For some problems, the Laplace transform can convert the problem into a more solvable form. The Laplace transform equation is defined as $L(f(t))=\int _{0}^{\infty }e^{-st}f(t)dt=F(s)$, for the values of $s$ such that the integral exists. There are many properties of the Laplace transform that make it desirable to work with, such as linearity, or in other words, $L(\alpha f(t)+\beta g(t))=\alpha L(f(t))+\beta L(g(t))$ .

To illustrate how to solve a differential equation using the Laplace transform, let's take the following equation: $y''+2y'+y=0,y(0)=1,y'(0)=1$ . The Laplace transform usually is suited for equations with initial conditions.

Take the Laplace transform of both sides ($L(y''+2y'+y)=L(0)=0$ ).

Use the associative property to split the left side into terms ($L(y'')+2L(y')+L(y)=0$ ).

Use the theorem $L(y')=sL(y)-y(0)$ , and by extension, $L(y'')=s^{2}L(y)-sy(0)-y'(0)$ to modify the terms into scalars and multiples of $L(y)$ ($s^{2}L(y)-sy(0)-y'(0)+2\left[sL(y)-y(0)\right]+L(y)=0$ ).

Solve for the Laplacian ($L(y)={\frac {s+3}{s^{2}+2s+1}}={\frac {s+3}{(s+1)^{2}}}$ ).

Take the inverse Laplace transform of both sides to get the solution, solving by method of partial fractions as needed: ($L(y)={\frac {s+1+2}{(s+1)^{2}}}={\frac {1}{s+1}}+{\frac {2}{(s+1)^{2}}},y(t)=e^{-t}+2te^{-t}$ ).

For reference, here are some basic Laplace transforms:

$L(1)={\frac {1}{s}}$

$L(t^{n})={\frac {n!}{s^{n+1}}},n=1,2,3,\cdots$

$L(e^{at})={\frac {1}{s-a}}$

$L(\sin at)={\frac {a}{s^{2}+a^{2}}}$

$L(\sinh at)={\frac {a}{s^{2}-a^{2}}}$

$L(\cos at)={\frac {s}{s^{2}+a^{2}}}$

$L(\cosh at)={\frac {s}{s^{2}-a^{2}}}$

For reference, here are some theorems for the Laplace transforms: