we are given a $2$nd order d.e.
$$y'+4y+5 \int_{0}^x y dx = e^{-x}$$ when $y(0)=0$.
if we take the derivative of both sides we get :
$y''+4y'+5y=-e^{-x} $
we take the reduced equation:
$y''+4y'+5y=0$
and get an auxiliary equation
$D^2+4D+5=0$
from here we can easily get our general solution.
$Y_g(x)= C_1 e^{-2x} \cos (x)+ C_2 e^{-2x} \sin (x)$
we can also solve for a particular solution and it ends up being
$Y_p(x)=\frac{1}{2} e^{-x}$
now we are asked for a unique solution. However, we only have $y(0)=0$, $1 $ initial condition. How would we solve for the second initial condition ?
plug zero into :
$y''(x)+4y'(x)+5 y(x)=-e^{-x}$ ?