Question

Laplace method of differential equation when initial condition is not zero?
y''+2y'+3y=0
y(3)=2 y'(3)=37
just out of curiosity if anyone could show me how to solve this using laplace

Answer 1

In this situation, you must define a new function:
\[v(t)=y(t+t_0)~~\Rightarrow~~\begin{cases}v(t)=y(t+t_0)\\v'(t)=y'(t+t_0)\\v''(t)=y''(t+t_0)\end{cases}\]
So if you were given the general equation \(ay''+by'+c=g(t)\) (with initial conditions \(y(t_0)=y_0\) and \(y'(t_0)={y_0}'\)), then the transformation gives you \(av''+bv'+c=g(t+t_0)\) (with initial conditions \(v(0)=y(0+t_0)=y_0\) and \(v'(0)=y'(0+t_0)={y_0}'\)). So,
\[y''+2y'+3y=0~~\iff~~v''+2v'+3v'=0\]
where \(g(t)=g(t+t_0)=0\).
\[\begin{align*}v''+2v'+3v&=0\\
\mathcal{L}\left\{v''+2v'+3v\right\}&=0\\
\left(s^2\mathcal{L}\left\{v\right\}-sv(0)-v'(0)\right)+2\left(s\mathcal{L}\left\{v\right\}-v(0)\right)+3\mathcal{L}\left\{v\right\}&=0\\
\left(s^2+2s+3\right)\mathcal{L}\left\{v\right\}-sv(0)-v'(0)-2v(0)&=0\\
\left(s^2+2s+3\right)\mathcal{L}\left\{v\right\}-2s-37-4&=0\\
\mathcal{L}\left\{v\right\}&=\frac{2s+41}{s^2+2s+3}\end{align*}\]
After you find \(v(t)\), keep in mind that you're looking for a solution for \(y(t)\), so remember to back-substitute:
\[v(t)=y(t+t_0))~~\iff~~v(t-t_0)=y(t)\]

Answer 2

could you elaborate on the substitution of sv(0) and -v'(0) where s(v(0)) becomes +2s and v'(0) to +3

Answer 3

\[sv(0)=sy(0+t_0)=sy(t_0)=sy(3)=2s\]
\[v'(0)=y'(0+t_0)=y'(t_0)=y'(3)=37\]
I was just plugging in the given conditions.

Answer 4

Or do you mean how I got the term \((\cdots)\mathcal{L}\{y\}\)?

Answer 5

ohh... that made sense, I miss read it.

thanks, I'm going to have to look more into the substitution you made