Use the Laplace transform and the inverses to solve the given initial value problem
$$y''-2y'+5y=0 ; y(0)=1, y'(0)=3$$

Question

Use the Laplace transform and the inverses to solve the given initial value problem
$$y''-2y'+5y=0 ; y(0)=1, y'(0)=3$$

swissgirl · Answer

So $ L\{y"\}-2L\{y'\}+5L\{y\}=L\{0\}$

$s^2Y(s)-sy(0)-y'(0)-2(sY(s)-y(0))+5Y(s)=0$

$Y(s)(s^2-2s+5)=s+1$

$Y(s)=\large \frac{s+1}{s^2-2s+5}$

How do I continue from here?

bahrom7893 · Answer

the bottom becomes:
$$Y(s)=\frac{s+1}{(s^2-2s+1+4)}=\frac{s-(-1)}{(s-1)^2+2^2}$$You might be off by a sign on top. If that was s-1, this would fit into $$e^{at}Cos(bt)$$

bahrom7893 · Answer

http://www.sosmath.com/diffeq/laplace/table/table.html

bahrom7893 · Answer

where a=1, and b=2

swissgirl · Answer

THANKKKKSSS :)