sirf tab admissible hai jab $y_0=1$; kisi bhi doosri value ka origin tak koi solution nahi pahunchta. (Dhyan do ki original equation $x=0$ pe $0+y=1$ banta hai, jo $y(0)=1$ force karta hai — consistent hai.)

• L1: Kya $\frac{dy}{dx}+y^2=x$ linear hai? ::: Nahi — $y^2$ $y$ ki second power hai.
• L2: $\frac{dy}{dx}+2xy=x$ ke liye $\mu$? ::: $e^{x^2}$.
• L3: $\cos x{y}^{\prime }+\sin xy=1$ ke liye pehla move? ::: Standardize karne ke liye $\cos x$ se divide karo.
• L4: $y^{\prime }+y=x{y}^2$ ke liye substitution? ::: $v=y^{-1}$ (Bernoulli, $n=2$).
• L5: $x{y}^{\prime }+y=1$ ke liye kaunsa $y(0)$ kaam karta hai? ::: Sirf $y(0)=1$.