4.6.28
Intuition Badi picture (WHY yeh exist karta hai)ODE solve karna matlab differentiation ko undo karna, jo ki kaafi jhanjhat ka kaam hai. Laplace transform ek aisi machine hai jo differentiation ko s . Toh t algebra equation ban jaata hai s f ′ ( t ) s F ( s ) f ( 0 ) − f ( 0 )
Definition Laplace transformEk function f ( t ) t ≥ 0 L { f ( t )} = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t s L { f } = F ( s )
Derive karne se pehle humein kya chahiye: yeh assumption ki f e − s t f ( t ) → 0 t → ∞ f
Derivation — integrate by parts.
L { f ′ ( t )} = ∫ 0 ∞ e − s t f ′ ( t ) d t
Yeh step kyun? Yeh sirf definition hai jo f ′
Use integration by parts ∫ u d v = uv − ∫ v d u u = e − s t ⟹ d u = − s e − s t d t , d v = f ′ ( t ) d t ⟹ v = f ( t ) .
Yeh choice kyun? Hum derivative ko f d v v = f f ′
L { f ′ ( t )} = [ e − s t f ( t ) ] 0 ∞ − ∫ 0 ∞ f ( t ) ( − s e − s t ) d t
Yeh step kyun? By-parts formula mein direct substitution hai.
Boundary term evaluate karo:
t → ∞ e − s t f ( t ) → 0 t = 0 e 0 f ( 0 ) = f ( 0 )
Toh [ e − s t f ( t ) ] 0 ∞ = 0 − f ( 0 ) = − f ( 0 )
L { f ′ ( t )} = − f ( 0 ) + s ∫ 0 ∞ e − s t f ( t ) d t = s F ( s ) − f ( 0 )
Last step kyun? Bacha hua integral exactly F ( s )
Intuition Formula ko ek sentence ki tarah padho"s − f ( 0 ) price hai jo aap lower limit t = 0 — aur convenient baat yeh hai ki initial conditions wahan hi rehti hain.
Derivation. f ′′ = ( f ′ ) ′ g = f ′ L { g ′ ( t )} = s L { g } − g ( 0 ) = s L { f ′ } − f ′ ( 0 )
Yeh step kyun? Rule kisi bhi nice function par kaam karta hai, including f ′ g ( 0 ) = f ′ ( 0 )
Ab L { f ′ } = s F ( s ) − f ( 0 ) L { f ′′ } = s ( s F ( s ) − f ( 0 ) ) − f ′ ( 0 ) = s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 )
y ′ + 3 y = 0 , y ( 0 ) = 2 Step 1 — dono sides transform karo. Maano Y = L { y } L { y ′ } + 3 L { y } = 0 ⟹ ( s Y − y ( 0 ) ) + 3 Y = 0 Yeh step kyun? y ′ s Y − y ( 0 )
Step 2 — initial condition daalo. y ( 0 ) = 2 s Y − 2 + 3 Y = 0 ⟹ ( s + 3 ) Y = 2 Yeh step kyun? "− f ( 0 )
Step 3 — algebra solve karo. Y = s + 3 2
Step 4 — invert karo. Kyunki L { e a t } = s − a 1 L − 1 { s + 3 2 } = 2 e − 3 t y ( t ) = 2 e − 3 t Check: y ′ = − 6 e − 3 t y ′ + 3 y = − 6 e − 3 t + 6 e − 3 t = 0 y ( 0 ) = 2
Worked example Second-order:y ′′ − y = 0 , y ( 0 ) = 0 , y ′ ( 0 ) = 1 Step 1. Transform karo: ( s 2 Y − sy ( 0 ) − y ′ ( 0 ) ) − Y = 0 Kyun? 2nd-derivative rule use karo; dono ICs ek saath aate hain.
Step 2. y ( 0 ) = 0 , y ′ ( 0 ) = 1 s 2 Y − 0 − 1 − Y = 0 ⟹ ( s 2 − 1 ) Y = 1
Step 3. Y = s 2 − 1 1
Step 4. Pehchano ki L { sinh t } = s 2 − 1 1 y ( t ) = sinh t Check: y ′′ = sinh t = y y ( 0 ) = 0 y ′ ( 0 ) = cosh 0 = 1
Worked example Forced ODE:y ′ + y = 1 , y ( 0 ) = 0 Forecast: RHS constant hai, system equilibrium y = 1 y ( t ) = 1 − e − t Laplace se verify karo. L { 1 } = s 1 ( s Y − 0 ) + Y = s 1 ⇒ ( s + 1 ) Y = s 1 Y = s ( s + 1 ) 1 = s 1 − s + 1 1 y ( t ) = 1 − e − t
− f ( 0 ) Galat belief: "L { f ′ } = s F ( s ) sahi lagta hai kyunki differentiation ↔ multiply by s Tempting kyun lagta hai: Fourier transform mein (t = 0 Fix: Laplace t = 0 t = 0 − f ( 0 ) s F ( s ) − f ( 0 )
f ′′ Galat: L { f ′′ } = s 2 F − f ′ ( 0 ) − s f ( 0 ) Tempting kyun lagta hai: Dono terms toh hain hi, toh order harmless lagta hai.
Fix: Higher power of s : concretely: s 2 F − s f ( 0 ) − f ′ ( 0 ) f ( k ) ( 0 ) s n − 1 − k n = 2 f ( 0 ) → s 1 f ′ ( 0 ) → s 0
t = 0 t = 0 y ( 1 ) = … − f ( 0 ) 0
Recall Feynman: 12-saal ke bachche ko samjhaoSocho ek recipe ek mushkil language mein likhi hai (calculus). Laplace transform Google Translate jaisi hai jo use easy language (algebra/multiplication) mein badal deti hai. Easy language mein "derivative lo" bas "s "tum kahan se shuru hue the" — woh hai − f ( 0 )
"s f ( 0 )
Jitne derivatives hatao → utne factors of s s s 2 F − s f ( 0 ) − f ′ ( 0 )
What does the Laplace transform turn differentiation into? Multiplication by s
State L { f ′ ( t )} s F ( s ) − f ( 0 )
State L { f ′′ ( t )} s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 )
General L { f ( n ) } s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f ′ ( 0 ) − ⋯ − f ( n − 1 ) ( 0 )
Which technique derives the first-derivative rule, and why does the boundary term reduce to − f ( 0 ) Integration by parts; the t → ∞ − e 0 f ( 0 ) = − f ( 0 )
Where do the − f ( 0 ) − f ′ ( 0 ) The boundary at the lower limit t = 0
In L { f ( n ) } f ( k ) ( 0 ) − s n − 1 − k
Why must initial conditions be given at t = 0 The derivative rule evaluates the boundary term exactly at t = 0
Solve y ′ + 3 y = 0 , y ( 0 ) = 2 Y = s + 3 2 ⇒ y = 2 e − 3 t
Why is dropping − f ( 0 ) You lose the initial condition; the solution won't satisfy y ( 0 )
turns diff into multiply by s
First derivative rule sF minus f0
General nth derivative rule
Initial conditions f0 f prime 0
Solve then invert transform
