Let [tex]\mathcal L_s\{y(t)\}=Y(s)[/tex] denote the Laplace transform of [tex]y(t)[/tex]. Recall that
[tex]\mathcal L_s\{y''(t)\}=s^2Y(s)-sy(0)-y'(0)[/tex]
[tex]\mathcal L_s\{y'(t)\}=sY(s)-y(0)[/tex]
[tex]\mathcal L_s\{\cos t\}=\dfrac s{s^2+1}[/tex]
[tex]\mathcal L_s\{\sin t\}=\dfrac1{s^2+1}[/tex]
Taking the transform of both sides yields
[tex]\bigg(s^2Y(s)-5s+4\bigg)-7\bigg(sY(s)-5\bigg)+10Y(s)=\dfrac{9s+7}{s^2+1}[/tex]
and solving for [tex]Y(s)[/tex] gives
[tex]Y(s)=\dfrac{\frac{9s+7}{s^2+1}+5s-39}{s^2-7s+10}[/tex]
[tex]Y(s)=\dfrac{5s^3-39s^2+14s-32}{(s^2+1)(s^2-7s+10)}[/tex]
[tex]Y(s)=\dfrac{5s^3-39s^2+14s-32}{(s^2+1)(s-5)(s-2)}[/tex]
[tex]Y(s)=-\dfrac4{s-5}+\dfrac8{s-2}+\dfrac s{s^2+1}[/tex]
Take the inverse transform and you're done:
[tex]y(t)=\mathcal L^{-1}_t\{Y(s)\}=-4e^{5t}+8e^{2t}+\cos t[/tex]
