# ODE with Laplace

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.

• Your $Y(s)$ is not correct ...Alberto Mar 24 '20 at 19:09

I got this for $$Y(s)$$: $$4y''(t)+6y'(t)+3y(t)=0$$ $$Y(s)(4s^2+6s+3)= 4sy(0)+4y'(0)+6y(0)$$ $$Y(s)=\dfrac { 0.8s-2.8}{(4s^2+6s+3)}$$
• @AlbertoLolli your procedure is correct you just have a mistake in the expression of $Y(s)$ Mar 25 '20 at 19:57