$$x''+x=e^{-λt}\cos(μt),~~x(0)=A, x'(0)=B$$ λ, μ, A, B, all are constants. There are so many unknown constants in this problem. I am facing difficulty in solving this. Using Laplace transformation, $$X(s)=\frac{(s+λ)}{((s+λ)^2+μ^2)(s^2+1)}+\frac{(As+B)}{(s^2+1)}$$ I tried to solve this using partial fraction but failed. $$\frac{(s+λ)}{((s+λ)^2+μ^2)(s^2+1)}=\frac{(Cs+D)}{((s+λ)^2+μ^2)}+\frac{(Es+F)}{(s^2+1)}$$ $$(s+λ)=(Cs+D)(s^2+1)+(Es+F)((s+λ)^2+μ^2)$$ By equating the coefficients i got, $$C+E=0~~.~.~.~.~.~.~.~.~.~(i)$$ $$D+Eλ+F=0~~.~.~.~.~.~.~.~.~.~(ii)$$ $$C+E(λ^2+μ^2)+Fλ=1~~.~.~.~.~.~.~.~.~.~(iii)$$ $$D+F(λ^2+μ^2)=λ~~.~.~.~.~.~.~.~.~.~(iv)$$
Any ideas or suggestions to find out the solution to the differential equation, I am stuck.
$$X(s)=\frac{(s+λ)}{((1+λ)^2+μ)(s^2+1)}+\frac{(As+B)}{(s^2+1)}$$ The last fraction is easy to decompose and take inverse Laplace transform. It's simply the solution to the homogeneous differential equation ($x''+x=0$) : $$\mathcal L^{-1}\left (\frac{As+B}{s^2+1} \right )=A \cos t + B\sin t$$
The first fraction should give you the particular solution to the inhomogeneous differential equation.
It seems to me you made a little mistake. You should have at the denominator $(s+\lambda )^2+\mu^2$.
You have to discuss. There is a special case where $\lambda =0$ and $\mu =1$ in this case the fraction decomposition is different. You will have to decompose: $$G(s)=\dfrac {s}{(s^2+1)^2}$$ $$\implies g(t)=\dfrac 12 t \sin t$$