Question

Solve the initial value problem using Laplace transforms.

Answer 1

Let $\mathcal L_s\{y(t)\}=Y(s)$ denote the Laplace transform of $y(t)$. Recall that

$\mathcal L_s\{y''(t)\}=s^2Y(s)-sy(0)-y'(0)$
$\mathcal L_s\{y'(t)\}=sY(s)-y(0)$
$\mathcal L_s\{\cos t\}=\dfrac s{s^2+1}$
$\mathcal L_s\{\sin t\}=\dfrac1{s^2+1}$

Taking the transform of both sides yields

$\bigg(s^2Y(s)-5s+4\bigg)-7\bigg(sY(s)-5\bigg)+10Y(s)=\dfrac{9s+7}{s^2+1}$

and solving for $Y(s)$ gives

$Y(s)=\dfrac{\frac{9s+7}{s^2+1}+5s-39}{s^2-7s+10}$
$Y(s)=\dfrac{5s^3-39s^2+14s-32}{(s^2+1)(s^2-7s+10)}$
$Y(s)=\dfrac{5s^3-39s^2+14s-32}{(s^2+1)(s-5)(s-2)}$
$Y(s)=-\dfrac4{s-5}+\dfrac8{s-2}+\dfrac s{s^2+1}$

Take the inverse transform and you're done:

$y(t)=\mathcal L^{-1}_t\{Y(s)\}=-4e^{5t}+8e^{2t}+\cos t$