Worked examples — Transforms of standard functions — proofs

4.6.26 · D3 · Maths › Ordinary Differential Equations › Transforms of standard functions — proofs

Neeche sab kuch ek hi machine par lean karta hai: $\mathcal{L}\{f(t)\}={\int }_0^{\infty }{e}^{-st}f(t)dt,$ aur usi se bani dictionary. Agar koi symbol unfamiliar lage, toh Laplace Transform — Definition and Existence aur Linearity and First Shifting Theorem notes unhe zero se build karte hain.

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Scenario matrix

Kuch bhi work karne se pehle, chalte hain har tarah ke cell list karein jisme yeh transforms rehte hain. Ek worked example tabhi useful hai jab woh kisi named cell mein land kare — isliye neeche har example ko Cell Xn tag kiya gaya hai.

Cell	Situation class	Kya galat ho sakta hai / kya dhyan rakhna hai
A	Constant / pure power $t^n$	$n!$ bhool jaana; $s$ ka galat power
B	Exponential $e^{at}$, positive $a$	convergence ko $s>a$ chahiye (strip shifts)
C	Exponential $e^{at}$, negative $a$	pole at $s=-
D	Trig $\sin /\cos$ (oscillating)	denominator $s^2+a^2$, plus sign
E	Hyperbolic $\sinh /\cosh$ (growing)	denominator $s^2-a^2$, minus, need $s>|a|$
F	Shift $e^{at}f(t)$ (First Shifting)	$s\to s-a$ har jagah replace karo
G	Zero / degenerate input ($a=0$, $n=0$)	formulas simpler ones mein collapse hone chahiye
H	Limiting behaviour ($s\to \infty$, $a\to 0$)	sanity checks, koi naya result nahi
I	Linear combination (kaafi terms ka sum)	linearity, term by term
J	Word problem (physics / ODE)	$f(t)$ set up karo, units rakho, interpret karo

Neeche ke das examples sab das cells cover karte hain.

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Example 1 — Pure power (Cell A)

Steps.

1. $\mathcal{L}\{t^n\}=\frac{n!}{s^{n+1}}$ apply karo with $n=3$. Yeh step kyun? Yeh direct dictionary entry hai — integral dobara karne ki zarurat nahi. $\mathcal{L}\{t^3\}=\frac{3!}{s^4}=\frac{6}{s^4},s>0.$
2. Constant ke liye, number bahar nikalo (linearity) aur $\mathcal{L}\{1\}=1/s$ use karo. Yeh step kyun? $\mathcal{L}$ linear hai: constants integral ke through slide karte hain. $\mathcal{L}\{5\}=5\cdot \frac{1}{s}=\frac{5}{s},s>0.$

Verify: check karo ki $n!$ drop toh nahi hua — $\mathcal{L}\{t^1\}$ ko $1!/s^2=1/s^2$ hona chahiye. Hai. Aur $t^3$ growing hai matlab phir bhi sirf $s>0$ chahiye (weight $e^{-st}$ kisi bhi polynomial ko beat karta hai). ✓

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Example 2 — Positive exponential (Cell B)

Steps.

1. Dictionary entry $\mathcal{L}\{e^{at}\}=\frac{1}{s-a}$ use karo with $a=3$. Yeh step kyun? Yeh exactly wahi $\mathcal{L}\{1\}$ proof hai lekin $s\to s-a$ ke saath. $\mathcal{L}\{e^{3t}\}=\frac{1}{s-3}.$
2. Convergence: humein $s-a>0$ chahiye tha, yani $s>3$. Yeh step kyun? Tabhi $e^{-(s-a)t}\to 0$ as $t\to \infty$.

Verify: pole $s=3$ par baithta hai, exactly abscissa of convergence, yeh match karta hai "f ki fastest growth threshold set karti hai." ✓

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Example 3 — Negative exponential (Cell C)

Steps.

1. $a=-2$ ko $\frac{1}{s-a}$ mein daalo. Yeh step kyun? Formula signed hai — $a$ negative ho sakta hai, koi special case nahi chahiye. $\mathcal{L}\{e^{-2t}\}=\frac{1}{s+2},s>-2.$

Verify: strip $s>-2$, Cell B ke $s>3$ se zyada loose hai kyunki decaying input "weigh karna easy" hai. Do exponential cells B aur C milkar $a$ ke dono signs cover karte hain. ✓

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Example 4 — Oscillating (Cell D)

Steps.

1. $\mathcal{L}\{\sin at\}=\frac{a}{s^2+a^2}$ apply karo, $a=4$. Yeh step kyun? Direct dictionary entry hai (proved via $e^{iat}$). $\mathcal{L}\{\sin 4t\}=\frac{4}{s^2+16},s>0.$
2. $\mathcal{L}\{\cos at\}=\frac{s}{s^2+a^2}$ apply karo, $a=4$. $\mathcal{L}\{\cos 4t\}=\frac{s}{s^2+16},s>0.$

Figure dekho: blue oscillating $\sin 4t$ aur orange decaying weight $e^{-st}$ ka product green curve deta hai, jiska net signed area exactly transform hai. Plus/minus lobes nearly cancel ho jaate hain — isliye answer ek modest fraction hai, koi badi cheez nahi.

Verify: $s\to \infty$ par dono transforms $\to 0$ (weight sab kuch crush kar deta hai). $a=4$, $s=3$ set karte hain: $\mathcal{L}\{\sin 4t\}=4/(9+16)=4/25=0.16$. ✓

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Example 5 — Growing hyperbolic (Cell E)

Steps.

1. $\mathcal{L}\{\cosh at\}=\frac{s}{s^2-a^2}$, $a=2$. Yeh step kyun? $\cosh at=\frac{1}{2}(e^{at}+e^{-at})$ plus linearity se bana hai. $\mathcal{L}\{\cosh 2t\}=\frac{s}{s^2-4},s>2.$
2. $\mathcal{L}\{\sinh at\}=\frac{a}{s^2-a^2}$, $a=2$. $\mathcal{L}\{\sinh 2t\}=\frac{2}{s^2-4},s>2.$

Figure plus (trig) aur minus (hyperbolic) denominators ko $s$ ke function ke roop mein contrast karta hai: hyperbolic curve mein $s=a$ par ek vertical wall (real pole) hai, jo exactly uske strip ki boundary $s>|a|$ hai. Trig curve smooth hai — koi real pole nahi, toh $s>0$ kaafi hai.

Verify: $s=3,a=2$ par: $\mathcal{L}\{\cosh 2t\}=3/(9-4)=3/5=0.6$ aur $\mathcal{L}\{\sinh 2t\}=2/5=0.4$. Note karo $s=3>2$ hume legal rakhta hai. ✓

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Example 6 — First Shifting Theorem (Cell F)

Steps.

1. Base transform lo: $\mathcal{L}\{\cos 3t\}=\frac{s}{s^2+9}=F(s)$. Yeh step kyun? Shifting ek already-known transform par act karta hai.
2. Har $s$ ko $s+1$ se replace karo. Yeh step kyun? Shift theorem: $\mathcal{L}\{e^{-t}\cos 3t\}=F(s+1)$. $\mathcal{L}\{e^{-t}\cos 3t\}=\frac{(s+1)}{(s+1{)}^2+9},s>-1.$

Verify: strip $s>0$ se $s>-1$ ho jaata hai (left mein $1$ shift, $a=-1$ se match karta hai). Agar $e^{+t}\cos 3t$ hota, toh strip $s>1$ hoti. ✓

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Example 7 — Degenerate / zero inputs (Cell G)

Steps.

1. Cosine at $a=0$: $\frac{s}{s^2+a^2}{|}_{a=0}=\frac{s}{s^2}=\frac{1}{s}$. Yeh step kyun? Seedha $a=0$ set karo aur simplify karo — yeh degenerate limit hai. $\mathcal{L}\{\cos 0\}=\frac{1}{s}=\mathcal{L}\{1\}.✓$
2. Sine at $a=0$: $\frac{a}{s^2+a^2}{|}_{a=0}=\frac{0}{s^2}=0$. $\mathcal{L}\{\sin 0\}=0=\mathcal{L}\{0\}.✓$
3. Power at $n=0$: $\frac{n!}{s^{n+1}}{|}_{n=0}=\frac{0!}{s^1}=\frac{1}{s}$ (yaad raho $0!=1$). $\mathcal{L}\{t^0\}=\frac{1}{s}.✓$

Verify: teeno degenerate cases ek hi anchor $\mathcal{L}\{1\}=1/s$ par fold ho jaate hain (ya $0$). Consistency confirm. ✓

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Example 8 — Limiting behaviour as a sanity engine (Cell H)

Steps.

1. $s\to \infty$: $\frac{a}{s^2+a^2}\to 0$. Yeh step kyun? Denominator bina ruke badhta hai — har transform ko $s$ mein decay karna chahiye.
2. $a\to 0$: $\frac{a}{s^2+a^2}\to \frac{0}{s^2}=0$, Cell G se agree karta hai.
3. Initial-value shape: $s\cdot \frac{s}{s^2+a^2}=\frac{s^2}{s^2+a^2}\stackrel{s\to \infty }{\to }1=\cos 0$. Yeh step kyun? ${\lim }_{s\to \infty }sF(s)=f(0^{+})$ ek free correctness check hai.

Verify: $a=5$ ke saath: ${\lim }_{s\to \infty }\frac{s^2}{s^2+25}=1$, aur $\cos (0)=1$. Match. ✓

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Example 9 — Linear combination (Cell I)

Steps.

1. $\mathcal{L}\{3\}=\frac{3}{s}$; $\mathcal{L}\{-2e^{4t}\}=-\frac{2}{s-4}$. Yeh step kyun? Linearity se har term alag handle kar sakte hain.
2. $\mathcal{L}\{t^2\}=\frac{2!}{s^3}=\frac{2}{s^3}$; $\mathcal{L}\{-5\sin 3t\}=-\frac{15}{s^2+9}$.
3. Add karo aur strip record karo. $\mathcal{L}\{\cdots \}=\frac{3}{s}-\frac{2}{s-4}+\frac{2}{s^3}-\frac{15}{s^2+9},s>4.$ $s>4$ kyun? Har term simultaneously converge karna chahiye; growing $e^{4t}$ sabse strict bound demand karta hai.

Verify: har numerator mein sahi factorial/coefficient hai; strip $\{0,4,0,0\}$ ka max $=4$ hai. ✓

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Example 10 — Word problem: ek damped spring (Cell J)

Steps.

1. Base transform: $\mathcal{L}\{\sin 2t\}=\frac{2}{s^2+4}=F(s)$. Yeh step kyun? Pehle exponential peel off karo; plain sine transform karo.
2. $a=-1$ ke saath First Shifting apply karo: $s\to s+1$ replace karo. Yeh step kyun? $\mathcal{L}\{e^{-t}\sin 2t\}=F(s+1)$. $X(s)=\frac{2}{(s+1{)}^2+4},s>-1.$
3. Initial-value check: ${\lim }_{s\to \infty }sX(s)={\lim }_{s\to \infty }\frac{2s}{(s+1{)}^2+4}=0=x(0)$. Yeh step kyun? ${\lim }_{s\to \infty }sX(s)=x(0^{+})$; ek released-from-rest sine ko $0$ dena chahiye.

Verify (units + number): $X(s)$ ke units metre·seconds hain (displacement integrated over time), $\int x{e}^{-st}dt$ se consistent. $s=1$ par: $X(1)=\frac{2}{(2{)}^2+4}=\frac{2}{8}=0.25$. Aur $\lim sX(s)=0$, $x(0)=e^0\sin 0=0$ se match karta hai. ✓ Yeh $X(s)$ exactly woh shape hai jo tum Inverse Laplace Transform — Partial Fractions se invert karke motion recover karoge.

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Coverage check

Recall Kya humne har cell hit kiya?

A (Ex 1) · B (Ex 2) · C (Ex 3) · D (Ex 4) · E (Ex 5) · F (Ex 6) · G (Ex 7) · H (Ex 8) · I (Ex 9) · J (Ex 10). $a$ ka har sign, dono zero cases, limiting checks, ek linear pile-up, aur ek physics setup — sab cover. Example 9 mein sabse tight convergence strip kis cell ne force ki? ::: Cell B ka $e^{4t}$ term, giving $s>4$. Hyperbolic strip $s>|a|$ kyun hai, sirf $s>a$ kyun nahi? ::: Dono $e^{at}$ aur $e^{-at}$ pieces converge karni chahiye, toh $s>a$ aur $s>-a$ dono chahiye.

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Connections

• 4.6.26 Transforms of standard functions — proofs (Hinglish) — parent proofs
• Laplace Transform — Definition and Existence
• Linearity and First Shifting Theorem — Cells F, I, J mein use hua
• Transform of Derivatives — solving ODEs — is dictionary ka payoff
• Inverse Laplace Transform — Partial Fractions — Example 10 undo karna
• Integration by Parts · Euler's Formula $e^{i\theta}$ · Improper Integrals

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