Laplace of derivatives — key property for solving ODEs: Worked examples

4.6.28 · D3 · Maths › Ordinary Differential Equations › Laplace of derivatives — key property for solving ODEs

Is page mein parent property ka derivative rule har tarah ke case ke against test kiya gaya hai. Pehle matrix padho, phir har worked example batata hai ki woh exactly kis cell mein fit hota hai.

Scenario matrix

Yahan $s$ Laplace variable hai, $Y = L {y}$ unknown ka transform hai, aur "IC" ka matlab hai initial condition (kisi chosen time par $y$ ya uske derivative ki value). Har row ek alag case class hai; last column us example ka naam deta hai jo use cover karta hai.

Case class	Ise alag kya banata hai	Covered by
1st order, nonzero IC	$- f (0)$ ka fee actually nonzero hai	Ex 1
1st order, zero IC + forcing	fee zero ho jaata hai, forcing $\frac{1}{s}$ enter karta hai	Ex 2
2nd order, two real roots ( $s^{2} - a^{2}$ )	denominator $(s - a) (s + a)$ mein factor hota hai	Ex 3
2nd order, complex roots (oscillation)	denominator $s^{2} + ω^{2}$ → sin/cos	Ex 4
2nd order, repeated root (critical)	denominator $(s + a)^{2}$ → $t e^{- a t}$	Ex 5
Degenerate: $n = 0$ / constant / pure decay limit	check karo ki rule edges par toot toh nahi raha	Ex 6
Word problem (real units)	RC circuit / cooling, physical meaning	Ex 7
Exam twist: IC $t = 0$ par nahi diya	formula ki boundary $0$ par baithi hai — shift karna padega	Ex 8

Neeche jo bhi numeric answer aata hai woh verify block mein machine-checked hai.

Worked example Ex 4 — 2nd order, complex roots (pure oscillation) ·

Case 4 Solve karo $y^{''} + 4 y = 0, y (0) = 3, y^{'} (0) = 0$ .

Forecast: $s^{2} + 4$ reals ke upar kabhi factor nahi hota — roots hain $\pm 2 i$ , matlab angular frequency $2$ par oscillation. $3$ se start, zero slope ke saath → pure cosine: guess karo $y = 3 cos 2 t$ .

Transform karo. $(s^{2} Y - sy (0) - y^{'} (0)) + 4 Y = 0$ .
ICs insert karo. $y (0) = 3, y^{'} (0) = 0$ : $s^{2} Y - 3 s - 0 + 4 Y = 0 \Rightarrow (s^{2} + 4) Y = 3 s$ . Yeh step kyun? Dhyaan do ki $y (0)$ ne numerator mein ek $s$ produce kiya — wahi $s$ sine ke bajaye cosine banata hai.
Solve karo. $Y = \frac{3 s}{s ^{2} + 4}$ .
Invert karo. $L {cos ω t} = \frac{s}{s ^{2} + ω ^{2}}$ with $ω = 2$ , toh $y (t) = 3 cos 2 t$ .

Verify: $y^{'} = - 6 sin 2 t$ , $y^{''} = - 12 cos 2 t$ ; $y^{''} + 4 y = - 12 cos 2 t + 12 cos 2 t = 0$ ✓, $y (0) = 3$ ✓, $y^{'} (0) = 0$ ✓.

Chalk-blue solution curve dekho: woh height $3$ se start hoti hai flat top ke saath (slope zero, $y^{'} (0) = 0$ se match karta hua) aur symmetrically swing karti hai — woh flat start pure cosine ka visual signature hai.

Worked example Ex 5 — 2nd order, repeated root (critical damping) ·

Case 5 Solve karo $y^{''} + 2 y^{'} + y = 0, y (0) = 0, y^{'} (0) = 1$ .

Forecast: Characteristic-style denominator hoga $s^{2} + 2 s + 1 = (s + 1)^{2}$ — ek repeated root $- 1$ . Repeated roots ek $t e^{- t}$ factor produce karte hain. Guess karo $y = t e^{- t}$ .

Transform karo. $(s^{2} Y - sy (0) - y^{'} (0)) + 2 (s Y - y (0)) + Y = 0$ . Yeh step kyun? $y^{''}$ par 2nd-derivative rule aur $y^{'}$ par 1st-derivative rule apply karo — property dono handle karti hai.
ICs insert karo. $y (0) = 0, y^{'} (0) = 1$ : $(s^{2} Y - 1) + 2 s Y + Y = 0 \Rightarrow (s^{2} + 2 s + 1) Y = 1$ .
Solve karo. $Y = \frac{1}{( s + 1 ) ^{2}}$ .
Invert karo. $L {t e^{a t}} = \frac{1}{( s - a ) ^{2}}$ with $a = - 1$ , toh $y (t) = t e^{- t}$ . Yeh step kyun? Squared denominator bilkul $t e^{a t}$ shape ka fingerprint hai — yeh woh case hai jisme koi plain exponential nahi hota.

Verify: $y^{'} = e^{- t} - t e^{- t}$ , $y^{''} = - 2 e^{- t} + t e^{- t}$ ; $y^{''} + 2 y^{'} + y = (- 2 e^{- t} + t e^{- t}) + 2 (e^{- t} - t e^{- t}) + t e^{- t} = 0$ ✓, $y (0) = 0$ ✓, $y^{'} (0) = 1$ ✓.

Worked example Ex 6 — Degenerate / edge cases ·

Case 6 Rule ki boundaries par teen sanity checks.

(a) $n = 0$ (koi derivative nahi). Convention ke hisaab se $f^{(0)} = f$ , aur general formula ki subtracted terms ki list empty hai, toh $L {f} = s^{0} F (s) = F (s)$ . Yeh kyun matter karta hai: pattern collapse nahi hota — yeh sirf identity mein reduce ho jaata hai, jo confirm karta hai ki descending-power list $s^{n - 1}$ se start hoti hai aur $n = 0$ contribute karne se pehle ruk jaati hai.

(b) Constant function $y \equiv c$ . Iska derivative $0$ hai, toh $L {y^{'}} = 0$ . Rule kehta hai $s Y - y (0) = s \cdot \frac{c}{s} - c = c - c = 0$ ✓ — fee exactly $s F (s)$ ko cancel kar deta hai. Yeh step kyun? Ek constant ki rate of change zero hoti hai; property ko $0$ return karna chahiye, aur woh karta hai.

(c) Ex 2 ka limiting behaviour. Jab $t \to \infty$ , $y = 2 (1 - e^{- 4 t}) \to 2$ ; jab $t \to 0^{+}$ , $y \to 0$ . Yeh Final Value aur Initial Value behaviour hai: $lim_{s \to 0} s Y = lim_{s \to 0} \frac{8}{s + 4} = 2$ aur $lim_{s \to \infty} s Y = lim_{s \to \infty} \frac{8}{s + 4} = 0$ . Yeh step kyun? $Y$ ko $s$ se multiply karna aur $s$ limits lena bina invert kiye $t$ -endpoints read off kar deta hai — ek free consistency check.

Worked example Ex 7 — Word problem with units ·

Case 7 Ek capacitor $C = 0.5 F$ ek resistor $R = 4 Ω$ se discharge hota hai. Voltage $v (t)$ yeh obey karta hai: $R C v^{'} + v = 0$ with $v (0) = 12 V$ . $v (t)$ nikalo aur woh time nikalo jab $12/ e V$ tak pahunche.

Forecast: $R C = 4 \cdot 0.5 = 2 s$ time constant $τ$ hai. Exponential decay: guess karo $v = 12 e^{- t /2}$ , $t = τ = 2 s$ par $12/ e$ tak pahunchega.

ODE cleanly likho. $2 v^{'} + v = 0 \Rightarrow v^{'} + \frac{1}{2} v = 0$ , $v (0) = 12$ . Yeh step kyun? $R C$ se divide karne par standard 1st-order form milta hai; coefficient $\frac{1}{2}$ yaani $1/ τ$ hai.
Transform karo. $(s V - 12) + \frac{1}{2} V = 0 \Rightarrow (s + \frac{1}{2}) V = 12$ .
Solve karo. $V = \frac{12}{s + \frac{1}{2}}$ .
Invert karo. $v (t) = 12 e^{- t /2} V$ .
$12/ e$ tak pahuncho. Set karo $12 e^{- t /2} = 12/ e \Rightarrow e^{- t /2} = e^{- 1} \Rightarrow t = 2 s$ .

Verify: units — $s$ carries $1/ s$ , $V$ carries $V \cdot s$ , toh $v$ carries $V$ ✓. $t = 2$ par: $v = 12 e^{- 1} \approx 4.415 V$ ✓, aur $12/ e \approx 4.415$ ✓.

Worked example Ex 8 — Exam twist: IC

$t = 0$ par nahi · Case 8 Solve karo $y^{'} - y = 0$ with yeh condition zero se door diya gaya: $y (1) = e$ .

Forecast: Property ka boundary term exactly $t = 0$ par rehta hai, toh hum $y (1)$ directly plug-in nahi kar sakte. Ek unknown $y (0) = c$ ke saath solve karo, phir $y (1) = e$ se $c$ fit karo. Equation $y^{'} = y$ deta hai $y = c e^{t}$ , aur $y (1) = ce = e \Rightarrow c = 1$ , toh $y = e^{t}$ .

Unknown start ke saath transform karo. Maano $y (0) = c$ . $(s Y - c) - Y = 0 \Rightarrow (s - 1) Y = c$ . Yeh step kyun? Hum honestly rule ke andar $y (1)$ use nahi kar sakte; hum true $t = 0$ value ko symbol $c$ ki tarah carry karte hain.
Solve karo. $Y = \frac{c}{s - 1}$ .
Invert karo. $y (t) = c e^{t}$ .
Di gayi condition fit karo. $y (1) = c e^{1} = e \Rightarrow c = 1$ . Is tarah $y (t) = e^{t}$ . Yeh step kyun? Yeh parent mein steel-manned mistake ka sahi handling hai: $0$ se door ka IC solve karne ke baad generally fit karna chahiye.

Verify: $y^{'} = e^{t} = y$ ✓, aur $y (1) = e^{1} = e$ ✓.

Recall Kaun sa cell kaun sa tha?

Ex1 nonzero-IC first order ko hit karta hai ::: $y^{'} + 2 y = 0 \Rightarrow y = 5 e^{- 2 t}$ . Ex4 vs Ex3 — $s$ mein fark kaise dekhte hain? ::: $s^{2} + ω^{2}$ (complex roots → oscillation) vs $s^{2} - a^{2}$ (real roots → $sinh / cosh$ ). Ex5 ka $s$ -domain mein fingerprint? ::: Squared denominator $(s + a)^{2}$ → ek $t e^{- a t}$ term. Ex8 rule mein $y (1)$ directly kyun plug-in nahi kar sakte? ::: Derivative property apna boundary term $t = 0$ par evaluate karta hai, toh IC $0$ par honi chahiye; warna generally solve karo aur fit karo.

Connections

4.6.28 Laplace of derivatives — key property for solving ODEs (Hinglish)
Solving Initial Value Problems with Laplace
Inverse Laplace Transform and Partial Fractions
Laplace Transform — Definition and Existence
Linearity of the Laplace Transform
Integration by Parts
Exponential Order and Convergence of Integrals
Fourier Transform — comparison (no boundary term)