Differential equations/Laplace transforms

Definition
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, $$L(\alpha f(t)+\beta g(t))=\alpha L(f(t))+\beta L(g(t))$$.

Solution
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.
 * 1) Take the Laplace transform of both sides ($$L(y''+2y'+y)=L(0)=0$$ ).
 * 2) Use the associative property to split the left side into terms ($$L(y'')+2L(y')+L(y)=0$$ ).
 * 3) Use the theorem $$L(y')=sL(y)-y(0)$$, and by extension, $$L(y'')=s^2L(y)-sy(0)-y'(0)$$ to modify the terms into scalars and multiples of $$L(y)$$ ($$s^2L(y)-sy(0)-y'(0)+2 \left [ sL(y)-y(0) \right ] +L(y)=0$$ ).
 * 4) Solve for the Laplacian ($$L(y)=\frac {s+3}{s^2+2s+1}=\frac {s+3}{(s+1)^2}$$ ).
 * 5) Take the inverse Laplace transform of both sides to get the solution, solving by method of partial fractions as needed: ($$L(y)=\frac {s+1+2}{(s+1)^2}=\frac {1}{s+1}+ \frac {2}{(s+1)^2}, y(t)=e^{-t}+2te^{-t}$$ ).
 * 6) For reference, here are some basic Laplace transforms:
 * 7) $$L(1)=\frac {1}{s}$$
 * 8) $$L(t^n)=\frac {n!}{s^{n+1}},n=1,2,3,\cdots$$
 * 9) $$L(e^{at})=\frac {1}{s-a}$$
 * 10) $$L(\sin at)=\frac {a}{s^2+a^2}$$
 * 11) $$L(\sinh at)=\frac {a}{s^2-a^2}$$
 * 12) $$L(\cos at)=\frac {s}{s^2+a^2}$$
 * 13) $$L(\cosh at)=\frac {s}{s^2-a^2}$$
 * 14) For reference, here are some theorems for the Laplace transforms:
 * 15) $$L(e^{at}f(t))=F(s-a)$$
 * 16) $$L(t \cdot f(t))=-F'(s)$$
 * 17) $$L(f^{(n)}(t))=s^n F(s)-s^{n-1}f(0)-s^{n-2}f(0)\cdots$$
 * 18) $$L(f(t-a) \cdot U(t-a)=e^{-as}F(s), a>0$$