Exercises — Laplace of derivatives — key property for solving ODEs

$\mathcal{L}\{f^{\prime }(t)\}=sF(s)-f(0)$ $\mathcal{L}\{f^{\prime \prime }(t)\}=s^{2}F(s)-sf(0)-f^{\prime }(0)$ Shape padho: har prime hatane par ek $s$ ka factor lagta hai, aur har derivative hatane par initial value ki "fee" deni padti hai. $f^{\prime \prime }$ ke liye do fees hain: $sf(0)$ (zyada $s$-power, chhoti derivative) aur $f^{\prime }(0)$ (chhoti $s$-power, badi derivative).

General rule yeh hai: $\mathcal{L}\{f^{(n)}\}=s^{n}F(s)-s^{n-1}f(0)-s^{n-2}{f}^{\prime }(0)-\cdots -f^{(n-1)}(0).$ Jo term $f^{(k)}(0)$ carry karti hai woh hai $-s^{n-1-k}{f}^{(k)}(0)$, toh coefficient hai $\boxed{-s^{n-1-k}}$. Check $n=2$ ke saath: $k=0$ deta hai $-s^1$ (matlab $-sf(0)$ term); $k=1$ deta hai $-s^0=-1$ (matlab $-f^{\prime }(0)$ term). ✓

Way A — pehle differentiate karo. $f^{\prime }(t)=2e^{2t}$, toh $\mathcal{L}\{f^{\prime }\}=\frac{2}{s-2}$. Way B — rule use karo. $f(0)=e^0=1$, toh $\mathcal{L}\{f^{\prime }\}=sF(s)-f(0)=\frac{s}{s-2}-1=\frac{s-(s-2)}{s-2}=\frac{2}{s-2}.$ Dono $\frac{2}{s-2}$ dete hain. ✓ $-f(0)=-1$ zaroori tha — uske bina hume $\frac{s}{s-2}$ milta, jo galat hai.

$f(0)=\cos 0=1$. Rule: $\mathcal{L}\{f^{\prime }\}=sF(s)-f(0)=\frac{s^2}{s^2+9}-1=\frac{s^2-(s^2+9)}{s^2+9}=\frac{-9}{s^2+9}.$ Direct check: $f^{\prime }(t)=-3\sin 3t$, aur $\mathcal{L}\{-3\sin 3t\}=-3\cdot \frac{3}{s^2+9}=\frac{-9}{s^2+9}$. ✓

Yahan $F(s)=\frac{2}{s^2+4}$, $f(0)=\sin 0=0$, $f^{\prime }(t)=2\cos 2t$ toh $f^{\prime }(0)=2$. $\mathcal{L}\{f^{\prime \prime }\}=s^{2}F(s)-sf(0)-f^{\prime }(0)=\frac{2s^2}{s^2+4}-0-2=\frac{2s^2-2(s^2+4)}{s^2+4}=\frac{-8}{s^2+4}.$ Direct check: $f^{\prime \prime }(t)=-4\sin 2t$, $\mathcal{L}\{-4\sin 2t\}=-4\cdot \frac{2}{s^2+4}=\frac{-8}{s^2+4}$. ✓

Boundary term $-f(0)$ hai, toh yeh exactly tab vanish hoga jab $\boxed{f(0)=0}$. Example: $f(t)=\sin t$ mein $f(0)=0$ hai, toh $\mathcal{L}\{f^{\prime }\}=sF(s)=\frac{s}{s^2+1}$. Direct check karo: $f^{\prime }=\cos t$, $\mathcal{L}\{\cos t\}=\frac{s}{s^2+1}$. ✓ Insight: "clean" formula universally galat nahi hai — yeh ek special case hai jo exactly tab hota hai jab function origin par zero ho. Jis pal tumhara function zero se door start kare, yeh fail ho jaata hai.

Integration by parts mein boundary term hai $[e^{-st}f(t){]}_0^{\infty }$. Figure dekho: orange curve $e^{-st}$ exponentially neeche crush hoti hai, jabki blue $f(t)$ zyada se zyada $e^{ct}$ ki tarah badhti hai. Unka product $e^{-st}f(t)$, $M{e}^{-(s-c)t}$ (green envelope) se bound hai, jo $t\to \infty$ par $\to 0$ hota hai jab bhi $s>c$ ho. Toh upar ki limit $0$ contribute karti hai. Sirf neeche ki limit $t=0$ bachi rahti hai, jo $-e^{0}f(0)=-f(0)$ deti hai. Yahi poora reason hai ki formula $sF(s)-f(0)$ kyun padha jaata hai aur koi surviving tail kyun nahi hoti. Agar $s\le c$ hota toh integral diverge ho jaata aur transform exist nahi karta — dekho Exponential Order and Convergence of Integrals.

Maano $g(t)=f^{\prime }(t)$. Pehla rule kisi bhi nice function par apply hota hai, toh $\mathcal{L}\{g^{\prime }\}=s\mathcal{L}\{g\}-g(0).$ Ab $g^{\prime }=f^{\prime \prime }$, $\mathcal{L}\{g\}=\mathcal{L}\{f^{\prime }\}=sF(s)-f(0)$, aur $g(0)=f^{\prime }(0)$. Substitute karo: $\mathcal{L}\{f^{\prime \prime }\}=s(sF(s)-f(0))-f^{\prime }(0)=s^{2}F(s)-sf(0)-f^{\prime }(0).$ ✓ Kyun kaam karta hai: rule "ek prime hatao toh ek $s$ aur ek fee lo." Do primes hatana matlab usi move ko do baar chalana — koi nayi integration nahi chahiye. Yahi woh inductive engine hai jo general $s^{n}F(s)-\cdots$ pattern ke peeche hai.

Transform: $(sY-y(0))-2Y=0\Rightarrow (s-2)Y=y(0)=5$. Solve: $Y=\frac{5}{s-2}$. Invert: ${\mathcal{L}}^{-1}\{\frac{1}{s-2}\}=e^{2t}$, toh $\boxed{y(t)=5e^{2t}}$. Check: $y^{\prime }=10e^{2t}$, $y^{\prime }-2y=10e^{2t}-10e^{2t}=0$ ✓; $y(0)=5$ ✓.

Transform: $(s^{2}Y-sy(0)-y^{\prime }(0))+4Y=0$. ICs daalo: $s^{2}Y-s-0+4Y=0\Rightarrow (s^2+4)Y=s$. Solve: $Y=\frac{s}{s^2+4}$. Invert: yeh exactly $\mathcal{L}\{\cos 2t\}$ hai, toh $\boxed{y(t)=\cos 2t}$. Check: $y^{\prime \prime }=-4\cos 2t=-4y$, toh $y^{\prime \prime }+4y=0$ ✓; $y(0)=1$ ✓, $y^{\prime }(0)=-2\sin 0=0$ ✓.

Transform: $\mathcal{L}\{6\}=\frac{6}{s}$, toh $(sY-0)+2Y=\frac{6}{s}\Rightarrow (s+2)Y=\frac{6}{s}$. $Y=\frac{6}{s(s+2)}.$ Partial fractions: $\frac{6}{s(s+2)}=\frac{A}{s}+\frac{B}{s+2}$. Multiply out: $6=A(s+2)+Bs$. $s=0$ set karo: $6=2A\Rightarrow A=3$. $s=-2$ set karo: $6=-2B\Rightarrow B=-3$. $Y=\frac{3}{s}-\frac{3}{s+2}.$ Invert: $\boxed{y(t)=3-3e^{-2t}}$. Check: $y^{\prime }=6e^{-2t}$, $y^{\prime }+2y=6e^{-2t}+6-6e^{-2t}=6$ ✓; $y(0)=3-3=0$ ✓; equilibrium $y\to 3$ as $t\to \infty$ (RHS/coeff $=6/2=3$ se match karta hai) ✓.

Har term ko transform karo rules use karke:

• $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-sy(0)-y^{\prime }(0)=s^{2}Y-2s-3$
• $\mathcal{L}\{-3y^{\prime }\}=-3(sY-y(0))=-3sY+6$
• $\mathcal{L}\{2y\}=2Y$

Sum $=0$: $(s^{2}Y-2s-3)+(-3sY+6)+2Y=0\Rightarrow (s^2-3s+2)Y=2s-3.$ Factor: $s^2-3s+2=(s-1)(s-2)$, toh $Y=\frac{2s-3}{(s-1)(s-2)}$. Partial fractions: $\frac{2s-3}{(s-1)(s-2)}=\frac{A}{s-1}+\frac{B}{s-2}$. Tab $2s-3=A(s-2)+B(s-1)$. $s=1$ set karo: $-1=A(-1)\Rightarrow A=1$. $s=2$ set karo: $1=B(1)\Rightarrow B=1$. $Y=\frac{1}{s-1}+\frac{1}{s-2}\Rightarrow \boxed{y(t)=e^t+e^{2t}}.$ Check: $y^{\prime }=e^t+2e^{2t}$, $y^{\prime \prime }=e^t+4e^{2t}$; $y^{\prime \prime }-3y^{\prime }+2y=(1-3+2)e^t+(4-6+2)e^{2t}=0$ ✓; $y(0)=1+1=2$ ✓; $y^{\prime }(0)=1+2=3$ ✓.

Transform: $s^{2}Y-sy(0)-y^{\prime }(0)+Y=0\Rightarrow s^{2}Y-0-2+Y=0\Rightarrow (s^2+1)Y=2$. $Y=\frac{2}{s^2+1}=2\cdot \frac{1}{s^2+1}\Rightarrow \boxed{y(t)=2\sin t}.$ ICs trace karna: $-sy(0)$ term vanish ho gaya kyunki $y(0)=0$ — isse koi $\cos t$ piece nahi aayi. $-y^{\prime }(0)=-2$ term akela survivor hai, aur usne $\sin t$ amplitude $2$ banaya. Toh sirf slope IC ne yeh answer shape kiya. Check: $y^{\prime \prime }=-2\sin t=-y$ ✓; $y(0)=0$ ✓; $y^{\prime }(0)=2\cos 0=2$ ✓.

Transform: $s^{2}Y-sy(0)-y^{\prime }(0)=0\Rightarrow s^{2}Y-s-4=0\Rightarrow Y=\frac{s+4}{s^2}=\frac{1}{s}+\frac{4}{s^2}.$ Invert: ${\mathcal{L}}^{-1}\{\frac{1}{s}\}=1$, ${\mathcal{L}}^{-1}\{\frac{1}{s^2}\}=t$, toh $\boxed{y(t)=1+4t}.$ Check: $y^{\prime }=4$, $y^{\prime \prime }=0$ ✓; $y(0)=1$ ✓; $y^{\prime }(0)=4$ ✓. Limiting insight: $y^{\prime \prime }+ky=0$ ke liye frequency $\sqrt{k}$ hai. Jaise $k\to 0$, $\sin (\sqrt{k}t)/\sqrt{k}\to t$ aur $\cos (\sqrt{k}t)\to 1$, toh oscillation "unroll" hokar straight line $y(0)+y^{\prime }(0)t$ ban jaati hai. $k=0$ case woh degenerate limit hai jahan do initial conditions seedha ek line ke intercept aur slope map ho jaate hain — bilkul bhi oscillation nahi.

Upar har solution teen checks se paas hua: khud ODE, value IC, aur (2nd order ke liye) slope IC. Agar koi check fail ho, to almost certainly tumne koi fee term $-s^{n-1-k}{f}^{(k)}(0)$ drop kar di.