Method of Laplace transform

The method of Laplace transform solves linear ODEs (especially IVPs) by transforming the time-domain ODE into an algebraic equation in the s-domain, solving algebraically, then inverting back to time.

The procedure

Given an IVP $a{y}^{\prime \prime }+b{y}^{\prime }+cy=f(t)$ with $y(0)=y_0$, $y^{\prime }(0)=y_1$:

1. Take the Laplace transform of both sides. Use linearity and the derivative properties:

   • $\mathcal{L}\{y^{\prime }\}=sY(s)-y(0)$
   • $\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y(s)-sy(0)-y^{\prime }(0)$
   • $\mathcal{L}\{f(t)\}=F(s)$ (look up or compute)

2. Substitute initial conditions into the resulting algebraic expression.

3. Solve algebraically for $Y(s)$. Often involves partial fractions.

4. Invert the Laplace transform: $y(t)={\mathcal{L}}^{-1}\{Y(s)\}$.

5. Verify (a posteriori): check continuity, differentiability, and that the initial conditions are satisfied.

Why this works

The derivative property converts $y^{(n)}$ into $s^{n}Y(s)$ minus initial-condition terms. So a linear ODE with constant coefficients becomes a polynomial equation in $Y(s)$ — which is just algebra.

The initial conditions are automatically incorporated as the constants you subtract during transformation. No need to apply them at the end like with general-solution methods.

Advantages

The Laplace transform method shines when:

1. You only want the particular solution to an IVP — no need to find the homogeneous general solution first.
2. Forcing is discontinuous — handles step functions and impulses cleanly.
3. Some linear ODEs with variable coefficients — though this is more delicate.
4. Integral equations — Laplace transforms convolutions to products.

Compared to other methods:

• vs Method of undetermined coefficients: Laplace handles arbitrary forcing without needing to guess; but undetermined coefficients is faster for nice forcings.
• vs Method of variation of parameters: Laplace handles initial conditions natively; variation of parameters gives a general particular solution that doesn’t depend on initial conditions.

Worked example: simple IVP

Find ${\mathcal{L}}^{-1}\left\{\frac{2}{s^2+3s-4}\right\}$.

Partial fractions: $s^2+3s-4=(s+4)(s-1)$.

$\frac{2}{(s+4)(s-1)}=\frac{A}{s+4}+\frac{B}{s-1}$

$2=A(s-1)+B(s+4)$. Set $s=1$: $2=5B$, $B=2/5$. Set $s=-4$: $2=-5A$, $A=-2/5$.

Inverse:

${\mathcal{L}}^{-1}\left\{\frac{-2/5}{s+4}+\frac{2/5}{s-1}\right\}=-\frac{2}{5}{e}^{-4t}+\frac{2}{5}{e}^t$

Worked example: shifted denominator

Find ${\mathcal{L}}^{-1}\left\{\frac{1}{(s-1{)}^2}\right\}$.

By the first shifting theorem: ${\mathcal{L}}^{-1}\{F(s-a)\}=e^{at}{\mathcal{L}}^{-1}\{F(s)\}$.

Identify $F(s)=1/s^2$ shifted by $a=1$: $F(s-1)=1/(s-1{)}^2$.

${\mathcal{L}}^{-1}\{1/s^2\}=t$. So:

${\mathcal{L}}^{-1}\left\{\frac{1}{(s-1{)}^2}\right\}=e^t\cdot t=t{e}^t$

Worked example: full IVP with time shift

$y^{\prime \prime }+2y^{\prime }+y=4e^{\pi -t}$, $y(\pi )=2$, $y^{\prime }(\pi )=-1$.

The initial conditions are at $t=\pi$, but Laplace expects them at $t=0$. Shift: let $\tau =t-\pi$. Then $\frac{d}{dt}=\frac{d}{d\tau }$, and the forcing becomes $4e^{\pi -(\tau +\pi )}=4e^{-\tau }$.

The IVP in $\tau$: $y^{\prime \prime }+2y^{\prime }+y=4e^{-\tau }$, $y(\tau =0)=2$, $y^{\prime }(\tau =0)=-1$.

Take Laplace ($Y(s)=\mathcal{L}\{y(\tau )\}$). With $y(0)=2$ and $y^{\prime }(0)=-1$:

$\mathcal{L}\{y^{\prime \prime }\}=s^{2}Y-sy(0)-y^{\prime }(0)=s^{2}Y-2s-(-1)=s^{2}Y-2s+1$

$\mathcal{L}\{y^{\prime }\}=sY-y(0)=sY-2$

Substituting into the ODE:

$(s^{2}Y-2s+1)+2(sY-2)+Y=\frac{4}{s+1}$

$(s^2+2s+1)Y-2s-3=\frac{4}{s+1}$

$(s+1{)}^{2}Y=\frac{4}{s+1}+2s+3$

$Y(s)=\frac{4}{(s+1{)}^3}+\frac{2s+3}{(s+1{)}^2}$

Inverse:

For $\frac{4}{(s+1{)}^3}$: identify $a=-1$, $F(s)=4/s^3\to f(t)=2t^2$. So ${\mathcal{L}}^{-1}\{4/(s+1{)}^3\}=e^{-\tau }\cdot 2{\tau }^2=2{\tau }^2{e}^{-\tau }$.

For $\frac{2s+3}{(s+1{)}^2}$: rewrite as $\frac{2(s+1)+1}{(s+1{)}^2}=\frac{2}{s+1}+\frac{1}{(s+1{)}^2}$. Inverses: $2e^{-\tau }$ and $\tau e^{-\tau }$.

Sum: $y(\tau )=2{\tau }^2{e}^{-\tau }+2e^{-\tau }+\tau e^{-\tau }=e^{-\tau }(2{\tau }^2+\tau +2)$.

Convert back to $t$: $\tau =t-\pi$, so:

$y(t)=e^{-(t-\pi )}\left(2(t-\pi {)}^2+(t-\pi )+2\right)$

When inversion is hard

Often the hardest step is the inverse Laplace transform. Strategies:

1. Match against a table — recognize standard forms.
2. Partial fractions — decompose into simpler pieces, each invertible.
3. Complete the square — turn $s^2+2s+5$ into $(s+1{)}^2+4$, useful for the shifted-denominator forms.
4. First / second shifting theorems — for shifted versions of standard forms.

For the inverse direction in detail, see Inverse Laplace transform.