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 $L(f(t))=\underset{0}{\overset{\mathrm{\infty }}{\int }}{e}^{-st}f(t)dt=F(s)$, 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: $y^{″}+2y^{\prime }+y=0,y(0)=1,y^{\prime }(0)=1$ . The Laplace transform usually is suited for equations with initial conditions.

1. Take the Laplace transform of both sides ($L(y^{″}+2y^{\prime }+y)=L(0)=0$ ).
2. Use the associative property to split the left side into terms ($L(y^{″})+2L(y^{\prime })+L(y)=0$ ).
3. Use the theorem $L(y^{\prime })=sL(y)-y(0)$ , and by extension, $L(y^{″})=s^{2}L(y)-sy(0)-y^{\prime }(0)$ to modify the terms into scalars and multiples of $L(y)$ ($s^{2}L(y)-sy(0)-y^{\prime }(0)+2[sL(y)-y(0)]+L(y)=0$ ).
   
   
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:
   ($L(y)=\frac{s+1+2}{(s+1{)}^2}=\frac{1}{s+1}+\frac{2}{(s+1{)}^2},y(t)=e^{-t}+2t{e}^{-t}$ ).
6. For reference, here are some basic Laplace transforms:
   1. $L(1)=\frac{1}{s}$
   2. $L(t^n)=\frac{n!}{s^{n+1}},n=1,2,3,\cdots$
   3. $L(e^{at})=\frac{1}{s-a}$
   4. $L(\sin at)=\frac{a}{s^2+a^2}$
   5. $L(\sinh at)=\frac{a}{s^2-a^2}$
   6. $L(\cos at)=\frac{s}{s^2+a^2}$
   7. $L(\cosh at)=\frac{s}{s^2-a^2}$
7. For reference, here are some theorems for the Laplace transforms:
   1. $L(e^{at}f(t))=F(s-a)$
   2. $L(t\cdot f(t))=-F^{\prime }(s)$
   3. $L(f^{(n)}(t))=s^{n}F(s)-s^{n-1}f(0)-s^{n-2}f(0)\cdots$
   4. $L(f(t-a)\cdot U(t-a)=e^{-as}F(s),a>0$