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

Where it pays off:

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.

See Inverse Laplace transform for the inverse direction in detail.