What I want to ask is to solve this ODE with Laplace transform because I'm in trouble in the sense that I've solved it but the results aren't correct so I want to know when I'm wrong.
$2y''(t)+3y'(t)+1.5y(t)=0$
where:
$y(0)=0.2$
$y'(0)=-1$
I post some passages of my solution:
Laplace polynomial obtained:
$Y(s) = \frac{0.2s-0.4}{2s^2+3s+1.5} $
with poles in $s_1$ and $s_2$ exploited below at the denominator of $Y(s)$
Applying partial fraction expansion, I get:
$Y(s)= \frac{A}{s-(-0.75-0.43i)} + \frac{B}{s-(-0.75+0.43i)}$
where:
$A=0.05-0.319i$
$B=0.05+0.319i$
Applying inverse Laplace transform one get:
$y(t) = Me^{Re(s_1)t}cos(Im(s_1)t + \phi)$
where:
$M=2|A|=0.647$
$\phi = \tan^{-1}[\frac {\operatorname{Im}(A)}{\operatorname{Re}(A)}] = 1.4153$
At the end the solution can be represented by:
$y(t) = 0.647e^{-0.75t}cos(0.43t + 1.4153)$
The answers I can choose are:
$1) y(t) =1.514e^{-0.75t}\cos(0.43t - 1.438) $,
$2) y(t) =0.8327e^{-0.75t}\cos(0.43t + 1.328) $,
$3) y(t) =1.973e^{-0.75t}\cos(0.43t - 1.469) $,
4) the equation can't be solved by Laplace transform.
I got this for $Y(s)$: $$4y''(t)+6y'(t)+3y(t)=0$$ $$Y(s)(4s^2+6s+3)= 4sy(0)+4y'(0)+6y(0)$$ $$Y(s)=\dfrac { 0.8s-2.8}{(4s^2+6s+3)}$$