Solving ODEs with Laplace (including discontinuous forcing): Question bank

4.6.30 · D5 · Maths › Ordinary Differential Equations › Solving ODEs with Laplace (including discontinuous forcing)

True or false — justify

True or false: kisi bhi twice-differentiable

y

ke liye

L {y^{''}} = s^{2} Y

hota hai.

False. Sahi rule hai

s^{2} Y - s y (0) - y^{'} (0)

; boundary terms silently drop karna initial conditions ko zero set karta hai, jo sirf tabhi sahi hai jab

y (0) = y^{'} (0) = 0

True or false: kisi sum ka Laplace transform, transforms ke sum ke barabar hota hai.

True. Transform ek integral hai, aur integration linear hota hai, isliye

L {a f + b g} = a F (s) + b G (s)

— yahi wajah hai ki hum ek ODE ko term-by-term transform kar sakte hain. (

F, G

f, g

ke transforms hain.)

True or false:

L {f (t) g (t)} = F (s) G (s)

False. Time mein product ka transform, transforms ke product nahi hota;

F (s) G (s)

jo invert hota hai woh convolution

\int_{0}^{t} f (τ) g (t - τ) d τ

mein (dekho Transfer Functions and Convolution).

True or false:

G (s)

ko multiply karne wala factor

e^{- a s}

ka matlab hai "answer ko

e^{- a}

se multiply karo."

False.

e^{- a s}

ek time-delay operator hai, koi numerical scale nahi: yeh

u (t - a) g (t - a)

mein invert hota hai,

g

a

se baad shift karta hai aur

t = a

se pehle gate karke band rakhta hai.

True or false:

L {e^{a t}} = \frac{1}{s - a}

har real

s

ke liye hold karta hai.

False. Defining integral

\int_{0}^{\infty} e^{- (s - a) t} d t

sirf tabhi converge karta hai jab

s > a

(region of convergence

Re (s) > a

);

s \leq a

ke liye integrand decay nahi karta aur transform undefined hai.

True or false:

F (s)

ka sabse daayein wala pole, region of convergence ke andar strictly ho sakta hai.

False. Poles exactly woh points hain jahan integral blow up hota hai, isliye woh convergence boundary line

Re (s) = c

par ya uske baayein baithe hain; sabse daayein wala pole usi boundary ko define karta hai.

True or false:

L {t f (t)} = t F (s)

False. Time mein

t

se multiply karna,

s

mein differentiate karna ban jaata hai:

L {t f (t)} = - \frac{d F}{d s}

(moment-generating property). Jaise

L {e^{- t}} = \frac{1}{s + 1}

se milta hai

L {t e^{- t}} = - \frac{d}{d s} \frac{1}{s + 1} = \frac{1}{( s + 1 ) ^{2}}

True or false:

L {f (2 t)} = F (2 s)

False. Time ko

a > 0

se scale karna transform ko scale aur shrink karta hai:

L {f (a t)} = \frac{1}{a} F (\frac{s}{a})

. Time mein signal ko squeeze karna

s

mein usse stretch aur lower karta hai — aap sirf

2 s

substitute nahi kar sakte.

True or false: kyunki

δ (t - a)

"infinitely tall" hai, iska Laplace transform bhi infinite hona chahiye.

False.

L {δ (t - a)} = e^{- a s}

hai, ek bilkul finite factor — delta sifting property

\int_{0}^{\infty} e^{- s t} δ (t - a) d t = e^{- a s}

se define hota hai, literal height se nahi.

True or false: Laplace method mein particular solution ke liye ek form guess karni padti hai.

False. Method of Undetermined Coefficients ke ulta, koi guessing nahi chahiye:

Y (s)

ki algebra poora answer automatically produce karta hai, forced part aur homogeneous part dono saath mein.

True or false: partial fractions optional hai — aap

Y (s)

ko directly invert kar sakte ho.

Practice mein zyaadatar false. Tables sirf simple pieces jaise

\frac{1}{s - a}

aur

\frac{ω}{s ^{2} + ω ^{2}}

list karti hain; partial fractions woh tool hai jo ==ek complicated

Y (s)

ko un recognisable pieces mein todta hai== (dekho Partial Fractions).

True or false: Laplace sirf constant coefficients wale equations ke liye kaam karta hai.

Essentially true jo clean algebra hum yahan use karte hain uske liye. Variable coefficients jaise

t y^{'}

, $s$ mein derivatives mein transform ho jaate hain (via

L {t f} = - F^{'} (s)

), algebraic problem ko wapas differential problem bana dete hain — toh Linear Constant-Coefficient ODEs iska natural home hai.

Spot the error

Ek student likhta hai

L^{- 1} {e^{- 2 s} / (s + 1)} = e^{- (t - 2)}

. Kya missing hai?

Gate. Sahi inverse hai

u (t - 2) e^{- (t - 2)}

; step

u (t - 2)

ke bina solution

t = 2

par switch se pehle nonzero (aur bada bhi) hoga, jo physically impossible hai.

y^{'} + y = u (t - 2), y (0) = 0

ke liye, ek student left side ko

s Y + Y - y (0) = s Y + Y

transform karta hai lekin phir

Y = \frac{e ^{- 2 s}}{s + 1}

likhta hai. Kahan galti hui?

Unhone bhool gaye ki

L {u (t - 2)} = \frac{e ^{- 2 s}}{s}

mein ek

\frac{1}{s}

hai;

(s + 1)

se divide karne par

Y = \frac{e ^{- 2 s}}{s ( s + 1 )}

milta hai,

\frac{e ^{- 2 s}}{s + 1}

nahi.

L {t u (t - 2)}

find karne ke liye ek student

e^{- 2 s} \cdot \frac{1}{s ^{2}}

likhta hai. Yeh kyun galat hai?

Second shift ko

g (t - a)

chahiye, lekin yahan

g (t) = t \neq = (t - 2)

hai.

t = (t - 2) + 2

likhkar rewrite karo, jisse milta hai

L {t u (t - 2)} = e^{- 2 s} (\frac{1}{s ^{2}} + \frac{2}{s})

Ek student

\frac{1}{( s + 1 ) ^{2}}

e^{- t}

mein invert karta hai, "same as

\frac{1}{s + 1}

just squared." Theek karo.

Repeated pole ke saath

t

ki extra power aati hai:

\frac{1}{( s + 1 ) ^{2}} \to t e^{- t}

. Yeh exactly moment-generating property

L {t f} = - F^{'} (s)

hai ulta padha, kyunki

- \frac{d}{d s} \frac{1}{s + 1} = \frac{1}{( s + 1 ) ^{2}}

Ek student

H (s) = \frac{1}{s + 1}

ka response input

f

ke liye calculate karta hai aur time mein

y (t) = h (t) f (t)

likhta hai. Galti pakdo.

s

mein transfer function ko

F (s)

se multiply karna time mein product nahi hota; yeh convolution

y (t) = \int_{0}^{t} h (τ) f (t - τ) d τ

se correspond karta hai (dekho Transfer Functions and Convolution).

Ek student

\frac{1}{s ( s + 1 )}

\frac{1}{s} - \frac{1}{s + 1}

mein invert karta hai lekin phir

h (t) = 1 - 1 = 0

likhta hai. Kya galat hua?

Unhone transforms ko invert karne ki bajaye evaluate kar diya.

\frac{1}{s} \to 1

aur

\frac{1}{s + 1} \to e^{- t}

hota hai, toh

h (t) = 1 - e^{- t}

hai, jo sirf

t = 0

par

0

hai.

Ek student

y^{''} + y = 0, y (0) = 0, y^{'} (0) = 2

solve karta hai aur

Y = \frac{2}{s ^{2} + 1}

paata hai, phir

y = 2 cos t

likhta hai. Theek karo.

Unhone galat pair match kiya.

\frac{ω}{s ^{2} + ω ^{2}} \to sin ω t

hota hai, toh

\frac{2}{s ^{2} + 1} \to 2 sin t

; cosine pair ke numerator mein

s

hota hai.

Ek student claim karta hai ki derivative rule mein boundary term

[e^{- s t} f (t)]_{0}^{\infty}

hamesha

- f (0)

hota hai. Yeh kab fail ho sakta hai?

Jab

f

e^{s t}

ke decay se zyada tez badhti hai, upper limit vanish nahi hoti aur transform (aur isliye rule bhi) exist nahi karta — yeh Laplace Transform — Definition and Existence mein region-of-convergence/existence condition hai.

Why questions

Time mein differentiation,

s

se multiply karna kyun ban jaata hai?

Integration by parts derivative ko

f

se hatakar

e^{- s t}

par le jaata hai, aur

\frac{d}{d t} e^{- s t} = - s e^{- s t}

ek factor

s

nikaal laata hai — yahi ek fact poore method ka engine hai.

$t$ se multiply karna, $s$ mein differentiate karna kyun ban jaata hai (moment property)?

F (s) = \int_{0}^{\infty} e^{- s t} f d t

s

ke respect mein differentiate karna integral ke andar ek factor

- t

le aata hai:

F^{'} (s) = \int_{0}^{\infty} (- t) e^{- s t} f d t = - L {t f}

, isliye

L {t f} = - F^{'} (s)

Time shift

a

, ek exponential factor

e^{- a s}

kyun produce karta hai, koi polynomial kyun nahi?

Integral mein

τ = t - a

substitute karne se

e^{- s (τ + a)} = e^{- s a} e^{- s τ}

alag ho jaata hai; constant

e^{- s a}

integral se factor out ho jaata hai, toh shifting hamesha is exponential stamp mein convert ho jaati hai.

Time ko

a

se scale karna,

L {f (a t)} = \frac{1}{a} F (s / a)

mein factor

\frac{1}{a}

kyun produce karta hai?

τ = a t

substitute karne se

d t

ki jagah

\frac{d τ}{a}

aa jaata hai (yahi

\frac{1}{a}

hai) aur

e^{- s t}

e^{- (s / a) τ}

ban jaata hai (yahi

s / a

hai) — stretch ka Jacobian hi transform ko rescale karta hai.

Laplace, impulses (

δ

) ko itni aasani se kyun handle karta hai jabki classical methods struggle karte hain?

Impulse algebra mein sirf ek finite factor

e^{- a s}

add karta hai, jabki classical methods ko impulse ke across solution ko slope mein jump conditions match karke haath se patch karna padta hai.

Initial conditions, end mein apply hone ki bajaye algebra ke andar kyun aate hain?

Derivative rule ka boundary term seedha transformed equation mein

- f (0)

(aur

- f^{'} (0)

) inject karta hai, toh ICs

Y (s)

mein start se hi bake in ho jaate hain bajaaye bad mein arbitrary constants fix karne ke.

Answer padhne se pehle partial fractions kyun chahiye?

Inversion table sirf standard shapes ki ek choti si list jaanti hai; partial fractions ek messy rational

Y (s)

ko exactly unhi shapes ke sum mein rewrite karta hai taaki har piece ko term-by-term invert kiya ja sake.

Hum

Y (s) = e^{- a s} H (s)

split karke pehle

H

invert kyun kar sakte hain, phir delay?

Kyunki

e^{- a s}

ek pure delay operator hai: chahe

H (s)

jo bhi invert kare, poora inverse wahi function hai shifted aur

u (t - a)

se gated — toh exponential ko khud kabhi invert nahi karna padta.

Edge cases

L {u (t - a) g (t - a)}

kya hoga jab

a = 0

Yeh

e^{0} G (s) = G (s)

mein reduce ho jaata hai; koi delay nahi toh step

u (t) = 1

for

t \geq 0

aur gate kuch nahi karta, ordinary transform recover ho jaata hai.

y^{'} + y = u (t - 2)

ke liye,

t = 2

par aur thoda pehle

y

exactly kya hai?

t < 2

ke liye forcing off hai aur

y \equiv 0

;

t = 2

par solution

0

se shuru hoti hai aur

1 - e^{- (t - 2)}

ki tarah

1

ki taraf relax honi lagti hai, toh switch par continuous value

0

hai.

y (t) = u (t - 2) (1 - e^{- (t - 2)})

long-term mein kis value ki taraf jaata hai, aur kyun?

Jab

t \to \infty

e^{- (t - 2)} \to 0

, toh

y \to 1

: system constant input level par settle ho jaata hai, woh steady state jahan

y^{'} = 0

y = u = 1

balance karta hai.

y^{''} + 3 y^{'} + 2 y = u (t - 1)

mein

H (s) = \frac{1}{s ( s + 1 ) ( s + 2 )}

ke saath,

s = 0

par residue final value

\frac{1}{2}

kyun deta hai?

\frac{1/2}{s}

term constant

\frac{1}{2}

mein invert hoti hai, aur kyunki exponential terms

e^{- t}, e^{- 2 t}

decay karte hain,

\frac{1}{2}

steady state hai — consistent with

2 y = 1 \Rightarrow y = \frac{1}{2}

Agar forcing

t = 0

se pehle turn on ho, yaani

a < 0

, toh method ka kya hoga?

One-sided transform sirf

t \geq 0

"dekhta" hai, toh negative

a

par switch

t = 0

par already fully on hai; uska effect

u (t - a)

term ki jagah initial conditions mein fold karna padega.

Agar system ka characteristic polynomial (denominator of

H

) ka ek repeated root ho, toh kya table directly apply hoti hai?

Directly nahi.

\frac{1}{( s + 1 ) ^{2}}

jaisa repeated factor

t e^{- t}

mein invert hota hai (moment property), toh partial fractions mein

\frac{1}{( s + 1 ) ^{2}}

term zaroor hona chahiye jiska inverse

t

ki extra power carry karta hai.

s

-plane mein

Re (s)

kahan hona chahiye ek aisi

y (t)

ke liye jo

e^{- t}

aur

e^{- 2 t}

pieces se bani ho?

Transform

Re (s) > - 1

ke liye converge karta hai (sabse daayein wala pole

s = - 1

par hai); dono poles left half-plane mein hain, exactly yahi wajah hai ki woh time-pieces decay karte hain badhte nahi.

Recall One-line self-test

Kya aap bina notes ke yeh bol sakte ho ki $e^{- a s}$ ek transform ke saath kya karta hai aur iske inverse mein kya do cheezein zaroor honi chahiye? Answer ::: $e^{- a s}$ ek signal ko time mein $a$ se delay karta hai; iske inverse mein shift $g (t - a)$ aur gate $u (t - a)$ dono zaroor hone chahiye.