Laplace transforms

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School:Mathematics > Topic:Differential_Equations > Ordinary 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 $L(f(t))=\int e^{-st}f(t) dt=F(s)$ . 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.

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, $L(y'')=s^2L(y)-sy(0)-y'(0)$ to modify the terms into scalars and multiples of $L(y)$ ( $s^2L(y)-sy(0)-y'(0)+2 \left [ sL(y)-y(0) \right ] +L(y)=0$ ).
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:
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. $L(\cosh at)=\frac {s}{s^2-a^2}$
For reference, here are some theorems for the Laplace transforms:
1. $L(e^{at}f(t))=F(s-a)$
2. $L(t \cdot f(t))=-F'(s)$
3. $L(f^{(n)}(t))=s^n F(s)-s^{n-1}f(0)-s^{n-2}f(0)\cdots$
4. $L(f(t-a) \cdot U(t-a)=e^{-as}F(s), a>0$

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