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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.

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  • $\begingroup$ Your $Y(s)$ is not correct ...Alberto $\endgroup$ Mar 24 '20 at 19:09
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I got this for $Y(s)$: $$4y''(t)+6y'(t)+3y(t)=0$$ $$Y(s)(4s^2+6s+3)= 4sy(0)+4y'(0)+6y(0)$$ $$Y(s)=\dfrac { 0.8s-2.8}{(4s^2+6s+3)}$$

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  • $\begingroup$ You turn my friend. I saw the defect but no enough time to edit. Thanks for the warn! $\endgroup$
    – Mikasa
    Mar 24 '20 at 19:06
  • $\begingroup$ You're welcomed mrs.... @mrs $\endgroup$ Mar 24 '20 at 19:07
  • $\begingroup$ thank you very much, the procedure is correct ? @LostInSpace $\endgroup$ Mar 25 '20 at 8:29
  • $\begingroup$ @AlbertoLolli your procedure is correct you just have a mistake in the expression of $Y(s)$ $\endgroup$ Mar 25 '20 at 19:57

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