Solving ODEs with Laplace (including discontinuous forcing)

1. The transform and its key property

WHY does this turn derivatives into algebra? Integrate by parts:

The boundary term $[e^{-st}f]_0^\infty$ vanishes at $\infty$ (decay) and gives $-f(0)$ at $0$ — that is where the initial conditions enter the algebra.

2. The recipe (HOW to solve any linear ODE)

Worked example Worked example 1 — smooth forcing

Solve $y''+y=0$ , $y(0)=0$ , $y'(0)=2$ .

Step 1 — transform. $\big(s^2Y - s\cdot0 - 2\big) + Y = 0$ . Why this step? The $-s\,y(0)-y'(0)$ pieces from $\mathcal{L}\{y''\}$ inject the ICs.

Step 2 — solve. $(s^2+1)Y = 2 \Rightarrow Y=\dfrac{2}{s^2+1}$ . Why this step? It's now pure algebra — divide.

Step 3 — invert. $\dfrac{2}{s^2+1}=2\cdot\dfrac{1}{s^2+1}\Rightarrow y(t)=2\sin t$ . Why this step? Match $\frac{\omega}{s^2+\omega^2}$ with $\omega=1$ . ✔ Check: $y(0)=0$ , $y'(0)=2$ . ✔

3. Discontinuous forcing: the Heaviside step

Special case $g=1$ : $\mathcal{L}\{u(t-a)\}=\dfrac{e^{-as}}{s}$ .

Figure — Solving ODEs with Laplace (including discontinuous forcing)

Worked example Worked example 2 — switched-on input

Solve $y'+y=u(t-2)$ , $y(0)=0$ .

Step 1. $sY+Y=\dfrac{e^{-2s}}{s}$ . Why? $\mathcal{L}\{u(t-2)\}=e^{-2s}/s$ , and $\mathcal{L}\{y'\}=sY-y(0)=sY$ .

Step 2. $Y=\dfrac{e^{-2s}}{s(s+1)}$ .

Step 3 — invert. First invert $H(s)=\dfrac{1}{s(s+1)}$ by partial fractions: $\dfrac{1}{s(s+1)}=\dfrac1s-\dfrac1{s+1}\Rightarrow h(t)=1-e^{-t}$ . Why? Cover-up: at $s=0$ residue $1$ , at $s=-1$ residue $-1$ . Then $Y=e^{-2s}H(s)\Rightarrow$ delay $h$ by $2$ and gate it: $\boxed{\,y(t)=u(t-2)\big(1-e^{-(t-2)}\big)\,}$ Meaning: nothing happens until $t=2$ , then $y$ relaxes toward $1$ . ✔

Worked example Worked example 3 — second order with a pulse

Solve $y''+3y'+2y=u(t-1)$ , $y(0)=0,\ y'(0)=0$ .

Step 1. $(s^2+3s+2)Y=\dfrac{e^{-s}}{s}$ . Step 2. $Y=\dfrac{e^{-s}}{s(s+1)(s+2)}$ . Step 3. Invert $H(s)=\dfrac{1}{s(s+1)(s+2)}=\dfrac{1/2}{s}-\dfrac{1}{s+1}+\dfrac{1/2}{s+2}$ Why these residues? Cover-up at $s=0,-1,-2$ gives $\frac1{(1)(2)},\frac1{(-1)(1)},\frac1{(-2)(-1)}=\tfrac12,-1,\tfrac12$ . So $h(t)=\tfrac12 - e^{-t}+\tfrac12 e^{-2t}$ , and $y(t)=u(t-1)\Big[\tfrac12 - e^{-(t-1)}+\tfrac12 e^{-2(t-1)}\Big].$

4. Common mistakes (Steel-man + fix)

5. Flashcards

What does the Laplace transform turn differentiation into?

Multiplication by

s

(with IC boundary terms):

\mathcal{L}\{f'\}=sF-f(0)

Why derive

\mathcal{L}\{f'\}=sF-f(0)

Integration by parts; the boundary term gives

-f(0)

, injecting initial conditions.

\mathcal{L}\{y''\}

in terms of

Y

s^2Y - s\,y(0) - y'(0)

\mathcal{L}\{u(t-a)g(t-a)\}=?

e^{-as}G(s)

(second shifting theorem).

\mathcal{L}\{u(t-a)\}=?

e^{-as}/s

Inverse of

e^{-as}G(s)

u(t-a)\,g(t-a)

— delay AND gate.

\mathcal{L}\{\delta(t-a)\}=?

e^{-as}

The three steps of the Laplace method?

Transform with ICs; solve algebraically for

Y(s)

; invert via partial fractions + table.

\mathcal{L}\{e^{at}\}=?

and why?

\frac{1}{s-a}

; from

\int_0^\infty e^{-(s-a)t}dt

Partial fraction of

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

\frac1s-\frac1{s+1}\Rightarrow 1-e^{-t}

Recall Feynman: explain to a 12-year-old

Solving a moving-equation by hand is like untying a tangled knot while it's still moving. The Laplace transform takes a photo that freezes the motion into a simple algebra picture (the land of " $s$ "). In picture-land, the knot becomes a straight line you can cut with normal arithmetic. When you're done, you "un-photo" it back into a moving answer. And if someone flips a switch at $t=2$ seconds, the photo just gets a tag $e^{-2s}$ that means "start this part 2 seconds late" — easy!

Concept Map

Hinglish (regional understanding)

Intuition Hinglish mein samjho

Dekho, Laplace transform ka basic idea simple hai: time-domain (variable $t$ ) mein jo differential equation solve karna mushkil hai, usko hum ek naye " $s$ " wale algebra-domain mein le jaate hain. Wahan derivative ka kaam sirf $s$ se multiply karna ho jaata hai — yaani $\mathcal{L}\{y'\}=sY-y(0)$ . Bonus ye ki initial conditions automatically formula ke andar ghus jaate hain ( $-y(0)$ jaise terms). Toh ODE ek normal algebraic equation ban jaati hai, jise hum $Y(s)$ ke liye solve karte hain, phir partial fractions laga ke table se wapas $t$ -domain mein le aate hain. Recipe yaad rakho: Stamp, Solve, Stick-back.

Sabse powerful part hai discontinuous forcing — jaise switch on hona at $t=a$ . Iske liye Heaviside step $u(t-a)$ use karte hain (jo $t<a$ pe $0$ , $t\ge a$ pe $1$ ). Second shifting theorem kehta hai: $\mathcal{L}\{u(t-a)g(t-a)\}=e^{-as}G(s)$ . Matlab time mein delay $a$ ka matlab $s$ -domain mein ek factor $e^{-as}$ . Inverse karte time do cheez yaad rakho: delay karo aur gate ( $u(t-a)$ ) lagao, warna answer switch se pehle bhi nonzero ho jayega — jo galat hai.

Classical methods (undetermined coefficients) sirf smooth forcing handle karte hain, lekin real life mein voltage on hota hai, hammer-blow ( $\delta$ impulse) lagta hai — yahin Laplace ka jaadu hai, kyunki impulse ka transform sirf $e^{-as}$ hai, bahut easy. Isliye engineering aur physics problems mein Laplace bahut zyada use hota hai.