Solving ODEs with Laplace (including discontinuous forcing): Exercises

4.6.30 · D4 · Maths › Ordinary Differential Equations › Solving ODEs with Laplace (including discontinuous forcing)

Shuru karne se pehle, ek shared toolbox — har woh symbol jo tum dekhoge, plain words mein:

Definition Woh symbols jo hum use karte hain (koi cheez bina explain ke nahi aayegi)

$t$ time hai, $0$ se upar ki taraf jaata hai. Hamara unknown answer ek function $y (t)$ hai: "system $t$ time pe kahan hai."
$s$ woh algebra-world variable hai jisme Laplace transform hamen bhejta hai. Isse ek knob ki tarah socho; bada $s$ matlab "shuru ke, fast behaviour pe focus karo."
$L {y (t)} = Y (s)$ ka matlab hai " $y$ ka transform." Capital letter = transformed (s-land mein); small letter = original (t-land mein).
$y^{'}$ woh slope hai $y$ ka (rate of change); $y^{''}$ slope ka slope hai (acceleration).
$u (t - a)$ ek switch hai: time $a$ se pehle $0$ , phir $1$ pe jump karta hai aur waheen rehta hai. Socho ek light jo $t = a$ pe on ki gayi.
$δ (t - a)$ ek hammer blow hai: bilkul $t = a$ pe deliver kiya gaya ek infinitely sharp kick.
$e^{- a s}$ ek delay tag hai: jab bhi yeh s-land mein kisi cheez ko multiply karta hai, matlab hai "t-land mein, is piece ko $a$ seconds baad start karo."

Teen rules jo hum baar baar reuse karte hain:

Level 1 — Recognition

L1.1 $L {y^{''} - 4 y^{'} + 3 y}$ ko general initial conditions $y (0) = y_{0}$ , $y^{'} (0) = y_{1}$ ke saath $Y = Y (s)$ ke terms mein likho.

Recall Solution L1.1

Term by term transform karo. Hum kya karte hain: har derivative ko uske rule se replace karo. Kyun: transform linear hai, isliye yeh $+$ aur constants se guzar jaata hai. $L {y^{''}} = s^{2} Y - s y_{0} - y_{1}, L {- 4 y^{'}} = - 4 (s Y - y_{0}), L {3 y} = 3 Y .$ Jodo: $L {y^{''} - 4 y^{'} + 3 y} = (s^{2} - 4 s + 3) Y - s y_{0} - y_{1} + 4 y_{0} .$

L1.2 $L {u (t - 3)}$ aur $L {δ (t - 3)}$ batao.

Recall Solution L1.2

Shift rule ka special case $g = 1$ : $L {u (t - 3)} = \frac{e ^{- 3 s}}{s}$ . Impulse: $L {δ (t - 3)} = e^{- 3 s}$ . Difference kyun hai? Step " $t = 3$ ke baad forever on" rehti hai (ek lasting $\frac{1}{s}$ ), impulse "sirf ek instant" ke liye hai (koi $\frac{1}{s}$ nahi).

L1.3 $\frac{2 s}{s ^{2} + 9}$ aur $\frac{4}{s ^{2} + 9}$ ko invert karo.

Recall Solution L1.3

$\frac{s}{s ^{2} + ω ^{2}} \leftrightarrow cos ω t$ ko $ω = 3$ ke saath match karo: $\frac{2 s}{s ^{2} + 9} \to 2 cos 3 t$ . $\frac{ω}{s ^{2} + ω ^{2}} \leftrightarrow sin ω t$ match karo: humein upar $ω = 3$ chahiye, lekin hamare paas $4$ hai. Likho $\frac{4}{s ^{2} + 9} = \frac{4}{3} \cdot \frac{3}{s ^{2} + 9} \to \frac{4}{3} sin 3 t$ .

Level 2 — Application

L2.1 $y^{'} + 3 y = 0$ , $y (0) = 5$ solve karo.

Recall Solution L2.1

Transform: $(s Y - 5) + 3 Y = 0$ . Solve: $(s + 3) Y = 5 \Rightarrow Y = \frac{5}{s + 3}$ . Invert: $\frac{1}{s - a} \leftrightarrow e^{a t}$ with $a = - 3$ , toh $y (t) = 5 e^{- 3 t}$ . Check: $y (0) = 5$ ✔, aur $y^{'} = - 15 e^{- 3 t} = - 3 y$ ✔.

L2.2 $y^{''} + 4 y = 0$ , $y (0) = 1$ , $y^{'} (0) = 0$ solve karo.

Recall Solution L2.2

Transform: $(s^{2} Y - s \cdot 1 - 0) + 4 Y = 0$ . Solve: $(s^{2} + 4) Y = s \Rightarrow Y = \frac{s}{s ^{2} + 4}$ . Invert: $\frac{s}{s ^{2} + ω ^{2}} \leftrightarrow cos ω t$ with $ω = 2$ , toh $y (t) = cos 2 t$ . Check: $y (0) = 1$ ✔, $y^{'} (0) = - 2 sin 0 = 0$ ✔.

L2.3 $y^{''} - y = 0$ , $y (0) = 0$ , $y^{'} (0) = 1$ solve karo.

Recall Solution L2.3

Transform: $(s^{2} Y - 0 - 1) - Y = 0 \Rightarrow (s^{2} - 1) Y = 1 \Rightarrow Y = \frac{1}{s ^{2} - 1}$ . Partial fractions: $\frac{1}{s ^{2} - 1} = \frac{1}{( s - 1 ) ( s + 1 )} = \frac{1/2}{s - 1} - \frac{1/2}{s + 1}$ . Invert: $y (t) = \frac{1}{2} e^{t} - \frac{1}{2} e^{- t} = sinh t$ . Check: $y (0) = 0$ ✔, $y^{'} (0) = cosh 0 = 1$ ✔.

L2.4 $y^{'} + 2 y = e^{- t}$ , $y (0) = 0$ solve karo.

Recall Solution L2.4

Transform: $s Y + 2 Y = \frac{1}{s + 1}$ . Solve: $Y = \frac{1}{( s + 1 ) ( s + 2 )}$ . Partial fractions: cover-up at $s = - 1$ gives $\frac{1}{( - 1 + 2 )} = 1$ ; at $s = - 2$ gives $\frac{1}{( - 2 + 1 )} = - 1$ . Toh $Y = \frac{1}{s + 1} - \frac{1}{s + 2}$ . Invert: $y (t) = e^{- t} - e^{- 2 t}$ . Check: $y (0) = 0$ ✔.

Level 3 — Analysis (switches aur shifts)

L3.1 $y^{'} + 2 y = u (t - 1)$ , $y (0) = 0$ solve karo.

Recall Solution L3.1

Transform: $s Y + 2 Y = \frac{e ^{- s}}{s}$ , toh $Y = \frac{e ^{- s}}{s ( s + 2 )}$ . Pehle un-delayed core invert karo. Maano $H (s) = \frac{1}{s ( s + 2 )}$ . Cover-up: at $s = 0$ residue $\frac{1}{2}$ , at $s = - 2$ residue $- \frac{1}{2}$ . Toh $H = \frac{1/2}{s} - \frac{1/2}{s + 2}$ aur $h (t) = \frac{1}{2} - \frac{1}{2} e^{- 2 t}$ . Delay tag $e^{- s}$ apply karo: $h$ ko $1$ se delay karo AUR gate karo: $y (t) = u (t - 1) [\frac{1}{2} - \frac{1}{2} e^{- 2 (t - 1)}] .$ Matlab: $t = 1$ tak kuch nahi, phir $y$ $\frac{1}{2}$ ki taraf badhta hai. Figure dekho.

Figure — Solving ODEs with Laplace (including discontinuous forcing)

L3.2 $L {t u (t - 2)}$ nikalo.

Recall Solution L3.2

Trick yeh hai: shift rule ke liye forcing ko $g (t - 2)$ ke roop mein likhna zaroori hai, yaani $(t - 2)$ ka function. Lekin hamare paas plain $t$ hai. Rewrite karo: $t = (t - 2) + 2$ . Toh $g (t - 2) = (t - 2) + 2$ , yaani $g (τ) = τ + 2$ . $G (s) = L {τ + 2} = \frac{1}{s ^{2}} + \frac{2}{s}$ . Delay tag se multiply karo: $L {t u (t - 2)} = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s}) .$

L3.3 Ek rectangular pulse: $y^{'} + y = u (t - 1) - u (t - 3)$ , $y (0) = 0$ solve karo. (Input sirf $t = 1$ aur $t = 3$ ke beech ON hai.)

Recall Solution L3.3

Transform: $s Y + Y = \frac{e ^{- s} - e ^{- 3 s}}{s}$ , toh $Y = \frac{e ^{- s} - e ^{- 3 s}}{s ( s + 1 )}$ . Core inverse: $H (s) = \frac{1}{s ( s + 1 )} = \frac{1}{s} - \frac{1}{s + 1} \Rightarrow h (t) = 1 - e^{- t}$ . Do delay tags → do gated copies (doosre ko subtract karo): $y (t) = u (t - 1) [1 - e^{- (t - 1)}] - u (t - 3) [1 - e^{- (t - 3)}] .$ Matlab: $t = 1$ tak flat $0$ ; $1$ ki taraf charge karta hai; $t = 3$ pe input switch off ho jaata hai aur doosra term ise wapas $0$ ki taraf relax karta hai. Figure dekho.

Level 4 — Synthesis (tools combine karo)

L4.1 $y^{''} + 3 y^{'} + 2 y = u (t - 1)$ , $y (0) = 0$ , $y^{'} (0) = 0$ solve karo, phir $y (2)$ evaluate karo.

Recall Solution L4.1

Transform: $(s^{2} + 3 s + 2) Y = \frac{e ^{- s}}{s}$ , toh $Y = \frac{e ^{- s}}{s ( s + 1 ) ( s + 2 )}$ . Core inverse: $H (s) = \frac{1}{s ( s + 1 ) ( s + 2 )}$ . Cover-up: at $s = 0$ : $\frac{1}{( 1 ) ( 2 )} = \frac{1}{2}$ ; at $s = - 1$ : $\frac{1}{( - 1 ) ( 1 )} = - 1$ ; at $s = - 2$ : $\frac{1}{( - 2 ) ( - 1 )} = \frac{1}{2}$ . $H = \frac{1/2}{s} - \frac{1}{s + 1} + \frac{1/2}{s + 2} \Rightarrow h (t) = \frac{1}{2} - e^{- t} + \frac{1}{2} e^{- 2 t} .$ Delay + gate: $y (t) = u (t - 1) [\frac{1}{2} - e^{- (t - 1)} + \frac{1}{2} e^{- 2 (t - 1)}]$ . $t = 2$ pe evaluate karo (toh $t - 1 = 1$ ): $y (2) = \frac{1}{2} - e^{- 1} + \frac{1}{2} e^{- 2} \approx 0.5 - 0.367879 + 0.067668 = 0.199789$ .

L4.2 $y^{''} + y = sin t$ , $y (0) = 0$ , $y^{'} (0) = 0$ solve karo (resonance — forcing natural frequency se match karti hai).

Recall Solution L4.2

Transform: $(s^{2} + 1) Y = \frac{1}{s ^{2} + 1}$ , toh $Y = \frac{1}{( s ^{2} + 1 ) ^{2}}$ . Yeh special kyun hai? Forcing $sin t$ ki frequency natural oscillation se same hai, isliye uska transform wahi denominator share karta hai — use square kar deta hai. Ek repeated quadratic factor resonance signal karta hai, $t$ mein growth. Invert karo known pair $\frac{1}{( s ^{2} + ω ^{2} ) ^{2}} \leftrightarrow \frac{1}{2 ω ^{3}} (sin ω t - ω t cos ω t)$ use karke, $ω = 1$ ke saath: $y (t) = \frac{1}{2} (sin t - t cos t) .$ $- \frac{1}{2} t cos t$ term bina bound ke badhta hai — amplitude linearly climb karta hai, resonance ki fingerprint. Check: $y (0) = 0$ ; $y^{'} (t) = \frac{1}{2} (t sin t)$ (simplification ke baad) toh $y^{'} (0) = 0$ ✔.

L4.3 $y^{''} + 4 y = u (t - π)$ , $y (0) = 0$ , $y^{'} (0) = 0$ solve karo; $y (\frac{3 π}{2})$ nikalo.

Recall Solution L4.3

Transform: $(s^{2} + 4) Y = \frac{e ^{- π s}}{s}$ , toh $Y = \frac{e ^{- π s}}{s ( s ^{2} + 4 )}$ . Core inverse: $H (s) = \frac{1}{s ( s ^{2} + 4 )} = \frac{A}{s} + \frac{B s + C}{s ^{2} + 4}$ . Multiply out: $1 = A (s^{2} + 4) + (B s + C) s$ . At $s = 0$ : $A = \frac{1}{4}$ . Coeff of $s^{2}$ : $A + B = 0 \Rightarrow B = - \frac{1}{4}$ . Coeff of $s$ : $C = 0$ . $H = \frac{1/4}{s} - \frac{1}{4} \cdot \frac{s}{s ^{2} + 4} \Rightarrow h (t) = \frac{1}{4} - \frac{1}{4} cos 2 t .$ Delay + gate: $y (t) = u (t - π) [\frac{1}{4} - \frac{1}{4} cos 2 (t - π)]$ . $t = \frac{3 π}{2}$ pe: $t - π = \frac{π}{2}$ , $cos 2 \cdot \frac{π}{2} = cos π = - 1$ . Toh $y = \frac{1}{4} - \frac{1}{4} (- 1) = \frac{1}{2}$ .

Level 5 — Mastery (impulses, reasoning, edge cases)

L5.1 $y^{''} + 2 y^{'} + y = δ (t - 2)$ , $y (0) = 0$ , $y^{'} (0) = 0$ solve karo. (Critically-damped system pe ek single hammer blow.)

Recall Solution L5.1

Transform: $(s^{2} + 2 s + 1) Y = e^{- 2 s}$ . Note karo $s^{2} + 2 s + 1 = (s + 1)^{2}$ . Toh $Y = \frac{e ^{- 2 s}}{( s + 1 ) ^{2}}$ . Core inverse: $\frac{1}{( s + 1 ) ^{2}} \leftrightarrow t e^{- t}$ (from $\frac{n !}{( s - a ) ^{n + 1}} \leftrightarrow t^{n} e^{a t}$ with $n = 1, a = - 1$ ). Toh $h (t) = t e^{- t}$ . Delay + gate: $y (t) = u (t - 2) (t - 2) e^{- (t - 2)}$ . Matlab: $t = 2$ pe kick tak kuch nahi, phir ek bump jo rise karta hai aur wapas $0$ pe decay ho jaata hai — system ka impulse response, delayed. Sanity check: $t = 3$ pe, $y = 1 \cdot e^{- 1} \approx 0.367879$ .

L5.2 Degenerate check: $y^{'} = δ (t)$ , $y (0) = 0$ solve karo. Answer physically kya kehta hai, aur chota sa $t = 0$ subtlety notice karna kyun zaroori hai?

Recall Solution L5.2

Transform: $s Y - 0 = e^{- 0 \cdot s} = 1$ , toh $Y = \frac{1}{s}$ aur $y (t) = 1$ for $t > 0$ . Physical matlab: $t = 0$ pe ek unit impulse instantly ek unit deposit karta hai; baad mein $y$ $1$ pe stuck rehta hai (koi restoring term nahi). Impulse ek instantaneous "step-in" ki tarah kaam karta hai initial condition ka. Subtlety: impulse bilkul lower limit $t = 0$ pe baitha hai. Standard one-sided transform ke saath kick fully capture ho jaata hai, giving $y = 1$ for all $t > 0$ ; exactly $t = 0$ pe value undefined hai (woh jump kar raha hai). Yeh woh boundary/degenerate case hai jahan " $0$ ke just baad" aur " $0$ se just pehle" alag hote hain — hamesha poochho ki tumhara matlab kaun sa side hai.

L5.3 Reasoning (poora inversion zaroori nahi): $y^{''} + 5 y^{'} + 6 y = f (t)$ ke liye zero initial conditions ke saath, transform factor hota hai $Y (s) = \frac{F ( s )}{( s + 2 ) ( s + 3 )}$ . Sirf roots use karke long-time behaviour predict karo jab $f (t) = u (t)$ ho (ek constant switch-on).

Recall Solution L5.3

Poles padho. System ke apne decay modes roots $s = - 2$ aur $s = - 3$ se aate hain, dono negative → dono terms $e^{- 2 t}, e^{- 3 t}$ dete hain jo khatam ho jaate hain. $f = u (t)$ ke saath, $F = \frac{1}{s}$ ek pole at $s = 0$ add karta hai, jiska residue ek constant term deta hai. Steady state directly compute karo us constant term ki value se: $t \to \infty$ ke baad decaying modes khatam ho jaate hain, $\frac{1}{s ( s + 2 ) ( s + 3 )}$ ka $s = 0$ residue bachta hai, jo hai $\frac{1}{( 2 ) ( 3 )} = \frac{1}{6}$ . Conclusion: $y (t) \to \frac{1}{6}$ . Tumhe poora invert karna hi nahi pada — poles ne sab bata diya. (ODE se cross-check: steady state pe $y^{''} = y^{'} = 0$ , toh $6 y = 1 \Rightarrow y = \frac{1}{6}$ ✔.)

Recap flashcards

Recall Quick self-test

$e^{- a s} H (s)$ ko tum kaise invert karte ho? ::: Delay AND gate: $u (t - a) h (t - a)$ . $t u (t - 2)$ transform karne se pehle $t = (t - 2) + 2$ kyun rewrite karte ho? ::: Shift rule ke liye forcing $(t - 2)$ ka function honi chahiye; phir $G (s) = \frac{1}{s ^{2}} + \frac{2}{s}$ aur $e^{- 2 s}$ se tag karo. Kaun sa denominator resonance signal karta hai? ::: Ek repeated quadratic jaise $(s^{2} + ω^{2})^{2}$ ; yeh ek $t cos ω t$ growing term deta hai. $L {δ (t - a)}$ vs $L {u (t - a)}$ ? ::: $e^{- a s}$ vs $\frac{e ^{- a s}}{s}$ — impulse mein koi $\frac{1}{s}$ nahi. Steady state of $\frac{1}{s ( s + 2 ) ( s + 3 )}$ jab $t \to \infty$ ? ::: $s = 0$ residue $\frac{1}{6}$ (decaying modes khatam ho jaate hain).