I have an ordinary differential equation:
$$ \frac{\mathrm{d}^2u}{\mathrm{d}t^2} + u = \mathrm{e}^{-t}\cos(t)$$
with $u(0) = u_0$ and $\dot{u}(0) = v_0$, when using the method of undetermined coefficients, I obtain the solution
$$u(t) = u_0\cos(t) + v_0\sin(t) + \frac{\cos(t) -2\sin(t)}{5e^t}$$
however using the laplace transform:
$$(p^2 + 1)(\mathcal{L}[u(t)]) - pu_0 - v_0 = \frac{p+1}{(p+1)^2 +1}$$
using partial fractions, we have,
$$ \mathcal{L}[u(t)] = \frac{3}{5(p^2+1)}-\frac{p}{5(p^2+1)}-\frac{1}{5(p^2+2p+2)} \\ \, \, \, \, \, \, \, \,+\frac{p}{(p^2+2p+2)}+\frac{pu_0}{(p^2+1)}+\frac{v_0}{(p^2+1)}$$
completing the square and with each $(p^2+1)$ in the denominator expressing $p$ in the numerator as $(p+1)-1$, we have
$$u(t) = \mathcal{L}^{-1}\bigg[\frac{3}{5p^2 +5}-\frac{p}{5p^2 +5}-\frac{2}{5(p^2+1)+5} \\ +\frac{p+1}{5(p^2+1)+5}+\frac{pu_0}{(p^2+1)}+\frac{v_0}{(p^2+1)}\bigg]$$
which yields:
$$u(t) = \frac{-\cos(t)}{5}+ \frac{\cos(t)}{5\mathrm{e}^t} + \frac{i}{5}(\cos((1+i))t) - \cosh((1+i)t) \\+ i\sin((1+i)t)+\sinh((1+i)t)) + u_0\cos(t) + v_0\sin(t)$$
Which is not the same as the solution for the method of coefficients, can anyone explain to me why this is?