ODE with Laplace transform

Question

I've found some difficulties solving the following ODEs, i think i'm following a wrong path when i do partial fraction expansion, so please show all passages.

$$\frac 3 2y''(t) - 2y'(t) + 4y(t) = u(t) \tag 1$$

$\newcommand{\eps}{\varepsilon}$ where : $u(t)=2\eps(t)$ ( $\eps$ = step function); $y(0)=0$; $y'(0)=0$

$$\frac 5 2 y''(t) - 2y'(t) + 5y(t) = u(t) \tag 2 $$

where : $u(t)=3\eps(t)$ ( $\eps$ = step function); $y(0)=0$; $y'(0)=0$

I hope someone can help me, tanks a lot

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This is my steps for the first equation :

Applying Laplace transform to the 1) i get the following polynomial:

$ Y(s) = \frac {2}{s(1.5s^2-2s+4)}$

it has 3 solutions(2 complex and one real)

$s_1=0$; $s_2=0.667+1.49i$; $s_3=0.667 - 1.49i$

i can now rewrite Y(s) as

$Y(s) = \frac {2}{s(s-0.667-1.49i)(s-0.667+1.49i)}$

Applying PFE i'm looking for solution of the form

$ \frac {A}{s} + \frac {B}{s-0.667-1.49i} + \frac {C}{s-0.667+1.49i} = \frac {2}{s(s-0.667-1.49i)(s-0.667+1.49i)}$

Substituting the solution s1,s2,s3 i have:

$s=s_1 \rightarrow A=0.7504 $

$s=s_2 \rightarrow B=-0.375-0.167i $

$s=s_3 \rightarrow C=-0.375+0.167i $

As an example, I report the calculation for B

$ B=(s-0.667+1.49i)Y(s) $

with $ s=0.667-1.49i $

so we have:

$ B= \frac{2}{(0.667-1.49i)(2*1.49i)} $

Now i have all my coefficients so i'm ready to apply the Inverse Laplace transform, as table suggest i've:

$ K=\lvert B\rvert = 0.4105$

$M=2K$

compute phase angle:

$\phi = \tan^{-1}(\frac {Im(B)}{Re(B)})$

finally i have :

$ y(t) = A + M e^{Re(s_2)t}cos(Im(s_2) + \phi) = 0.75 +0.821e^{0.667t}cos(1.49t+ 0.418) $

Answer 1

$$ay''+by'+cy=k\varepsilon(t)$$

Take the Laplace transform of both sides, recalling that $\mathcal L_s\{y(t)\}=\frac1s$ for $y(t)=\varepsilon(t)$, and $\mathcal L_s\{y''(t)\}=s^2\mathcal L_s\{y(t)\}$ if $y'(0)=y(0)=0$. Doing so gives

$$(as^2+bs+c)Y(s)=\frac ks\implies Y(s)=\frac k{s(as^2+bs+c)}$$

where $Y(s)=\mathcal L_s\{y(t)\}$ is the Laplace transform of $y(t)$.

Partial fraction decomposition yields

$$\begin{align*} \frac k{s(as^2+bs+c)}&=\frac{\theta_1}s+\frac{\theta_2s+\theta_3}{as^2+bs+c}\\[1ex] k&=\theta_1(as^2+bs+c)+(\theta_2s+\theta_3)s\\[1ex] k&=(\theta_1a+\theta_2)s^2+(\theta_1b+\theta_3)s+\theta_1c \end{align*}$$

$$\implies\begin{cases}\theta_1a+\theta_2=0\\\theta_1b+\theta_3=0\\\theta_1=\frac kc\end{cases}\implies\theta_1=\frac kc,\theta_2=-\frac{ak}c,\theta_3=-\frac{bk}c$$

and hence

$$Y(s)=\frac k{cs}-\frac{k(as+b)}{c(as^2+bs+c)}$$

Now take the inverse transform to solve for $y(t)$.

Answer 2

$$\frac 3 2y''(t) - 2y'(t) + 4y(t) = u(t) \tag 1$$ Take Laplace Transform: $$Y(s)(3s^2-4s+8)=\dfrac 4 s$$ $$Y(s)=\dfrac 4 {s(3s^2-4s+8)}$$ Decompose the fraction : $$2Y(s)=\dfrac {1}{s} -\dfrac {(3s-4)}{3s^2-4s+8}$$ For the denominator: $$\phi (s)=3s^2-4s+8$$ $$\phi (s)=3(s^2-\dfrac 4 3s+\dfrac 8 3)$$ $$\phi (s)=3((s- \dfrac2 3)^2+\dfrac {20} 9)$$ So that we have: $$2Y(s)=\dfrac {1}{s} -\dfrac {(3s-2-2)}{3((s- \dfrac2 3)^2+\dfrac {20} 9)}$$ $$2Y(s)=\dfrac {1}{s} -\dfrac {(s-2/3)}{((s- \dfrac2 3)^2+\dfrac {20} 9)}+\dfrac {2}{3((s- \dfrac2 3)^2+\dfrac {20} 9)}$$ Finally, taking inverse Laplace Transform gives us: $$y(t)=\dfrac 1 2 -\dfrac {e^{2t/3}}2 \left (\cos (\dfrac {2\sqrt {5}}{3}t)-\dfrac 1 {\sqrt 5}\sin (\dfrac {2\sqrt {5}}{3}t)\right )$$