Solving ODEs with Laplace (including discontinuous forcing): Worked examples

4.6.30 · D3 · Maths › Ordinary Differential Equations › Solving ODEs with Laplace (including discontinuous forcing)

Scenario matrix

Har linear-ODE-with-Laplace problem ko ek grid mein ek point ki tarah socho. Do axes hain jo decide karti hain ki algebra kitna mushkil hoga:

$Y (s)$ ka denominator kaisa dikhta hai — alag real roots, ek repeated root, ya ek complex pair (jo oscillation ban jaata hai).
Forcing kya karti hai — kuch nahi (homogeneous), ek smooth push, ek single switch $u (t - a)$ , ek pulse (switch on phir off), ya ek hammer-blow $δ (t - a)$ .

Neeche, har cell ek class of behaviour hai. Har cell ko kam se kam ek worked example se cover kiya gaya hai.

Cell	$Y (s)$ ka denominator type	Forcing type	Behaviour jo tumhe pehchanni chahiye	Example
A	distinct real roots	smooth (ya koi nahi)	pure decaying exponentials	Ex 1
B	complex pair $s^{2} + ω^{2}$	koi nahi	pure oscillation	Ex 2
C	complex pair, root = forcing frequency	smooth sinusoid	resonance, amplitude $t$ ki tarah badhti hai	Ex 3
D	distinct real roots	single switch $u (t - a)$	gated relaxation, delay tag $e^{- a s}$	Ex 4
E	distinct real roots	pulse $u (t - a) - u (t - b)$	on-then-off, do delay tags	Ex 5
F	forcing ko $g (t - a)$ ki tarah rewrite karna padega	ramp $\times$ step	shifting trap $t = (t - a) + a$	Ex 6
G	distinct real roots	Dirac $δ (t - a)$	instant kick, koi $1/ s$ nahi	Ex 7
H	repeated real root	smooth	$t e^{r t}$ factor wale terms	Ex 8

Figure — Solving ODEs with Laplace (including discontinuous forcing)

Cell A — distinct real roots, smooth forcing

Worked example Example 1 ·

$y^{''} + 3 y^{'} + 2 y = e^{- 3 t}$ , $y (0) = 0, y^{'} (0) = 0$ Forecast: kaam shuru karne se pehle andaza lagao. Homogeneous roots $s^{2} + 3 s + 2 = (s + 1) (s + 2)$ se aate hain, toh $e^{- t}$ aur $e^{- 2 t}$ . Forcing ek $e^{- 3 t}$ add karti hai. Teen decaying exponentials ka sum expect karo, sab $\to 0$ .

Step 1 — Stamp. Dono ICs zero hain, toh $L {y^{''}} = s^{2} Y$ , $L {y^{'}} = s Y$ , aur $L {e^{- 3 t}} = \frac{1}{s + 3}$ : $(s^{2} + 3 s + 2) Y = \frac{1}{s + 3} .$ Ye step kyun? Recipe demand karti hai ki hum term-by-term transform karein; zero ICs matlab koi boundary terms survive nahi karti.

Step 2 — Solve. Divide karo (pure algebra): $Y = \frac{1}{( s + 1 ) ( s + 2 ) ( s + 3 )} .$ Ye step kyun? $Y (s)$ ab ek rational function hai — Partial Fractions ke liye ready.

Step 3 — Stick-back. Cover-up se split karo (factor se multiply karo, $s$ ko uski root par set karo): $\frac{1}{( s + 1 ) ( s + 2 ) ( s + 3 )} = \frac{A}{s + 1} + \frac{B}{s + 2} + \frac{C}{s + 3},$ $A = \frac{1}{( 1 ) ( 2 )} = \frac{1}{2}, B = \frac{1}{( - 1 ) ( 1 )} = - 1, C = \frac{1}{( - 2 ) ( - 1 )} = \frac{1}{2} .$ Har ek ko table $\frac{1}{s - r} \leftrightarrow e^{r t}$ se padho: $y (t) = \frac{1}{2} e^{- t} - e^{- 2 t} + \frac{1}{2} e^{- 3 t}$ Cover-up kyun kaam karta hai: dono sides ko $(s + 1)$ se multiply karke $s = - 1$ set karo — baaki har term mein abhi bhi $(s + 1)$ factor hai toh sab zero ho jaate hain.

Verify karo: $t = 0$ par, $y (0) = \frac{1}{2} - 1 + \frac{1}{2} = 0$ ✔. Differentiate karo: $y^{'} = - \frac{1}{2} e^{- t} + 2 e^{- 2 t} - \frac{3}{2} e^{- 3 t}$ , toh $y^{'} (0) = - \frac{1}{2} + 2 - \frac{3}{2} = 0$ ✔. Dono ICs recover ho gaye.

Cell B — complex pair, no forcing (pure oscillation)

Worked example Example 2 ·

$y^{''} + 4 y = 0$ , $y (0) = 3, y^{'} (0) = - 2$ Forecast: $s^{2} + 4$ ki roots $\pm 2 i$ hain → $ω = 2$ par oscillation. Non-zero ICs dono cosine (position se) aur sine (velocity se) feed karte hain. Amplitude hamesha constant rahega (koi decay nahi).

Step 1 — Stamp. Ab ICs zero nahi hain, toh har boundary term rakho: $(s^{2} Y - s \cdot 3 - (- 2)) + 4 Y = 0.$ Ye step kyun? $L {y^{''}} = s^{2} Y - s y (0) - y^{'} (0)$ — inhe drop karna classic galti hai.

Step 2 — Solve. $(s^{2} + 4) Y = 3 s - 2 \Rightarrow Y = \frac{3 s - 2}{s ^{2} + 4} .$

Step 3 — Stick-back. Do table entries se match karo: $\frac{s}{s ^{2} + ω ^{2}} \leftrightarrow cos ω t$ aur $\frac{ω}{s ^{2} + ω ^{2}} \leftrightarrow sin ω t$ , $ω = 2$ ke saath: $Y = 3 \cdot \frac{s}{s ^{2} + 4} - 1 \cdot \frac{2}{s ^{2} + 4} .$ $- 2$ ko $- 1 \cdot 2$ kyun likha? Sine table entry ke liye numerator bilkul $ω = 2$ chahiye; hum factor karke woh visible banate hain. $y (t) = 3 cos 2 t - sin 2 t$ Verify karo: $y (0) = 3 \cdot 1 - 0 = 3$ ✔. $y^{'} = - 6 sin 2 t - 2 cos 2 t$ , toh $y^{'} (0) = - 2$ ✔.

Cell C — resonance (forcing frequency = natural frequency)

Worked example Example 3 ·

$y^{''} + y = cos t$ , $y (0) = 0, y^{'} (0) = 0$ Forecast: natural frequency $ω = 1$ ( $s^{2} + 1$ se), forcing frequency bhi $1$ . Resonance! Expect karo ki amplitude $t$ ki tarah badhe.

Step 1 — Stamp. $L {cos t} = \frac{s}{s ^{2} + 1}$ : $(s^{2} + 1) Y = \frac{s}{s ^{2} + 1} .$

Step 2 — Solve. $Y = \frac{s}{( s ^{2} + 1 ) ^{2}} .$ Ye alag kyun hai: factor $s^{2} + 1$ do baar aata hai. Plain $cos t, sin t$ ka koi sum yeh produce nahi kar sakta; humein growing pair chahiye.

Step 3 — Stick-back. Pair yaad karo (derivable hai, lekin yaad karo): $L {t sin ω t} = \frac{2 ω s}{( s ^{2} + ω ^{2} ) ^{2}} .$ $ω = 1$ ke saath: $L {t sin t} = \frac{2 s}{( s ^{2} + 1 ) ^{2}}$ , toh $\frac{s}{( s ^{2} + 1 ) ^{2}} = \frac{1}{2} L {t sin t}$ . $y (t) = \frac{1}{2} t sin t$ Aage ka $t$ growing envelope hai — resonance ki physical pehchan.

Verify karo: $y (0) = 0$ ✔. $y^{'} = \frac{1}{2} sin t + \frac{1}{2} t cos t$ , $y^{'} (0) = 0$ ✔. ODE mein plug karo: $y^{''} = cos t - \frac{1}{2} t sin t$ , toh $y^{''} + y = cos t - \frac{1}{2} t sin t + \frac{1}{2} t sin t = cos t$ ✔.

Cell D — distinct real roots, single switch

Worked example Example 4 ·

$y^{'} + 2 y = 3 u (t - 1)$ , $y (0) = 0$ Forecast: $t = 1$ tak kuch nahi; phir system apni steady value $3/2$ ki taraf relax karta hai ( $y^{'} = 0$ set karo: $2 y = 3$ ). Root $- 2$ → time constant $\frac{1}{2}$ .

Step 1 — Stamp. $L {u (t - 1)} = \frac{e ^{- s}}{s}$ : $(s + 2) Y = \frac{3 e ^{- s}}{s} .$

Step 2 — Solve. $Y = e^{- s} \cdot H (s) \frac{3}{s ( s + 2 )} .$ $e^{- s}$ ko factor out kyun kiya? Delay tag algebra ke through untouched ride karta hai; pehle plain $H (s)$ invert karo, phir delay-and-gate karo.

Step 3 — Stick-back. $H$ par cover-up: $s = 0$ par residue $\frac{3}{2}$ ; $s = - 2$ par residue $\frac{3}{- 2} = - \frac{3}{2}$ : $H (s) = \frac{3/2}{s} - \frac{3/2}{s + 2} \Rightarrow h (t) = \frac{3}{2} - \frac{3}{2} e^{- 2 t} .$ Phir $Y = e^{- s} H (s)$ ka matlab hai $h$ ko $1$ se delay karo AUR $u (t - 1)$ se gate karo: $y (t) = u (t - 1) [\frac{3}{2} - \frac{3}{2} e^{- 2 (t - 1)}]$ Verify karo: $t < 1$ ke liye, $y = 0$ ✔. Jab $t \to \infty$ , $y \to \frac{3}{2}$ = steady state ✔. $t = 1$ ke thoda baad: $y (1^{+}) = \frac{3}{2} - \frac{3}{2} = 0$ , continuous ✔.

Cell E — ek pulse (on phir off)

Cell F — shifting trap (pehle forcing rewrite karo)

Worked example Example 6 ·

$y^{'} + y = t u (t - 2)$ , $y (0) = 0$ Forecast: $t = 2$ tak silent; phir ramp se driven. Kyunki switch par ramp ki value already $2$ hai, zero slope se start hone ki bajay jump-like build-up expect karo.

Step 1 — Forcing rewrite karo. $τ = t - 2$ lo, toh $t = τ + 2$ : $t u (t - 2) = [(t - 2) + 2] u (t - 2) = g_{1} (t - 2) u (t - 2) + g_{2} 2 u (t - 2) .$ Ye step kyun? Dono pieces ab genuine functions of $(t - 2)$ hain, toh shift theorem cleanly apply hota hai. $L {(t - 2) u (t - 2)} = e^{- 2 s} \cdot \frac{1}{s ^{2}}, L {2 u (t - 2)} = 2 \cdot \frac{e ^{- 2 s}}{s} .$ Toh $L {t u (t - 2)} = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s})$ .

Step 2 — Stamp & Solve. $(s + 1) Y = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s}) \Rightarrow Y = e^{- 2 s} H (s) \frac{2 s + 1}{s ^{2} ( s + 1 )} .$ (Combine karo: $\frac{1}{s ^{2}} + \frac{2}{s} = \frac{1 + 2 s}{s ^{2}}$ , phir $\div (s + 1)$ .)

Step 3 — Stick-back. Partial-fraction $H (s) = \frac{2 s + 1}{s ^{2} ( s + 1 )} = \frac{A}{s} + \frac{B}{s ^{2}} + \frac{C}{s + 1}$ . Denominators clear karo: $2 s + 1 = A s (s + 1) + B (s + 1) + C s^{2}$ .

$s = 0$ : $1 = B$ .
$s = - 1$ : $- 1 = C$ .
$s^{2}$ coefficient compare karo: $0 = A + C \Rightarrow A = 1$ .

Toh $H (s) = \frac{1}{s} + \frac{1}{s ^{2}} - \frac{1}{s + 1} \Rightarrow h (t) = 1 + t - e^{- t}$ . $2$ se delay-and-gate karo: $y (t) = u (t - 2) [1 + (t - 2) - e^{- (t - 2)}] = u (t - 2) [t - 1 - e^{- (t - 2)}]$ Verify karo: $t < 2$ ke liye, $y = 0$ ✔. $t = 2^{+}$ par: $y = 1 + 0 - 1 = 0$ , continuous ✔. $t > 2$ ke liye ODE check karo: $y = t - 1 - e^{- (t - 2)}$ , $y^{'} = 1 + e^{- (t - 2)}$ , toh $y^{'} + y = (1 + e^{- (t - 2)}) + (t - 1 - e^{- (t - 2)}) = t$ ✔ (forcing $t$ se match karta hai).

Cell G — Dirac impulse (ek hammer blow)

Worked example Example 7 ·

$y^{''} + 2 y^{'} + 2 y = δ (t - 1)$ , $y (0) = 0, y^{'} (0) = 0$ Forecast: rest par rakha system $t = 1$ par kick hota hai. $s^{2} + 2 s + 2$ ki roots: complete the square $(s + 1)^{2} + 1$ → $- 1 \pm i$ → decaying oscillation $e^{- t}$ envelope, $ω = 1$ . Toh $t = 1$ ke baad ek damped wiggle expect karo.

Step 1 — Stamp. $L {δ (t - 1)} = e^{- s}$ : $(s^{2} + 2 s + 2) Y = e^{- s} .$

Step 2 — Solve. $Y = e^{- s} H (s) \frac{1}{( s + 1 ) ^{2} + 1} .$

Step 3 — Stick-back. Shifted-sine table entry $L {e^{- t} sin t} = \frac{1}{( s + 1 ) ^{2} + 1}$ se match karo (yeh standard $\frac{ω}{( s - α ) ^{2} + ω ^{2}}$ hai $α = - 1, ω = 1$ ke saath). Toh $h (t) = e^{- t} sin t$ . $1$ se delay-and-gate karo: $y (t) = u (t - 1) e^{- (t - 1)} sin (t - 1)$ Verify karo: $t < 1$ ke liye, $y = 0$ ✔. Kick ke thoda baad, $y (1^{+}) = 0$ lekin velocity jump karti hai: $y^{'} (1^{+}) = e^{0} (cos 0) = 1$ — unit impulse ek unit velocity inject karta hai ek aisa system mein jo rest se shuru ho raha ho ✔ (consistent hai $δ$ ke ek unit "kick" deliver karne ke saath).

Cell H — repeated real root

Poori matrix ek picture mein

Recall Quick self-test

Kis cell mein linearly-growing amplitude hoti hai aur kyun? ::: Cell C (resonance): forcing frequency ek natural frequency ke barabar hai, jo ek repeated complex factor $(s^{2} + ω^{2})^{2}$ deta hai → ek $t sin ω t$ term. $t u (t - 2)$ ko shift theorem se pehle rewrite kyun karna padta hai? ::: Theorem ko $(t - 2)$ ka function chahiye; $t = (t - 2) + 2$ likho taaki dono pieces shifted hon, jo $e^{- 2 s} (1/ s^{2} + 2/ s)$ deta hai. $L {δ (t - a)}$ kya hai aur impulse forcing aasaan kyun hai? ::: $e^{- a s}$ — bas bare delay tag, koi $1/ s$ nahi, toh yeh directly transfer factor se multiply hota hai. Answer mein repeated real root $r$ ki pehchan kya hai? ::: Ek factor $t e^{r t}$ (double root) ya $t^{2} e^{r t}$ (triple), $L {t^{n} e^{r t}} = n! / (s - r)^{n + 1}$ se.