Worked examples — Inverse Laplace transform — partial fractions, tables

4.6.29 · D3 · Maths › Ordinary Differential Equations › Inverse Laplace transform — partial fractions, tables

Shuru karne se pehle, un do worlds ki ek reminder jinke beech hum translate karte hain. Picture ko left-to-right padho: real time $t$ mein ek differential equation transform variable $s$ mein ek algebra problem ban jaata hai, aur inverse Laplace transform ${\mathcal{L}}^{-1}$ return arrow hai.

---

First-shift theorem — ek baar state karo, har jagah use karo

Neeche ke kai examples is ek result par depend karte hain, isliye kaam aane se pehle ise saaf se pin down karte hain.

---

Scenario matrix

Har rational function $F(s)=\frac{P(s)}{Q(s)}$ jo hum mil sakti hai, usse classify kiya jaata hai denominator $Q(s)$ kaise factor hota hai aur fraction proper hai ya nahi ke basis par. Yahan proper ka matlab hai ki upar ki degree neeche ki degree se choti hai — yahi woh condition hai jo table silently assume karti hai.

Cell	Case class	$Q(s)$ ki signature	Inverse kaisi dikhti hai	Example
L1	Distinct real poles	$(s-a)(s-b)$, $a\ne b$	$e^{at}$ ka sum	Ex 1
L2	Real poles, ek $s=0$ par	$s(s-a)$	constant $+$ $e^{at}$	Ex 2
R1	Repeated real pole	$(s-a{)}^2$ ya $(s-a{)}^3$	$e^{at}(c_0+c_{1}t+\dots )$	Ex 3
Q1	Irreducible quadratic, $a=0$	$s^2+{\omega }^2$	pure $\sin /\cos$	Ex 4
Q2	Irreducible quadratic, $a\ne 0$	$(s-a{)}^2+{\omega }^2$	damped/growing $e^{at}\sin$	Ex 5
H1	Real poles $\pm b$ (sign flip!)	$s^2-b^2$	$\sinh /\cosh$, not $\sin$	Ex 6
D1	Degenerate / improper	$\mathrm{deg}P\ge \mathrm{deg}Q$	$\delta (t)$ $+$ proper part	Ex 7
S1	$t$ mein shifted (Heaviside)	$e^{-cs}F(s)$	delayed switch-on	Ex 8
W1	Real-world word problem	ek ODE se aaya	physical curve	Ex 9
X1	Exam twist (mixed)	quadratic $\times$ linear	$e^{at}+$ damped osc.	Ex 10
RQ1	Repeated irreducible quadratic	$(s^2+{\omega }^2{)}^2$	$t\sin \omega t$, $t\cos \omega t$	Ex 11

Page ka baaki hissa har cell ke liye ek example walkthrough karta hai. Har ek apne cell ke label ke saath hai taaki tum dekh sako pura grid cover ho raha hai.

---

Ex 1 — Cell L1 · distinct real poles

Forecast: $s=3$ aur $s=-1$ par do alag real poles. Aage padhne se pehle guess karo — answer mein $e^{3t}$ (growing) aur $e^{-t}$ (decaying) ka mix hona chahiye, bilkul bhi koi wiggle nahi.

1. Atoms mein split karo. $\frac{2s-1}{(s-3)(s+1)}=\frac{A}{s-3}+\frac{B}{s+1}$ likho. Yeh step kyun? Do distinct linear factors → table sirf ek pole ek baar jaanti hai, isliye hume inhe alag karna hoga.
2. $A$ ke liye cover-up. $(s-3)$ se multiply karo aur $s=3$ set karo: $A=\frac{2(3)-1}{3+1}=\frac{5}{4}$. Yeh step kyun? $s=3$ par $B$-term mein abhi bhi $(s-3)$ ka factor hai isliye woh zero ho jaata hai; sirf $A$ bachta hai.
3. $B$ ke liye cover-up. $(s+1)$ se multiply karo aur $s=-1$ set karo: $B=\frac{2(-1)-1}{-1-3}=\frac{-3}{-4}=\frac{3}{4}$. Yeh step kyun? Doosre pole par wohi trick.
4. Har atom look up karo. ${\mathcal{L}}^{-1}\{1/(s-a)\}=e^{at}$, isliye $f(t)=\frac{5}{4}{e}^{3t}+\frac{3}{4}{e}^{-t}.$ Yeh step kyun? Linearity hume pieces ko alag-alag invert karne aur add karne deti hai.

Verify: $t=0$ plug karo: $f(0)=\frac{5}{4}+\frac{3}{4}=2$. Sanity: initial-value theorem kehta hai $f(0^{+})={\lim }_{s\to \infty }sF(s)=\lim \frac{2s^2-s}{s^2-2s-3}=2$. ✓ Match karta hai.

---

Ex 2 — Cell L2 · $s=0$ par baitha ek pole

Forecast: Ek pole $s=0$ par hai. Yaad karo ${\mathcal{L}}^{-1}\{1/s\}=1$, ek constant hai. Isliye expect karo "ek constant plus $e^{2t}$" — ek horizontal offset ke saath ek growing exponential.

1. Split karo. $\frac{s+4}{s(s-2)}=\frac{A}{s}+\frac{B}{s-2}$. Yeh step kyun? $s=s-0$ bilkul ek ordinary linear factor hai; $0$ ko kisi bhi doosre pole ki tarah treat karo.
2. $A$ ke liye cover-up ($0$ par pole). $s$ se multiply karo, $s=0$ set karo: $A=\frac{0+4}{0-2}=-2$. Yeh step kyun? Origin par residue isolate karta hai.
3. $B$ ke liye cover-up. $(s-2)$ se multiply karo, $s=2$ set karo: $B=\frac{2+4}{2}=3$. Yeh step kyun? Doosre pole par residue.
4. Invert karo. ${\mathcal{L}}^{-1}\{1/s\}=1$ aur ${\mathcal{L}}^{-1}\{1/(s-2)\}=e^{2t}$: $f(t)=-2+3e^{2t}.$

Verify: $f(0)=-2+3=1$. Initial-value check: ${\lim }_{s\to \infty }sF(s)=\lim \frac{s^2+4s}{s^2-2s}=1$. ✓

---

Ex 3 — Cell R1 · repeated real pole (cubed)

Forecast: Ek triple pole. Parent note ne warning di thi: ek factor $(s-a{)}^k$ $k$ terms contribute karta hai, har power ke liye ek. Isliye expect karo $e^{-t}$ times $t$ mein ek quadratic — $t$ ki powers precisely repetition ki wajah se aati hain.

1. Teeno powers mein split karo. $\frac{s^2}{(s+1{)}^3}=\frac{A}{s+1}+\frac{B}{(s+1{)}^2}+\frac{C}{(s+1{)}^3}$. Yeh step kyun? Sirf highest power se underfitting karna real information kho deta hai; numerator degree $<3$ ka koi bhi polynomial ho sakta hai.
2. Denominators clear karo. $(s+1{)}^3$ se multiply karo: $s^2=A(s+1{)}^2+B(s+1)+C$. Yeh step kyun? Fraction-matching ko polynomial-matching mein badal deta hai, jise hum coefficient by coefficient compare kar sakte hain.
3. $C$ ke liye $s=-1$ substitute karo. $(-1{)}^2=C\Rightarrow C=1$. Yeh step kyun? Highest power par cover-up $A$ aur $B$ ko ek saath khatam kar deta hai.
4. Expand aur match karo. $A(s+1{)}^2+B(s+1)+C=A{s}^2+(2A+B)s+(A+B+C)$. Match $s^2$: $A=1$. Match $s^1$: $2A+B=0\Rightarrow B=-2$. (Constant check: $A+B+C=1-2+1=0=$ coeff of $s^0$ in $s^2$. ✓) Yeh step kyun? Like powers compare karna remaining coefficients ke liye ek neat system deta hai.
5. Power rule + shift ke saath invert karo. ${\mathcal{L}}^{-1}\{1/(s+1{)}^{n+1}\}=\frac{t^n}{n!}{e}^{-t}$ use karte hue: $f(t)=e^{-t}\left(1-2t+\frac{1}{2}{t}^2\right).$ Yeh step kyun? $1/(s+1{)}^2\to t{e}^{-t}$ aur $1/(s+1{)}^3\to \frac{t^2}{2}{e}^{-t}$ — $t$ ki extra powers repeated pole ki signature hain.

Verify: $f(0)=1$. Initial-value: ${\lim }_{s\to \infty }s\cdot \frac{s^2}{(s+1{)}^3}=1$. ✓

---

Ex 4 — Cell Q1 · pure oscillation ($a=0$)

Forecast: Denominator $s^2+9=(s^2+{\omega }^2)$ mein koi $s$-shift nahi, isliye koi exponential envelope nahi — frequency $\omega =\sqrt{9}=3$ par pure undying $\sin$ aur $\cos$.

1. $\omega$ read off karo. $s^2+9\Rightarrow {\omega }^2=9\Rightarrow \omega =3$. Yeh step kyun? Table ko $\omega$ chahiye, ${\omega }^2$ nahi; ek classic slip $9$ use karna hai.
2. "Upar kya baitha hai" ke basis par split karo. $\frac{4s+6}{s^2+9}=4\cdot \frac{s}{s^2+9}+2\cdot \frac{3}{s^2+9}$. Yeh step kyun? $s$-on-top → $\cos$; constant upar hona chahiye $\omega =3$ → $\sin$. Hum ne $6=2\cdot 3$ likha taaki woh $\omega$ manufacture ho sake.
3. Invert karo. $f(t)=4\cos 3t+2\sin 3t.$

Verify: $f(0)=4$; initial-value ${\lim }_{s\to \infty }s\frac{4s+6}{s^2+9}=4$. ✓ Ek second check final-value theorem se yahan valid nahi hai (imaginary axis par poles hain) — yeh notice karna achha hai.

---

Ex 5 — Cell Q2 · damped oscillation (shifted quadratic)

Yeh topic ka geometric core hai: completing the square poles ko imaginary axis se slide kar deta hai, aur woh slide time mein ek exponential envelope ban jaata hai.

Forecast: Quadratic ke real roots nahi hain (discriminant $16-52<0$). Completing the square se $(s-2{)}^2+9$ milega: andar ka "$-2$" matlab poles $s=2\pm 3i$ par hain, imaginary axis ke right mein → ek growing $e^{2t}$ envelope times $\omega =3$ par oscillation. Guess: $e^{2t}(\dots \cos 3t+\dots \sin 3t)$.

1. Square complete karo. $s^2-4s+13=(s-2{)}^2+9$, isliye $a=2$, $\omega =3$. Yeh step kyun? Shift table ko $(s-a{)}^2+{\omega }^2$ shape chahiye; tabhi first-shift theorem $e^{at}$ factor de sakta hai.
2. Numerator ko $(s-2)$ ke around rewrite karo. $3s-2=3(s-2)+4$. Yeh step kyun? Hume $\frac{s-a}{(s-a{)}^2+{\omega }^2}$ ($e^{at}\cos$ deta hai) match karna hai. Numerator split karo taaki "shifted $s$" explicitly dikhe; $3(s-2)=3s-6$, aur $-2-(-6)=+4$ leftover constant hai.
3. Do table atoms mein group karo. $\frac{3(s-2)}{(s-2{)}^2+9}+\frac{4}{(s-2{)}^2+9}=3\cdot \frac{s-2}{(s-2{)}^2+9}+\frac{4}{3}\cdot \frac{3}{(s-2{)}^2+9}.$ Yeh step kyun? Sine atom ko upar $\omega =3$ chahiye, isliye $4=\frac{4}{3}\cdot 3$ factor karte hain.
4. Invert karo (first-shift theorem). $f(t)=e^{2t}\left(3\cos 3t+\frac{4}{3}\sin 3t\right).$ Yeh step kyun? Upar state kiye gaye first-shift theorem ke according, denominator mein $(s-2)$ $e^{2t}$ envelope ban jaata hai; upar ki picture poles ko $s=2\pm 3i$ par baithte aur resulting growing wave dikhati hai.

Verify: $f(0)=e^0(3+0)=3$; initial-value ${\lim }_{s\to \infty }s\frac{3s-2}{s^2-4s+13}=3$. ✓

---

Ex 6 — Cell H1 · sign trap ($s^2-b^2$)

Forecast: Dhyan se dekho — sign minus hai: $s^2-1$, $s^2+1$ nahi. $\pm 1$ par real poles, isliye yeh $\sinh /\cosh$ (hyperbolic) hai, $\sin /\cos$ kabhi nahi. Agar tumne yahan $\sin t$ likha toh tum classic trap mein pad gaye.

1. Real linear poles mein factor karo. $s^2-1=(s-1)(s+1)$. Yeh step kyun? Minus sign ka matlab hai roots real hain, isliye hum distinct linear factors ki tarah split karte hain — koi completing-the-square nahi, koi imaginary poles nahi.
2. Split karo. $\frac{2s+1}{(s-1)(s+1)}=\frac{A}{s-1}+\frac{B}{s+1}$. Yeh step kyun? Do distinct real linear factors → do simple-pole atoms, har factor ke liye ek, taaki har ek ko ${\mathcal{L}}^{-1}\{1/(s-a)\}=e^{at}$ entry se invert kiya ja sake.
3. Cover-up. $A=\frac{2(1)+1}{1+1}=\frac{3}{2}$; $B=\frac{2(-1)+1}{-1-1}=\frac{-1}{-2}=\frac{1}{2}$. Yeh step kyun? Do real poles par residues.
4. Invert karo (exponential form — hamesha safe). $f(t)=\frac{3}{2}{e}^t+\frac{1}{2}{e}^{-t}.$ Yeh step kyun? ${\mathcal{L}}^{-1}\{1/(s-a)\}=e^{at}$ har pole par apply kiya. Yeh exponential form final answer hai.

Verify: $f(0)=\frac{3}{2}+\frac{1}{2}=2$; initial-value ${\lim }_{s\to \infty }s\frac{2s+1}{s^2-1}=2$. ✓

---

Ex 7 — Cell D1 · degenerate / improper fraction

Forecast: Top degree $=$ bottom degree $=2$. Yeh improper hai — table directly apply nahi hoti. Expect karo ki leftover $\delta (t)$ (Dirac spike from Heaviside Step & Dirac Delta) plus ek proper oscillation mein invert hoga.

1. Pehle long-divide karo. $\frac{s^2+3s+5}{s^2+4}=1+\frac{3s+1}{s^2+4}$. Yeh step kyun? Jab $\mathrm{deg}P\ge \mathrm{deg}Q$, tumhe polynomial part extract karna hi hoga; table sirf proper fractions invert karti hai.
2. Constant part invert karo. ${\mathcal{L}}^{-1}\{1\}=\delta (t)$. Yeh step kyun? $s$ mein ek pure constant $t=0$ par impulse correspond karta hai — $\delta (t)$ ka transform $1$ hota hai.
3. Proper remainder split karo. $\omega =2$ ke saath: $\frac{3s+1}{s^2+4}=3\cdot \frac{s}{s^2+4}+\frac{1}{2}\cdot \frac{2}{s^2+4}$. Yeh step kyun? $s$-on-top → $\cos 2t$; constant $1=\frac{1}{2}\cdot 2$ → $\omega =2$ par $\sin 2t$.
4. Assemble karo. $f(t)=\delta (t)+3\cos 2t+\frac{1}{2}\sin 2t.$ Yeh step kyun? ${\mathcal{L}}^{-1}$ ki linearity hume har piece alag-alag invert karne aur results simply add karne deti hai — step 2 ka impulse aur step 3 ke do oscillation atoms bina kisi cross-term ke combine hote hain, kyunki ek sum ko invert karna sum ke inverses ka sum hota hai.

Verify: $t>0$ ke liye (impulse ignore karte hue), $f(0^{+})=3$. Initial-value proper part $\frac{3s+1}{s^2+4}$ par apply: ${\lim }_{s\to \infty }s\cdot \frac{3s+1}{s^2+4}=3$. ✓

---

Ex 8 — Cell S1 · time-delayed switch-on (Heaviside)

Forecast: $e^{-2s}$ factor second-shift theorem ki signature hai — yeh kehta hai "kuch nahi hota jab tak $t=2$ nahi aata, phir answer start hota hai." Expect karo ki jo bhi $\frac{1}{s^2+s}$ mein invert hota hai, woh Heaviside step $u(t-2)$ se multiply hoga, jisme $t$ ki jagah $t-2$ hoga.

1. Pehle un-shifted core handle karo. $\frac{1}{s^2+s}=\frac{1}{s(s+1)}=\frac{1}{s}-\frac{1}{s+1}$. Yeh step kyun? $e^{-2s}$ ko mentally peel karo; pehle $g(t)={\mathcal{L}}^{-1}\{1/(s^2+s)\}$ nikalo, kyunki second-shift theorem ko woh base function $g$ chahiye.
2. Core invert karo. $g(t)=1-e^{-t}$. Yeh step kyun? ${\mathcal{L}}^{-1}\{1/s\}=1$ aur ${\mathcal{L}}^{-1}\{1/(s+1)\}=e^{-t}$, linearity ke zariye term-by-term apply kiya.
3. Second-shift theorem $c=2$ ke saath apply karo. $t$ ko $t-2$ se replace karo aur $t=2$ par switch on karo: $f(t)=(1-e^{-(t-2)})u(t-2).$ Yeh step kyun? $e^{-2s}$ factor $g(t)$ ko $2$ se delay karta hai aur Heaviside step $u(t-2)$ se multiply karta hai taaki "$t=2$ se pehle off" enforce ho — yeh precisely upar state kiya gaya second-shift theorem hai.

Verify: $t=2^{+}$ par: $f=1-e^0=0$, isliye yeh continuously $0$ se start hota hai (koi jump nahi). Jaise $t\to \infty$: $f\to 1$. Dono ek switched-on rising exponential ke liye sensible hain.

---

Ex 9 — Cell W1 · real-world word problem

Forecast: Quadratic denominator mein real roots nahi hain → damped oscillation. Complete the square dekho ki woh kitni tez decay karti hai aur kis frequency par wobble karti hai.

1. Square complete karo. $s^2+2s+10=(s+1{)}^2+9$, isliye $a=-1$, $\omega =3$. Yeh step kyun? Andar ka "$+1$" matlab poles $s=-1\pm 3i$ par hain — imaginary axis ke left mein → ek decaying $e^{-t}$ envelope (ek stable, settling system).
2. Sine atom match karo. $\frac{10}{(s+1{)}^2+9}=\frac{10}{3}\cdot \frac{3}{(s+1{)}^2+9}$. Yeh step kyun? Sine atom ko upar $\omega =3$ chahiye; $10=\frac{10}{3}\cdot 3$ factor karo.
3. Invert karo (first-shift theorem). $x(t)=\frac{10}{3}{e}^{-t}\sin 3t.$ Yeh step kyun? First-shift theorem ke according, shifted denominator $(s+1{)}^2+9$ → ek $e^{-t}$ envelope times $\sin 3t$.

Verify (units & physics): $x(0)=0$ (mass rest position se start karta hai — ek impulse nudge se match karta hai). Envelope $\frac{10}{3}{e}^{-t}$ shrink hoti hai → oscillations khatam hote hain → stable system, exactly wahi jo ek real damper karta hai. Frequency $\omega =3$ rad/s. Initial-value: ${\lim }_{s\to \infty }sX(s)=0$. ✓

---

Ex 10 — Cell X1 · exam twist (linear × quadratic mix)

Forecast: Ek real pole ($s=-1$) aur ek irreducible quadratic ($s^2+4$). Expect karo ek blend: ek $e^{-t}$ term plus ek undamped $\cos 2t/\sin 2t$ pair.

1. Mixed decomposition set up karo. $\frac{2s^2+s+3}{(s+1)(s^2+4)}=\frac{A}{s+1}+\frac{Bs+C}{s^2+4}$. Yeh step kyun? Linear factor → constant top; irreducible quadratic → ek full linear top $Bs+C$ (isko $\cos$ aur $\sin$ dono piece ki zaroorat ho sakti hai).
2. $A$ ke liye cover-up. $(s+1)$ se multiply karo, $s=-1$ set karo: $A=\frac{2(1)+(-1)+3}{1+4}=\frac{4}{5}$. Yeh step kyun? Real-pole residue ko saaf se isolate karta hai.
3. Denominators clear karo & match karo. $2s^2+s+3=A(s^2+4)+(Bs+C)(s+1)$. Match $s^2$: $2=A+B\Rightarrow B=2-\frac{4}{5}=\frac{6}{5}$. Match constant: $3=4A+C\Rightarrow C=3-\frac{16}{5}=-\frac{1}{5}$. (Check $s^1$: $1=B+C=\frac{6}{5}-\frac{1}{5}=1$ ✓.) Yeh step kyun? Powers compare karna $A$ ke known hone ke baad $B,C$ pin down karta hai.
4. Quadratic atom split karo & invert karo. $\omega =2$ ke saath: $\frac{\frac{6}{5}s-\frac{1}{5}}{s^2+4}=\frac{6}{5}\cdot \frac{s}{s^2+4}-\frac{1}{10}\cdot \frac{2}{s^2+4}.$ Isliye $f(t)=\frac{4}{5}{e}^{-t}+\frac{6}{5}\cos 2t-\frac{1}{10}\sin 2t.$ Yeh step kyun? $s$-on-top → $\cos 2t$; constant $-\frac{1}{5}=-\frac{1}{10}\cdot 2$ → $\sin 2t$.

Verify: $f(0)=\frac{4}{5}+\frac{6}{5}+0=\frac{10}{5}=2$. Initial-value: ${\lim }_{s\to \infty }s\frac{2s^2+s+3}{(s+1)(s^2+4)}=2$. ✓

---

Ex 11 — Cell RQ1 · repeated irreducible quadratic

Forecast: Quadratic $s^2+4$ squared hai — yeh repeated real pole ka complex-pole analogue hai. Jaise ek repeated real pole ne $t$ ka ek factor diya tha (Ex 3), ek repeated complex pole ek oscillation times $t$ ka factor deta hai. Isliye expect karo kuch jisme $t\sin 2t$ ya $t\cos 2t$ ho.

1. $\omega$ identify karo aur sahi table atom pick karo. Yahan $\omega =2$. Repeated complex poles ka standard result hai ${\mathcal{L}}^{-1}\left\{\frac{2{\omega }^3}{(s^2+{\omega }^2{)}^2}\right\}=\sin \omega t-\omega t\cos \omega t,{\mathcal{L}}^{-1}\left\{\frac{2\omega s}{(s^2+{\omega }^2{)}^2}\right\}=t\sin \omega t.$ Yeh step kyun? Yeh do atoms hain jinmein $(s^2+{\omega }^2{)}^2$ shape decompose hoti hai; $t$-factor repeated (squared) pole ka fingerprint hai, bilkul Ex 3 ke parallel.
2. Apna fraction pehle atom se match karo. Hume $\frac{1}{(s^2+4{)}^2}$ chahiye. $\omega =2$ ke saath, $2{\omega }^3=2\cdot 8=16$, isliye $\frac{1}{(s^2+4{)}^2}=\frac{1}{16}\cdot \frac{16}{(s^2+4{)}^2}=\frac{1}{16}\cdot \frac{2{\omega }^3}{(s^2+{\omega }^2{)}^2}.$ Yeh step kyun? Humara numerator $1$ ko atom ke numerator $2{\omega }^3=16$ tak scale karo $\frac{1}{16}$ pull out karke.
3. Invert karo. $f(t)=\frac{1}{16}(\sin 2t-2t\cos 2t).$ Yeh step kyun? $\omega =2$ ke saath pehle atom ka direct read-off, phir humara $\frac{1}{16}$ multiply karo.

Verify: $f(0)=\frac{1}{16}(0-0)=0$. Initial-value: ${\lim }_{s\to \infty }s\frac{1}{(s^2+4{)}^2}=0$ ✓. $-2t\cos 2t$ term woh tell-tale growing-in-amplitude oscillation hai jo repeated complex poles hamesha produce karte hain (resonance!).

---

Recall Self-test — har shape kis cell mein jaata hai?

$\frac{1}{s^2-9}$ kis cell mein hai? ::: H1 — real poles $\pm 3$, isliye $\sinh /\cosh$, $\sin$ nahi. $\frac{5}{(s-2{)}^3}$ kis cell mein hai? ::: R1 — repeated pole, $\frac{5}{2}{t}^2{e}^{2t}$ deta hai. $\frac{s^3}{s^2+1}$ kis cell mein hai? ::: D1 — improper, pehle long-divide karo ($\delta$ ke derivative ka term deta hai). $\frac{e^{-3s}}{s}$ kis cell mein hai? ::: S1 — delayed step $u(t-3)$. $\frac{s+1}{s^2+2s+2}$ kis cell mein hai? ::: Q2 — square complete karo $(s+1{)}^2+1$ mein, $e^{-t}\cos t$ deta hai. $\frac{s}{(s^2+1{)}^2}$ kis cell mein hai? ::: RQ1 — repeated complex pole, $\frac{1}{2}t\sin t$ deta hai.

---

Parent par wapas: Inverse Laplace transform — partial fractions & tables. Dekhe jaane layak prerequisites: Partial Fraction Decomposition (Algebra), Laplace Transform — definition and existence, Convolution Theorem.