Intuition The big idea (WHY this exists)Solving an ODE means undoing differentiation, which is annoying. The Laplace transform is a machine that turns differentiation into multiplication by s s s . So a differential equation in t t t algebra equation in s s s f ′ ( t ) f'(t) f ′ ( t ) s F ( s ) sF(s) s F ( s ) f ( 0 ) f(0) f ( 0 ) − f ( 0 ) -f(0) − f ( 0 )
Definition Laplace transformFor a function f ( t ) f(t) f ( t ) t ≥ 0 t\ge 0 t ≥ 0 L { f ( t ) } = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t \mathcal{L}\{f(t)\} = F(s) = \int_0^{\infty} e^{-st} f(t)\, dt L { f ( t )} = F ( s ) = ∫ 0 ∞ e − s t f ( t ) d t s s s L { f } = F ( s ) \mathcal{L}\{f\}=F(s) L { f } = F ( s )
WHAT we need before deriving: the assumption that f f f e − s t f ( t ) → 0 e^{-st}f(t)\to 0 e − s t f ( t ) → 0 t → ∞ t\to\infty t → ∞ f f f
Derivation — integrate by parts.
L { f ′ ( t ) } = ∫ 0 ∞ e − s t f ′ ( t ) d t \mathcal{L}\{f'(t)\} = \int_0^\infty e^{-st} f'(t)\, dt L { f ′ ( t )} = ∫ 0 ∞ e − s t f ′ ( t ) d t
Why this step? This is just the definition applied to f ′ f' f ′
Use integration by parts ∫ u d v = u v − ∫ v d u \int u\,dv = uv - \int v\,du ∫ 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 ) . u = e^{-st} \implies du = -s\,e^{-st}\,dt, \qquad dv = f'(t)\,dt \implies v = f(t). u = e − s t ⟹ d u = − s e − s t d t , d v = f ′ ( t ) d t ⟹ v = f ( t ) .
Why this choice? We want to move the derivative off f f f d v dv d v v = f v=f v = f f ′ f' f ′
L { f ′ ( t ) } = [ e − s t f ( t ) ] 0 ∞ − ∫ 0 ∞ f ( t ) ( − s e − s t ) d t \mathcal{L}\{f'(t)\} = \Big[e^{-st} f(t)\Big]_0^\infty - \int_0^\infty f(t)\,(-s\,e^{-st})\,dt L { f ′ ( t )} = [ e − s t f ( t ) ] 0 ∞ − ∫ 0 ∞ f ( t ) ( − s e − s t ) d t
Why this step? Direct substitution into the by-parts formula.
Evaluate the boundary term:
At t → ∞ t\to\infty t → ∞ e − s t f ( t ) → 0 e^{-st}f(t)\to 0 e − s t f ( t ) → 0
At t = 0 t=0 t = 0 e 0 f ( 0 ) = f ( 0 ) e^{0}f(0) = f(0) e 0 f ( 0 ) = f ( 0 )
So [ e − s t f ( t ) ] 0 ∞ = 0 − f ( 0 ) = − f ( 0 ) \big[e^{-st}f(t)\big]_0^\infty = 0 - f(0) = -f(0) [ 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 ) \mathcal{L}\{f'(t)\} = -f(0) + s\int_0^\infty e^{-st} f(t)\, dt = sF(s) - f(0) L { f ′ ( t )} = − f ( 0 ) + s ∫ 0 ∞ e − s t f ( t ) d t = s F ( s ) − f ( 0 )
Why the last step? The remaining integral is exactly the definition of F ( s ) F(s) F ( s )
Intuition Read the formula like a sentence"s s s − f ( 0 ) -f(0) − f ( 0 ) price you pay for the lower limit t = 0 t=0 t = 0 — and conveniently it's where initial conditions live.
Derivation. Treat f ′ ′ = ( f ′ ) ′ f''=(f')' f ′′ = ( f ′ ) ′ g = f ′ g=f' g = f ′ L { g ′ ( t ) } = s L { g } − g ( 0 ) = s L { f ′ } − f ′ ( 0 ) \mathcal{L}\{g'(t)\} = s\,\mathcal{L}\{g\} - g(0) = s\,\mathcal{L}\{f'\} - f'(0) L { g ′ ( t )} = s L { g } − g ( 0 ) = s L { f ′ } − f ′ ( 0 )
Why this step? The rule works for ANY nice function, including f ′ f' f ′ g ( 0 ) = f ′ ( 0 ) g(0)=f'(0) g ( 0 ) = f ′ ( 0 )
Now substitute L { f ′ } = s F ( s ) − f ( 0 ) \mathcal{L}\{f'\} = sF(s)-f(0) 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 ) \mathcal{L}\{f''\} = s\big(sF(s) - f(0)\big) - f'(0) = s^2 F(s) - s f(0) - 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 y' + 3y = 0,\ \ y(0)=2 y ′ + 3 y = 0 , y ( 0 ) = 2 Step 1 — transform both sides. Let Y = L { y } Y=\mathcal{L}\{y\} Y = L { y } L { y ′ } + 3 L { y } = 0 ⟹ ( s Y − y ( 0 ) ) + 3 Y = 0 \mathcal{L}\{y'\} + 3\mathcal{L}\{y\} = 0 \implies \big(sY - y(0)\big) + 3Y = 0 L { y ′ } + 3 L { y } = 0 ⟹ ( s Y − y ( 0 ) ) + 3 Y = 0 Why this step? Replace y ′ y' y ′ s Y − y ( 0 ) sY-y(0) s Y − y ( 0 )
Step 2 — plug in the initial condition. y ( 0 ) = 2 y(0)=2 y ( 0 ) = 2 s Y − 2 + 3 Y = 0 ⟹ ( s + 3 ) Y = 2 sY - 2 + 3Y = 0 \implies (s+3)Y = 2 s Y − 2 + 3 Y = 0 ⟹ ( s + 3 ) Y = 2 Why this step? The "− f ( 0 ) -f(0) − f ( 0 )
Step 3 — solve algebra. Y = 2 s + 3 Y = \dfrac{2}{s+3} Y = s + 3 2
Step 4 — invert. Since L { e a t } = 1 s − a \mathcal{L}\{e^{at}\}=\frac{1}{s-a} L { e a t } = s − a 1 L − 1 { 2 s + 3 } = 2 e − 3 t \mathcal{L}^{-1}\{\frac{2}{s+3}\}=2e^{-3t} L − 1 { s + 3 2 } = 2 e − 3 t y ( t ) = 2 e − 3 t \boxed{y(t)=2e^{-3t}} y ( t ) = 2 e − 3 t Check: y ′ = − 6 e − 3 t y'=-6e^{-3t} y ′ = − 6 e − 3 t y ′ + 3 y = − 6 e − 3 t + 6 e − 3 t = 0 y'+3y=-6e^{-3t}+6e^{-3t}=0 y ′ + 3 y = − 6 e − 3 t + 6 e − 3 t = 0 y ( 0 ) = 2 y(0)=2 y ( 0 ) = 2
Worked example Second-order:y ′ ′ − y = 0 , y ( 0 ) = 0 , y ′ ( 0 ) = 1 y'' - y = 0,\ \ y(0)=0,\ y'(0)=1 y ′′ − y = 0 , y ( 0 ) = 0 , y ′ ( 0 ) = 1 Step 1. Transform: ( s 2 Y − s y ( 0 ) − y ′ ( 0 ) ) − Y = 0 \big(s^2Y - s y(0) - y'(0)\big) - Y = 0 ( s 2 Y − sy ( 0 ) − y ′ ( 0 ) ) − Y = 0 Why? Use the 2nd-derivative rule; both ICs enter at once.
Step 2. Insert y ( 0 ) = 0 , y ′ ( 0 ) = 1 y(0)=0,\ y'(0)=1 y ( 0 ) = 0 , y ′ ( 0 ) = 1 s 2 Y − 0 − 1 − Y = 0 ⟹ ( s 2 − 1 ) Y = 1 s^2 Y - 0 - 1 - Y = 0 \implies (s^2-1)Y = 1 s 2 Y − 0 − 1 − Y = 0 ⟹ ( s 2 − 1 ) Y = 1
Step 3. Y = 1 s 2 − 1 Y = \dfrac{1}{s^2-1} Y = s 2 − 1 1
Step 4. Recognise L { sinh t } = 1 s 2 − 1 \mathcal{L}\{\sinh t\} = \frac{1}{s^2-1} L { sinh t } = s 2 − 1 1 y ( t ) = sinh t y(t)=\sinh t y ( t ) = sinh t Check: y ′ ′ = sinh t = y y''=\sinh t = y y ′′ = sinh t = y y ( 0 ) = 0 y(0)=0 y ( 0 ) = 0 y ′ ( 0 ) = cosh 0 = 1 y'(0)=\cosh 0=1 y ′ ( 0 ) = cosh 0 = 1
Worked example Forced ODE:y ′ + y = 1 , y ( 0 ) = 0 y' + y = 1,\ y(0)=0 y ′ + y = 1 , y ( 0 ) = 0 Forecast: RHS is constant, system decays to equilibrium y = 1 y=1 y = 1 y ( t ) = 1 − e − t y(t)=1-e^{-t} y ( t ) = 1 − e − t Verify with Laplace. L { 1 } = 1 s \mathcal{L}\{1\}=\frac1s L { 1 } = s 1 ( s Y − 0 ) + Y = 1 s ⇒ ( s + 1 ) Y = 1 s (sY - 0) + Y = \frac1s \Rightarrow (s+1)Y=\frac1s ( s Y − 0 ) + Y = s 1 ⇒ ( s + 1 ) Y = s 1 Y = 1 s ( s + 1 ) = 1 s − 1 s + 1 Y = \frac{1}{s(s+1)} = \frac{1}{s} - \frac{1}{s+1} Y = s ( s + 1 ) 1 = s 1 − s + 1 1 y ( t ) = 1 − e − t y(t) = 1 - e^{-t} y ( t ) = 1 − e − t
Common mistake Forgetting the− f ( 0 ) -f(0) − f ( 0 ) Wrong belief: "L { f ′ } = s F ( s ) \mathcal{L}\{f'\}=sF(s) L { f ′ } = s F ( s ) feels right because differentiation ↔ multiply by s s s Why it's tempting: In the Fourier transform (no lower limit at 0) the boundary term truly vanishes, so people import that habit.
Fix: Laplace integrates from t = 0 t=0 t = 0 t = 0 t=0 t = 0 − f ( 0 ) -f(0) − f ( 0 ) s F ( s ) − f ( 0 ) sF(s)-f(0) s F ( s ) − f ( 0 )
Common mistake Mixing up the order of initial conditions inf ′ ′ f'' f ′′ Wrong: L { f ′ ′ } = s 2 F − f ′ ( 0 ) − s f ( 0 ) \mathcal{L}\{f''\}=s^2F - f'(0) - s f(0) L { f ′′ } = s 2 F − f ′ ( 0 ) − s f ( 0 ) Why tempting: Both terms are there, so order seems harmless.
Fix: Pair the higher power of s s s : s 2 f ( 0 ) s^2 f(0) s 2 f ( 0 ) s 2 F − s f ( 0 ) − f ′ ( 0 ) s^2F - s\,f(0) - f'(0) s 2 F − s f ( 0 ) − f ′ ( 0 ) f ( k ) ( 0 ) f^{(k)}(0) f ( k ) ( 0 ) s n − 1 − k s^{\,n-1-k} s n − 1 − k n = 2 n=2 n = 2 f ( 0 ) → s 1 f(0)\to s^1 f ( 0 ) → s 1 f ′ ( 0 ) → s 0 f'(0)\to s^0 f ′ ( 0 ) → s 0
Common mistake Forgetting ICs must be att = 0 t=0 t = 0 t = 0 t=0 t = 0 y ( 1 ) = … y(1)=\dots y ( 1 ) = … − f ( 0 ) -f(0) − f ( 0 ) 0 0 0
Recall Feynman: explain to a 12-year-oldImagine a recipe written in a hard language (calculus). The Laplace transform is like Google Translate that turns it into easy language (algebra/multiplication). In the easy language, "take a derivative" just becomes "multiply by s s s "where you started" — that's the − f ( 0 ) -f(0) − f ( 0 )
"s s s f ( 0 ) f(0) f ( 0 )
Each derivative you remove → one factor of s s s s s s s 2 F − s f ( 0 ) − f ′ ( 0 ) s^2F - s f(0) - f'(0) s 2 F − s f ( 0 ) − f ′ ( 0 )
What does the Laplace transform turn differentiation into? Multiplication by
s s s (plus subtracting initial values).
State L { f ′ ( t ) } \mathcal{L}\{f'(t)\} L { f ′ ( t )} s F ( s ) − f ( 0 ) sF(s) - f(0) s F ( s ) − f ( 0 ) .
State L { f ′ ′ ( t ) } \mathcal{L}\{f''(t)\} L { f ′′ ( t )} s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 ) s^2 F(s) - s\,f(0) - f'(0) s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 ) .
General L { f ( n ) } \mathcal{L}\{f^{(n)}\} L { f ( n ) } s n F ( s ) − s n − 1 f ( 0 ) − s n − 2 f ′ ( 0 ) − ⋯ − f ( n − 1 ) ( 0 ) s^nF(s) - s^{n-1}f(0) - s^{n-2}f'(0) - \cdots - f^{(n-1)}(0) 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 ) -f(0) − f ( 0 ) Integration by parts; the
t → ∞ t\to\infty t → ∞ end vanishes (exponential order) leaving
− e 0 f ( 0 ) = − f ( 0 ) -e^{0}f(0)=-f(0) − e 0 f ( 0 ) = − f ( 0 ) .
Where do the − f ( 0 ) -f(0) − f ( 0 ) − f ′ ( 0 ) -f'(0) − f ′ ( 0 ) The boundary at the lower limit
t = 0 t=0 t = 0 — they inject the initial conditions into the algebra.
In L { f ( n ) } \mathcal{L}\{f^{(n)}\} L { f ( n ) } f ( k ) ( 0 ) f^{(k)}(0) f ( k ) ( 0 ) − s n − 1 − k -s^{\,n-1-k} − s n − 1 − k .
Why must initial conditions be given at t = 0 t=0 t = 0 The derivative rule evaluates the boundary term exactly at
t = 0 t=0 t = 0 .
Solve y ′ + 3 y = 0 , y ( 0 ) = 2 y'+3y=0,\ y(0)=2 y ′ + 3 y = 0 , y ( 0 ) = 2 Y = 2 s + 3 ⇒ y = 2 e − 3 t Y=\frac{2}{s+3}\Rightarrow y=2e^{-3t} Y = s + 3 2 ⇒ y = 2 e − 3 t .
Why is dropping − f ( 0 ) -f(0) − f ( 0 ) You lose the initial condition; the solution won't satisfy
y ( 0 ) y(0) 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
Intuition Hinglish mein samjhoDekho, Laplace transform ka asli magic yahi hai: jab tum kisi function ka derivative ka Laplace lete ho, to differentiation "s s s L { f ′ ( t ) } = s F ( s ) − f ( 0 ) \mathcal{L}\{f'(t)\} = sF(s) - f(0) L { f ′ ( t )} = s F ( s ) − f ( 0 ) s F ( s ) sF(s) s F ( s ) s s s − f ( 0 ) -f(0) − f ( 0 ) − f ( 0 ) -f(0) − f ( 0 ) t = 0 t=0 t = 0 f ( 0 ) f(0) f ( 0 )
Iska fायda kya hai? ODE solve karna calculus me mushkil hai, par algebra easy hai. To hum poori differential equation ka Laplace le lete hain — woh ek simple algebra equation ban jaati hai s s s Y ( s ) Y(s) Y ( s ) y ( t ) y(t) y ( t ) − f ( 0 ) -f(0) − f ( 0 ) − f ′ ( 0 ) -f'(0) − f ′ ( 0 ) s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 ) s^2F(s) - s f(0) - f'(0) s 2 F ( s ) − s f ( 0 ) − f ′ ( 0 )
Sabse common galti: students − f ( 0 ) -f(0) − f ( 0 ) s F ( s ) sF(s) s F ( s ) f ( 0 ) f(0) f ( 0 ) s s s f ( 0 ) f(0) f ( 0 ) s s s
