Inverse Laplace transform — partial fractions, tables: Exercises

4.6.29 · D4 · Maths › Ordinary Differential Equations › Inverse Laplace transform — partial fractions, tables

Ek table jis par hum poore waqt rely karte hain — ise visible rakho:

Yahan $s$ frequency-domain variable hai, $t \geq 0$ time hai, $a$ ek real growth/decay rate hai, $ω$ angular frequency hai, aur $n$ ek whole-number power hai.

Level 1 — Recognition

Goal: ek single atom ko table mein dekh ke directly match karo.

Problem 1.1

$L^{- 1} {\frac{5}{s}}$ find karo.

Recall Solution 1.1

LOOK-UP. Atom $\frac{1}{s}$ invert hokar constant $1$ deta hai. Linearity $5$ ko seedha bahar le aati hai: $L^{- 1} {\frac{5}{s}} = 5 \cdot L^{- 1} {\frac{1}{s}} = 5 \cdot 1 = 5 .$

Problem 1.2

$L^{- 1} {\frac{1}{s + 4}}$ find karo.

Recall Solution 1.2

LOOK-UP. Atom $\frac{1}{s - a}$ invert hokar $e^{a t}$ deta hai. Yahan denominator $s - (- 4)$ hai, toh $a = - 4$ : $L^{- 1} {\frac{1}{s + 4}} = e^{- 4 t} .$ Negative $a$ ka matlab decay hai — function $t$ badhne ke saath shrink karta hai.

Problem 1.3

$L^{- 1} {\frac{6}{s ^{2} + 9}}$ find karo.

Recall Solution 1.3

LOOK-UP. Sine atom hai $\frac{ω}{s ^{2} + ω ^{2}}$ . Yahan $ω^{2} = 9 \Rightarrow ω = 3$ , toh sine numerator hona chahiye $3$ . Hamare paas $6 = 2 \cdot 3$ hai, toh $2$ factor out karo: $L^{- 1} {\frac{6}{s ^{2} + 9}} = 2 \cdot L^{- 1} {\frac{3}{s ^{2} + 9}} = 2 sin 3 t .$

Level 2 — Application

Goal: table se pehle ek chhota sa preparation step (scale, shift, ya do mein split) karo.

Problem 2.1

$L^{- 1} {\frac{s}{s ^{2} + 16}} + L^{- 1} {\frac{2}{s ^{2} + 16}}$ find karo.

Recall Solution 2.1

Do atoms LOOK-UP karo. Yahan $ω^{2} = 16 \Rightarrow ω = 4$ .

$s$ -on-top → cosine: $\frac{s}{s ^{2} + 16} \to cos 4 t$ .
constant-on-top → sine, lekin humein $ω = 4$ upar chahiye aur sirf $2 = \frac{1}{2} \cdot 4$ hai: $\frac{2}{s ^{2} + 16} = \frac{1}{2} \cdot \frac{4}{s ^{2} + 16} \to \frac{1}{2} sin 4 t$ . $cos 4 t + \frac{1}{2} sin 4 t .$

Problem 2.2

$L^{- 1} {\frac{1}{( s - 3 ) ^{2}}}$ find karo.

Recall Solution 2.2

First-shift theorem. Base atom $\frac{1}{s ^{2}} \to t$ se shuru karo (yeh $\frac{1}{s ^{n + 1}} \to \frac{t ^{n}}{n !}$ hai jab $n = 1$ ). $s \to s - 3$ replace karne se answer $e^{3 t}$ se multiply ho jaata hai: $L^{- 1} {\frac{1}{( s - 3 ) ^{2}}} = t e^{3 t} .$

Problem 2.3

$L^{- 1} {\frac{2 s + 5}{s ^{2} + 4}}$ find karo.

Recall Solution 2.3

SPLIT, phir LOOK-UP. $ω^{2} = 4 \Rightarrow ω = 2$ . Numerator ke pieces se break karo: $\frac{2 s + 5}{s ^{2} + 4} = 2 \cdot \frac{s}{s ^{2} + 4} + \frac{5}{s ^{2} + 4} .$

$2 \cdot \frac{s}{s ^{2} + 4} \to 2 cos 2 t$ .
$\frac{5}{s ^{2} + 4} = \frac{5}{2} \cdot \frac{2}{s ^{2} + 4} \to \frac{5}{2} sin 2 t$ . $2 cos 2 t + \frac{5}{2} sin 2 t .$

Level 3 — Analysis

Goal: real partial fractions, completing the square, aur $sin$ aur $sinh$ mein distinguish karna.

Problem 3.1

$L^{- 1} {\frac{s - 1}{( s + 1 ) ( s - 3 )}}$ find karo.

Recall Solution 3.1

SPLIT (distinct linear poles). Likho $\frac{s - 1}{( s + 1 ) ( s - 3 )} = \frac{A}{s + 1} + \frac{B}{s - 3}$ . Cover-up at $s = - 1$ : $A = \frac{- 1 - 1}{- 1 - 3} = \frac{- 2}{- 4} = \frac{1}{2}$ . Cover-up at $s = 3$ : $B = \frac{3 - 1}{3 + 1} = \frac{2}{4} = \frac{1}{2}$ . LOOK-UP har $\frac{1}{s - a} \to e^{a t}$ : $f (t) = \frac{1}{2} e^{- t} + \frac{1}{2} e^{3 t} = \frac{1}{2} (e^{- t} + e^{3 t}) .$

Problem 3.2

Pehle form predict karo, phir $L^{- 1} {\frac{3}{s ^{2} - 1}}$ find karo. Figure dekho: real poles → exponential/hyperbolic, na ki oscillation.

Figure — Inverse Laplace transform — partial fractions, tables

Recall Solution 3.2

Predict: $s^{2} - 1 = (s - 1) (s + 1)$ ke real roots $\pm 1$ hain. Real poles ka matlab growing/decaying exponentials hain, jo combine hokar $sinh / cosh$ banaate hain — koi wiggling sine nahin. Compare karo $s^{2} + 1$ se (roots real axis se door) jo sach mein $sin / cos$ deta. Figure dono dikhata hai. SPLIT: $\frac{3}{( s - 1 ) ( s + 1 )} = \frac{A}{s - 1} + \frac{B}{s + 1}$ . Cover-up: $A = \frac{3}{1 + 1} = \frac{3}{2}$ , $B = \frac{3}{- 1 - 1} = - \frac{3}{2}$ . LOOK-UP: $f (t) = \frac{3}{2} e^{t} - \frac{3}{2} e^{- t} = 3 \cdot \frac{e ^{t} - e ^{- t}}{2} = 3 sinh t .$ Confirmed: hyperbolic sine, exactly as predicted.

Problem 3.3

$L^{- 1} {\frac{s + 7}{s ^{2} + 4 s + 13}}$ find karo.

Recall Solution 3.3

Square COMPLETE karo: $s^{2} + 4 s + 13 = (s + 2)^{2} + 9$ , toh $a = - 2$ , $ω^{2} = 9 \Rightarrow ω = 3$ . Numerator ko $(s + 2)$ ke around SPLIT karo: humein ek $(s + 2)$ chahiye (cosine ke liye) aur ek leftover (sine ke liye): $\frac{s + 7}{( s + 2 ) ^{2} + 9} = \frac{( s + 2 ) + 5}{( s + 2 ) ^{2} + 9} = \frac{s + 2}{( s + 2 ) ^{2} + 9} + \frac{5}{( s + 2 ) ^{2} + 9} .$ LOOK-UP shift atoms ke saath:

$\frac{s + 2}{( s + 2 ) ^{2} + 9} \to e^{- 2 t} cos 3 t$ .
$\frac{5}{( s + 2 ) ^{2} + 9} = \frac{5}{3} \cdot \frac{3}{( s + 2 ) ^{2} + 9} \to \frac{5}{3} e^{- 2 t} sin 3 t$ . $f (t) = e^{- 2 t} (cos 3 t + \frac{5}{3} sin 3 t) .$

Level 4 — Synthesis

Goal: ek hi problem mein repeated poles, quadratics, aur improper-fraction handling combine karo.

Problem 4.1

$L^{- 1} {\frac{2 s + 3}{( s + 1 ) ^{2} ( s + 2 )}}$ find karo.

Recall Solution 4.1

SPLIT. Double pole $(s + 1)^{2}$ ko do terms chahiye; simple pole $(s + 2)$ ko ek: $\frac{2 s + 3}{( s + 1 ) ^{2} ( s + 2 )} = \frac{A}{s + 1} + \frac{B}{( s + 1 ) ^{2}} + \frac{C}{s + 2} .$ $(s + 1)^{2} (s + 2)$ se multiply karo: $2 s + 3 = A (s + 1) (s + 2) + B (s + 2) + C (s + 1)^{2} .$

$s = - 1$ set karo: $2 (- 1) + 3 = 1 = B (1) \Rightarrow B = 1$ .
$s = - 2$ set karo: $2 (- 2) + 3 = - 1 = C (1) \Rightarrow C = - 1$ .
$s^{2}$ -coefficients compare karo: $0 = A + C \Rightarrow A = 1$ . LOOK-UP: $\frac{1}{s + 1} \to e^{- t}$ , $\frac{1}{( s + 1 ) ^{2}} \to t e^{- t}$ ( $\frac{1}{s ^{2}} \to t$ ka shift), $\frac{1}{s + 2} \to e^{- 2 t}$ : $f (t) = e^{- t} + t e^{- t} - e^{- 2 t} = (1 + t) e^{- t} - e^{- 2 t} .$

Problem 4.2

$L^{- 1} {\frac{s ^{2} + 2 s + 3}{s ^{2} + 1}}$ find karo.

Recall Solution 4.2

Pehle proper check karo. Numerator ki degree (2) $=$ denominator ki degree (2): yeh improper hai. Polynomial long division karo: $\frac{s ^{2} + 2 s + 3}{s ^{2} + 1} = 1 + \frac{2 s + 2}{s ^{2} + 1} .$ LOOK-UP:

$L^{- 1} {1} = δ (t)$ (Dirac delta — ek constant ka inverse).
$\frac{2 s}{s ^{2} + 1} \to 2 cos t$ ( $ω = 1$ ).
$\frac{2}{s ^{2} + 1} \to 2 sin t$ . $f (t) = δ (t) + 2 cos t + 2 sin t .$

Level 5 — Mastery

Goal: full ODE round-trip, aur convolution route jab table cleanly split na kare.

Problem 5.1

$y^{''} + 4 y = 0$ ko $y (0) = 1, y^{'} (0) = 6$ ke saath Laplace transforms use karke solve karo. $y (t)$ report karo.

Recall Solution 5.1

ODE transform karo (yaad karo $L {y^{''}} = s^{2} Y - s y (0) - y^{'} (0)$ , Solving ODEs with Laplace Transforms se): $s^{2} Y - s (1) - 6 + 4 Y = 0 \Rightarrow (s^{2} + 4) Y = s + 6 \Rightarrow Y = \frac{s + 6}{s ^{2} + 4} .$ SPLIT / LOOK-UP with $ω = 2$ : $Y = \frac{s}{s ^{2} + 4} + \frac{6}{s ^{2} + 4} = \frac{s}{s ^{2} + 4} + 3 \cdot \frac{2}{s ^{2} + 4} .$ $y (t) = cos 2 t + 3 sin 2 t .$ Sanity check: $y (0) = cos 0 + 3 sin 0 = 1$ ✓, aur $y^{'} (t) = - 2 sin 2 t + 6 cos 2 t$ deta hai $y^{'} (0) = 6$ ✓.

Problem 5.2

$L^{- 1} {\frac{1}{s ^{2} ( s ^{2} + 1 )}}$ partial fractions se find karo, phir confirm karo ki $\frac{1}{s ^{2}} \cdot \frac{1}{s ^{2} + 1}$ structure Convolution Theorem se agree karta hai.

Recall Solution 5.2

SPLIT. Sirf even powers ke saath, likho $\frac{1}{s ^{2} ( s ^{2} + 1 )} = \frac{A}{s} + \frac{B}{s ^{2}} + \frac{C s + D}{s ^{2} + 1}$ . Multiply out karo: $1 = A s (s^{2} + 1) + B (s^{2} + 1) + (C s + D) s^{2} .$ Coefficients match karo: $s^{0} : B = 1$ ; $s^{1} : A = 0$ ; $s^{2} : B + D = 0 \Rightarrow D = - 1$ ; $s^{3} : A + C = 0 \Rightarrow C = 0$ . Toh $\frac{1}{s ^{2} ( s ^{2} + 1 )} = \frac{1}{s ^{2}} - \frac{1}{s ^{2} + 1}$ . LOOK-UP: $\frac{1}{s ^{2}} \to t$ , $\frac{1}{s ^{2} + 1} \to sin t$ : $f (t) = t - sin t .$ Convolution cross-check. $L^{- 1} {\frac{1}{s ^{2}}} = t$ aur $L^{- 1} {\frac{1}{s ^{2} + 1}} = sin t$ , toh product transform invert hota hai convolution mein: $(t) * (sin t) = \int_{0}^{t} τ sin (t - τ) d τ = t - sin t,$ jo match karta hai. Do raaste, ek jawaab — yahi reassurance theorem deta hai.