Recipe

Solve an IVP with impulse forcing for Y(s)

Use when: the forcing term $g(t)$ in $ay''+by'+cy=g(t)$ contains a Dirac delta $\delta(t-t_0)$ , alone or added to other forcing

Goal: the algebraic equation solved for $Y(s)$ , with every delta term correctly transformed, ready for whichever inversion technique fits its shape

Taught in The Dirac delta and impulse forcing

Quick recipe

Transform every delta term in the forcing
Transform the rest of the equation with the derivative rule
Combine into one algebraic equation and solve for $Y(s)$

1
For each term of the form $a\,\delta(t-t_0)$ in $g(t)$ , replace it with $a\,e^{-t_0s}$ — pull the constant $a$ out by linearity, then apply $\mathcal{L}\{\delta(t-t_0)\}=e^{-t_0s}$ .

Example

For $3\delta(t-2)$ : pull out the constant $3$ , then apply the rule with $t_0=2$ : $3\mathcal{L}\{\delta(t-2)\}=3e^{-2s}$ .

⚡ Shortcut

An impulse at $t_0=0$ (forcing is just $\delta(t)$ , or $a\delta(t)$ ) transforms to $a\cdot e^{0}=a$ — no exponential factor to carry through the rest of the algebra at all.

Why: $\delta(t-t_0)$ isn't an ordinary function, so none of the earlier chapters' rules (power, exponential, sine/cosine) apply to it — it needs its own transform rule, applied directly.
2
Transform every other term of the ODE using §6's derivative rule, which builds the initial conditions $y(0)$ and $y'(0)$ directly into the transforms of $y'$ and $y''$ .

Example

For $y''+4y$ with $y(0)=0,y'(0)=0$ : $\mathcal{L}\{y''\}=s^2Y(s)-sy(0)-y'(0)=s^2Y(s)$ , and $\mathcal{L}\{4y\}=4Y(s)$ .
- both initial conditions are $0$ (rest initial conditions, the usual setup for an impulse response) — every derivative term simplifies to a bare power of $s$ times $Y(s)$ , with nothing extra to carry.
- at least one initial condition is nonzero — the extra constant terms from §6's rule survive and must be carried through Step 3 along with the transformed delta term.
Why: the delta term needs its own rule from Step 1, but the equation's other terms are ordinary derivatives, transformed exactly as every earlier IVP in this course was.
3
Add the transformed pieces together to get one equation with only $Y(s)$ , $s$ , and constants — no $t$ anywhere — then solve for $Y(s)$ by ordinary algebra.

Example

Combining Steps 1-2: $s^2Y(s)+4Y(s) = 3e^{-2s}$ , so $(s^2+4)Y(s)=3e^{-2s}$ , giving $Y(s)=\dfrac{3e^{-2s}}{s^2+4}$ .

Why: every term on both sides is now an algebraic expression in $s$ — exactly the "differential equation becomes algebra" promise this whole chapter is built on.

Common mistakes

Trying to transform $\delta(t-t_0)$ with the exponential or table rules from earlier sections instead of its own transform rule.
Forgetting the $y(0)$ and $y'(0)$ terms that §6's derivative rule attaches to $\mathcal{L}\{y'\}$ and $\mathcal{L}\{y''\}$ , especially when they're nonzero.
Dropping the $e^{-t_0s}$ factor's exponent sign, or its $t_0$ value, while combining terms in Step 3.
Trying to invert $Y(s)$ back to $y(t)$ inside this recipe — that's a separate step (§12) this recipe deliberately stops short of.