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