Polynomial annihilator method $y''+8y=5x+2e^{-x}$ $y''+8y=5x+2e^{-x}$
It is asked to solve the equation by this method.
My result was $c_1\sin(\sqrt8x)+c_2\cos(\sqrt8x)+5x+\frac{2}{9}e^{-x}$
Then it is aked a particular solution so that $y(0)=0$ and $y'(0)=-{1/15}$
I arrived to $c_2=-\frac{2}{9}$ and to $c_1=-\frac{218}{45}$
By the last result I think something might be wrong.
Thanks
 A: We are given
$$\tag 1 y'' + 8y = 5x + 2e^{-x}, y(0)=0, y'(0)=-\dfrac 1{15}$$
We are asked to use the Annihilator Method, but here are some clearer notes that I will follow.
For the homogeneous part of $(1)$, we have:
$$(D^2+8)y = 5x + 2e^{-x} \implies D^2 + 8 = 0 \implies D = \pm 2 \sqrt{2}~i$$
This gives us a homogeneous solution of:
$$y_h(x) = c_1 \cos(2 \sqrt{2} x) + c_2 \sin(2 \sqrt{2} x)$$
For the Right-Hand-Side (RHS) of $(1)$, we have the annihilator of $D^2(D+1)$, see notes if that is not clear. Applying this to both sides of the original differential 
equation we get:
$$D^2(D + 1)(D^2 + 8) y = 0 \implies D = \pm 2 \sqrt{2}~i, -1, 0, 0$$
The general solution to this homogeneous equation is:
$$y(x) = c_1 \cos(2 \sqrt{2} x) + c_2 \sin(2 \sqrt{2} x) + c_3 e^{-x} + c_4 + c_5 x$$
We eliminate the original homogeneous solution from this solution and that leaves our particular solution as:
$$y_p(x) = c_3 e^{-x} + c_4 + c_5 x$$
We now want to substitute this into $(1)$ to determine the coefficients and have:
$$y_p(x) = c_3 e^{-x} + c_4 + c_5 x, ~~y'_p(x) = -c_3 e^{-x} + c_5, ~~y''_p(x) = c_3e^{-x}$$
Substituting these into $(1)$ and equating terms yields:
$$c_3 e^{-x} + 8(c_3 e^{-x} + c_4 + c_5 x) = 9 c_3 e^{-x} + 8c_4 + 8 c_5 x = 2 e^{-x} + 5x$$
This gives $c_3 = \dfrac 29, c_4 = 0, c_5 = \dfrac 58$.
Thus our solution is:
$$y(x) = c_1 \cos(2 \sqrt{2} x) + c_2 \sin(2 \sqrt{2} x) + \dfrac 29 e^{-x} + \dfrac 58 x$$
Now, we are given the two initial conditions in $(1)$ to solve for $c_1$ and $c_2$.
From $y(0) = 0$, we have:
$$y(0) = c_1 + \dfrac 29 = 0 \implies c_1 = -\dfrac 29$$
From $y'(0)=-\dfrac 1{15}$, we need the derivative of the solution, which is:
$$y'(x) = -2 \sqrt{2} c_1 \sin(2 \sqrt{2} x) + 2 \sqrt{2} c_2 \cos(2 \sqrt{2} x) -\dfrac 29 e^{-x} + \dfrac 58$$
So, we have:
$$y'(0) = 2 \sqrt{2} c_2 -\dfrac 29 + \dfrac 58 = -\dfrac 1{15} \implies 2 \sqrt{2} c_2 = - \dfrac {169}{360} \implies c_2 = -\dfrac {169}{720 \sqrt{2}}$$
Thus, our final solution is:
$$y(x) = -\dfrac 29 \cos(2 \sqrt{2} x) -\dfrac {169}{720 \sqrt{2}} \sin(2 \sqrt{2} x) + \dfrac 29 e^{-x} + \dfrac 58 x$$
