Ordinary Differential Equations

Laplace Transform

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Definition

The Laplace transform converts a function $f(x)$ into a function $F(p)$ in the $p$ domain:

$$\mathcal{L}[f(x)] = \int_0^\infty e^{-px} f(x)\, dx = F(p)$$

Its utility is to convert ODEs into algebraic equations, which are easier to solve.

1 · General procedure
How to solve an ODE with Laplace
Method
Step by step
1
Apply $\mathcal{L}$ to both sides of the ODE
2
Substitute initial conditions $y(0)$, $y'(0)$
3
Isolate $\mathcal{L}[y]$ algebraically
4
Decompose into partial fractions if needed
5
Apply $\mathcal{L}^{-1}$ using the table to obtain $y(x)$
2 · Key transforms
Transforms of derivatives
First derivative
$\mathcal{L}[y']$
$$\mathcal{L}[y'] = p\,\mathcal{L}[y] - y(0)$$
!
The derivative becomes multiplication by $p$ and the initial condition $y(0)$ is subtracted
Second derivative
$\mathcal{L}[y'']$
$$\mathcal{L}[y''] = p^2\,\mathcal{L}[y] - p\,y(0) - y'(0)$$
!
Each derivative adds a power of $p$ and subtracts an initial condition
Table of basic function transforms
$f(x)$ $F(p) = \mathcal{L}[f(x)]$ Validity condition
$1$$\dfrac{1}{p}$$p > 0$
$x$$\dfrac{1}{p^2}$$p > 0$
$x^n$$\dfrac{n!}{p^{n+1}}$$p > 0$
$e^{ax}$$\dfrac{1}{p-a}$$p > a$
$\sin ax$$\dfrac{a}{p^2+a^2}$$p > 0$
$\cos ax$$\dfrac{p}{p^2+a^2}$$p > 0$
$\sinh ax$$\dfrac{a}{p^2-a^2}$$p > a$
$\cosh ax$$\dfrac{p}{p^2-a^2}$$p > a$
Initial conditions not at $x=0$
Time shift
!
When to use it: when initial conditions are at $x = x_0$ instead of $x = 0$
1
Make the substitution $t = x - x_0$. Initial conditions shift to $t = 0$:
$y(x = x_0)$  →  $y(t = 0)$
$y'(x = x_0)$  →  $y'(t = 0)$
2
Solve the ODE for $t$ with Laplace (the ODE does not change, only the variable)
3
Substitute $t = x - x_0$ to return to the original variable $x$
3 · Shift theorem
Theorem
If $\mathcal{L}[f(x)] = F(p)$, then:
$$\mathcal{L}\left[e^{ax}f(x)\right] = F(p-a)$$
↔
Practical use: if you see $(p-a)^n$ where you would expect $p^n$, the inverse transform is the same function multiplied by $e^{ax}$
Examples
→
$\mathcal{L}^{-1}\!\left[\dfrac{1}{(p-3)^2}\right] = x\,e^{3x}$
→
$\mathcal{L}^{-1}\!\left[\dfrac{1}{p-3}\right] = e^{3x}$
4 · Partial fractions
When $\mathcal{L}[y]$ is a ratio of polynomials, it must be decomposed into simpler fractions that appear directly in the transform table.
How to decompose $\mathcal{L}[y]$ to apply $\mathcal{L}^{-1}$
Case 1
Simple factors
$$\frac{N(p)}{(p-a)(p-b)} = \frac{A}{p-a} + \frac{B}{p-b}$$
1
Multiplying by $(p-a)(p-b)$ we get:
$N(p) = A(p-b) + B(p-a)$
2
Substituting specific values of $p$ we get the constants:
$p = a$ → find $A$
$p = b$ → find $B$
Case 2
Repeated factor
$$\frac{N(p)}{(p-a)^2(p-b)} = \frac{A}{p-a} + \frac{B}{(p-a)^2} + \frac{C}{p-b}$$
1
Multiplying by $(p-a)^2(p-b)$ we get:
$N(p) = A(p-a)(p-b) + B(p-b) + C(p-a)^2$
2
Substituting specific values of $p$ we get the constants:
$p = a$ → find $B$
$p = b$ → find $C$
$p = 0$ (or any other) → find $A$