Transforms of standard functions — proofs: Exercises

4.6.26 · D4 · Maths › Ordinary Differential Equations › Transforms of standard functions — proofs

Do symbols baar baar aate hain, toh pehle unhe words mein pin kar lete hain, kisi bhi exercise se pehle.

Mechanics ke liye hum Linearity and First Shifting Theorem, Integration by Parts, Euler's Formula $e^{i\theta}$, aur Improper Integrals pe rely karte hain. Core table (parent mein prove ki gayi):

Neeche number line dikhata hai ki sabse chhota allowed $s$ — yaani abscissa of convergence — har function ke liye alag kaise hota hai. Ise paas rakhna.

Figure — Transforms of standard functions — proofs

Level 1 — Recognition

Yeh toh table se seedha padhna chahiye.

Exercise 1.1. $L {1}$ , $L {t}$ , aur $L {t^{4}}$ conditions ke saath likho.

Exercise 1.2. $L {cos 3 t}$ aur $L {sin 3 t}$ likho.

Exercise 1.3. $L {e^{5 t}}$ aur $L {cosh 2 t}$ likho, har ek apne region of convergence ke saath.

Recall Solution 1.1

$L {1} = \frac{1}{s}$ , $s > 0$ ke liye.
$L {t} = \frac{1 !}{s ^{2}} = \frac{1}{s ^{2}}$ , $s > 0$ ke liye.
$L {t^{4}} = \frac{4 !}{s ^{5}} = \frac{24}{s ^{5}}$ , $s > 0$ ke liye.

Factorial kyun? Recursion $I_{n} = \frac{n}{s} I_{n - 1}$ har step mein ek naya factor jodhta hai, isliye $t^{4}$ mein $4! = 24$ aata hai, sirf $1$ nahi.

Recall Solution 1.2

Yahan $a = 3$ hai, aur $a^{2} = 9$ : $L {cos 3 t} = \frac{s}{s ^{2} + 9}, L {sin 3 t} = \frac{3}{s ^{2} + 9}, s > 0.$ $cos$ ka numerator $s$ hai (kyunki $t = 0$ par value $1$ hoti hai); $sin$ ka numerator $a = 3$ hai.

Recall Solution 1.3

$L {e^{5 t}} = \frac{1}{s - 5}$ , $s > 5$ ke liye (zaroorat hai $s - 5 > 0$ taaki $e^{- (s - 5) t}$ decay kare).
$L {cosh 2 t} = \frac{s}{s ^{2} - 4}$ , $s > 2$ ke liye (yaani $s > ∣ a ∣$ jab $a = 2$ ; real poles $\pm 2$ par hain).

Level 2 — Application

Ab linearity use karke entries combine karo: $L {α f + β g} = α L {f} + β L {g}$ .

Exercise 2.1. $L {3 t^{2} - 2 sin 4 t + 5}$ nikalo.

Exercise 2.2. $L {(t + 1)^{2}}$ pehle expand karke nikalo.

Exercise 2.3. Identity $cos^{2} t = \frac{1 + c o s 2 t}{2}$ use karke $L {cos^{2} t}$ nikalo.

Recall Solution 2.1

Term by term (linearity se hum har piece ko transform karke add kar sakte hain): $L {3 t^{2}} = 3 \cdot \frac{2 !}{s ^{3}} = \frac{6}{s ^{3}}, L {2 sin 4 t} = 2 \cdot \frac{4}{s ^{2} + 16} = \frac{8}{s ^{2} + 16}, L {5} = \frac{5}{s} .$ $L {3 t^{2} - 2 sin 4 t + 5} = \frac{6}{s ^{3}} - \frac{8}{s ^{2} + 16} + \frac{5}{s} s > 0.$

Recall Solution 2.2

$(t + 1)^{2} = t^{2} + 2 t + 1$ expand karo taaki har piece ek table entry ho: $L {t^{2}} = \frac{2}{s ^{3}}, L {2 t} = \frac{2}{s ^{2}}, L {1} = \frac{1}{s} .$ $L {(t + 1)^{2}} = \frac{2}{s ^{3}} + \frac{2}{s ^{2}} + \frac{1}{s} s > 0.$ Expand kyun karte hain? $L$ ka koi "product rule" nahi hai — tum $(t + 1)^{2}$ ko ek product ki tarah transform nahi kar sakte. Expand karne se yeh sum ban jaata hai, jahan linearity kaam karti hai.

Recall Solution 2.3

$cos^{2} t = \frac{1}{2} + \frac{1}{2} cos 2 t$ . $a = 2$ ke saath: $L {cos^{2} t} = \frac{1}{2} \cdot \frac{1}{s} + \frac{1}{2} \cdot \frac{s}{s ^{2} + 4} = \frac{1}{2 s} + \frac{s}{2 ( s ^{2} + 4 )} s > 0.$ Agar chaaho toh common denominator pe combine karo: $\frac{( s ^{2} + 4 ) + s ^{2}}{2 s ( s ^{2} + 4 )} = \frac{2 s ^{2} + 4}{2 s ( s ^{2} + 4 )} = \frac{s ^{2} + 2}{s ( s ^{2} + 4 )}$ .

Level 3 — Analysis

Yahan tum machine ko ulta chalate ho, ya convergence ke baare mein sochte ho.

Exercise 3.1. $L^{- 1} {\frac{2 s + 5}{s ^{2} + 9}}$ nikalo (transform undo karo).

Exercise 3.2. $L^{- 1} {\frac{1}{s ^{2} - 4}}$ nikalo.

Exercise 3.3. $L {e^{3 t} cosh 2 t}$ kis $s$ ke liye converge karta hai, aur yeh kya hai? (Abscissa of convergence dhyan se dekho.)

Recall Solution 3.1

Fraction ko table shapes se match karne ke liye split karo (numerators $s$ aur constant alag alag): $\frac{2 s + 5}{s ^{2} + 9} = 2 \cdot \frac{s}{s ^{2} + 9} + \frac{5}{3} \cdot \frac{3}{s ^{2} + 9} .$ Pehla $cos 3 t$ ka transform hai, doosra $sin 3 t$ ka ( $a = 3$ ke saath): $L^{- 1} {\frac{2 s + 5}{s ^{2} + 9}} = 2 cos 3 t + \frac{5}{3} sin 3 t$ $5$ ko $\frac{5}{3} \cdot 3$ kyun likhte hain? Sine entry ko upar bare $a = 3$ chahiye; hum use manufacture karte hain aur $\frac{5}{3}$ se rebalance karte hain.

Recall Solution 3.2

Yeh $\frac{a}{s ^{2} - a ^{2}}$ se match karta hai jab $a = 2$ , lekin us entry mein upar $a = 2$ hota hai. Humare paas $1$ hai, toh $\frac{1}{2}$ bahar nikalo: $\frac{1}{s ^{2} - 4} = \frac{1}{2} \cdot \frac{2}{s ^{2} - 4} \Rightarrow L^{- 1} {\frac{1}{s ^{2} - 4}} = \frac{1}{2} sinh 2 t$ $s^{2} - 4$ mein minus sign hyperbolic indicate karta hai, trig nahi. (General splitting method ke liye Inverse Laplace Transform — Partial Fractions note dekho.)

Recall Solution 3.3

First shifting theorem use karo: $e^{3 t}$ se multiply karna $s \to s - 3$ shift karta hai. Hum jaante hain $L {cosh 2 t} = \frac{s}{s ^{2} - 4}$ $s > 2$ ke liye. Isliye $L {e^{3 t} cosh 2 t} = \frac{( s - 3 )}{( s - 3 ) ^{2} - 4} .$ Convergence bhi shift hoti hai: originally $s > 2$ chahiye tha; shift ke baad $s - 3 > 2$ chahiye, yaani $s > 5$ .

Level 4 — Synthesis

Kai tools ko ek saath jodo.

Exercise 4.1. Shifting theorem use karke $L {e^{- 2 t} sin 3 t}$ nikalo.

Exercise 4.2. $L {t e^{2 t}}$ do tareekon se nikalo: (a) shifting theorem, (b) ise $t^{1}$ shifted ke roop mein pehchaankar.

Exercise 4.3. $e^{ia t} = cos a t + i sin a t$ use karke $L {t e^{ia t}}$ ka real part lekar $L {t cos a t}$ compute karo.

Recall Solution 4.1

$L {sin 3 t} = \frac{3}{s ^{2} + 9}$ ( $s > 0$ ) se shuru karo. $e^{- 2 t}$ se multiply karna matlab $s \to s - (- 2) = s + 2$ shift karna: $L {e^{- 2 t} sin 3 t} = \frac{3}{( s + 2 ) ^{2} + 9} s > - 2.$ Abscissa $0$ se $- 2$ par aa gayi (shift by $a = - 2$ ).

Recall Solution 4.2

$L {t} = \frac{1}{s ^{2}}$ ( $s > 0$ ). Time function ko $e^{2 t}$ se multiply karo → $s \to s - 2$ shift: $L {t e^{2 t}} = \frac{1}{( s - 2 ) ^{2}} s > 2.$ Dono readings (" $t$ transform shift karo" aur " $t^{1}$ times $e^{2 t}$ ") ek hi theorem hain; dono agree karte hain.

Recall Solution 4.3

Pehle, $L {t e^{ia t}}$ : yeh $t^{1}$ hai $e^{ia t}$ se multiply kiya hua, toh $L {t} = \frac{1}{s ^{2}}$ ko $s \to s - ia$ shift karo: $L {t e^{ia t}} = \frac{1}{( s - ia ) ^{2}} .$ Upar aur neeche $(s + ia)^{2}$ se multiply karke rationalize karo: $\frac{1}{( s - ia ) ^{2}} = \frac{( s + ia ) ^{2}}{( ( s - ia ) ( s + ia ) ) ^{2}} = \frac{( s + ia ) ^{2}}{( s ^{2} + a ^{2} ) ^{2}} .$ Upar expand karo: $(s + ia)^{2} = s^{2} - a^{2} + 2 ia s$ . Toh $L {t e^{ia t}} = \frac{( s ^{2} - a ^{2} ) + i ( 2 a s )}{( s ^{2} + a ^{2} ) ^{2}} .$ Kyunki $t e^{ia t} = t cos a t + i t sin a t$ hai, real part se $L {t cos a t}$ milta hai: $L {t cos a t} = \frac{s ^{2} - a ^{2}}{( s ^{2} + a ^{2} ) ^{2}} (aur L {t sin a t} = \frac{2 a s}{( s ^{2} + a ^{2} ) ^{2}}) .$

Level 5 — Mastery

Raw integral se kuch prove karo, table nahi.

Exercise 5.1. Definition se, Integration by Parts use karke $L {t} = \frac{1}{s ^{2}}$ prove karo, aur exactly batao kahan $s > 0$ use hota hai.

Exercise 5.2. $\int_{0}^{\infty} e^{- (s - a) t} d t$ se directly $L {e^{a t}} = \frac{1}{s - a}$ derive karo, aur exact convergence condition batao.

Exercise 5.3. Do IBP equations $S = \frac{a}{s} C$ aur $C = \frac{1}{s} - \frac{a}{s} S$ (jahan $S = L {sin a t}$ , $C = L {cos a t}$ ) use karke, linear system solve karo aur dono transforms recover karo.

Recall Solution 5.1

$I = \int_{0}^{\infty} e^{- s t} t d t$ lo. $u = t$ , $d v = e^{- s t} d t$ choose karo, toh $d u = d t$ , $v = \frac{e ^{- s t}}{- s}$ : $I = [t \cdot \frac{e ^{- s t}}{- s}]_{0}^{\infty} - \int_{0}^{\infty} \frac{e ^{- s t}}{- s} d t .$ Boundary term. $t = 0$ par yeh $0$ hai. $t \to \infty$ par, $t e^{- s t} \to 0$ kyunki $s > 0$ hai — exponential linear factor ko beat kar deta hai. (Agar $s \leq 0$ hota toh yeh blow up karta: yahi woh jagah hai jahan condition janam leti hai.) Bacha hua integral. $- \int_{0}^{\infty} \frac{e ^{- s t}}{- s} d t = \frac{1}{s} \int_{0}^{\infty} e^{- s t} d t = \frac{1}{s} \cdot \frac{1}{s} = \frac{1}{s ^{2}}$ . $L {t} = \frac{1}{s ^{2}}, s > 0.$

Recall Solution 5.2

$L {e^{a t}} = \int_{0}^{\infty} e^{- s t} e^{a t} d t = \int_{0}^{\infty} e^{- (s - a) t} d t = [\frac{e ^{- (s - a) t}}{- ( s - a )}]_{0}^{\infty} .$ $t \to \infty$ par exponential $\to 0$ tabhi jab exponent negative ho, yaani $s - a > 0$ . Tab upper limit $0$ contribute karta hai aur lower $- \frac{1}{- ( s - a )} = \frac{1}{s - a}$ deta hai: $L {e^{a t}} = \frac{1}{s - a}, s > a .$

Recall Solution 5.3

Doosri equation pehli mein substitute karo: $S = \frac{a}{s} (\frac{1}{s} - \frac{a}{s} S) = \frac{a}{s ^{2}} - \frac{a ^{2}}{s ^{2}} S .$ $S$ collect karo: $S (1 + \frac{a ^{2}}{s ^{2}}) = \frac{a}{s ^{2}} \Rightarrow S \cdot \frac{s ^{2} + a ^{2}}{s ^{2}} = \frac{a}{s ^{2}}$ , toh $S = \frac{a}{s ^{2} + a ^{2}} .$ $C = \frac{1}{s} - \frac{a}{s} S$ mein back-substitute karo: $C = \frac{1}{s} - \frac{a}{s} \cdot \frac{a}{s ^{2} + a ^{2}} = \frac{( s ^{2} + a ^{2} ) - a ^{2}}{s ( s ^{2} + a ^{2} )} = \frac{s ^{2}}{s ( s ^{2} + a ^{2} )} = \frac{s}{s ^{2} + a ^{2}} .$ $L {sin a t} = \frac{a}{s ^{2} + a ^{2}}, L {cos a t} = \frac{s}{s ^{2} + a ^{2}} .$

Recall Self-check summary

Main kis level par comfortable hoon? ::: L1 table padhna · L2 sums par linearity · L3 invert + convergence track karna · L4 shifting + complex exponentials · L5 integral se prove karna. L3–L4 mein jo ek theorem dominate karta hai ::: First shifting: $L {e^{a t} f (t)} = F (s - a)$ , aur abscissa bhi $+ a$ se shift hoti hai.

Connections

4.6.26 Transforms of standard functions — proofs (Hinglish)
Laplace Transform — Definition and Existence
Linearity and First Shifting Theorem
Transform of Derivatives — solving ODEs
Inverse Laplace Transform — Partial Fractions
Integration by Parts · Euler's Formula $e^{i\theta}$ · Improper Integrals