Tried solving those for the past few days and could not get far enough to find the answers anyone can help with those?
Using the Laplace transform, find the the solution $y(t)$ of the ODE
$$2y''(t)+20y'(t)+32y(t)=2u(t)$$
where $y'(0)=0$, $y(0)=1$, and $u(t)$ is the unit step function.
Expected solution: $$y(t) = 0.0625 + 1.25e^{-2t} - 0.3125e^{-8t}.$$
We want to solve the IVP $$2y''(t)+20y'(t)+32y(t)=2H(t)$$ with conditions $y'(0)=0$ and $y(0)=1$ using Laplace transformations.
Begin by removing the factor of $2$ and taking the Laplace transform to get \begin{align*} \mathcal L\{y''(t)\}(s)+10\mathcal L\{y'(t)\}(s)+16\mathcal L\{y(t)\}(s)&=\mathcal L\{H(t)\}(s)\\ \left[s^2Y(s)-sy(0)-y'(0)\right]+10\left[sY(s)-y(0)\right]+16[Y(s)]&=\frac1s\\ s^2Y(s)-s+10sY(s)-10+16Y(s)&=\frac1s\\ \left(s^2+10s+16\right)Y(s)&=\frac1s+s+10\\ Y(s)&=\frac{s^2+10s+1}{s(s+8)(s+2)} \end{align*} Upon this result we can perform PFD to get $$Y(s)=\frac{s^2+10s+1}{s(s+8)(s+2)}=\frac1{16s}+\frac5{4(2+s)}-\frac5{16(8+s)}$$ Now, take the inverse Laplace transform to get \begin{align*} \mathcal{L}^{-1}\{Y(s)\}&=\frac1{16}\mathcal{L}^{-1}\left\{\frac1{s}\right\}+\frac{5}{4}\mathcal{L}^{-1}\left\{\frac1{2+s}\right\}-\frac5{16}\mathcal{L}^{-1}\left\{\frac1{8+s}\right\}\\ y(t)&=\boxed{\frac1{16}+\frac5{4}e^{-2t}-\frac{5}{16}e^{-8t}} \end{align*}