Trial solution for in-homogenous differential equation with complex numbers I am trying to solve following:
Now I have to find trial solutions for the particular solutions, however, I struggle here:
The solutions are given in the right order
I have tried to define the homogenous solution first and get to the solution: $y_h=c_0cos(2x)+c_1sin(2x)+c_3xcos(2x)+c_4sin(2x)$. And then I tried to determine the trial solutions taking the multiplicity of solutions into account by comparing. However, I don't get the solution provided above. Especially for the third one.
 A: $$y^{4} + 4y'' + 16y = b(x)$$
Since the complementary solution is:
$$y_h=c_0\cos(2x)+c_1\sin(2x)+c_3x\cos(2x)+c_4\sin(2x)$$
And given that: 
$$b(x)=x\sin 2x$$
We have to first choose a particular solution of the form:
$$y_p=(Ax+B)\sin 2x+ (Cx+D)\cos 2x$$
But since $\sin 2x $ is already part of the homogeneous solution we multiply the solution by x. Then we have $Bx \sin 2x $ and this is also part of the homogeneous solution so we have to multiply our guess by $x^2$. Hence we have:
$$y_p=x^2(Ax+B)\sin 2x+ x^2(Cx+D)\cos 2x$$
For your second question. The guess should be $y_p=at^2+bt+c$. No need to multpily by $t^2$ the guess. It's maybe just a mistake. Since we have:
$$(r^2+1)(r-1)^2=0$$
The differential equation is:
$$y^{(4)}-2y'''+2y''-2y'+y=0$$
The inhomogeneous equation is:
$$y^{(4)}-2y'''+2y''-2y'+y=2t^2+1$$
Here is the solution provided by Wolfram WA solution 
So that the particular solution is:
$$y_p=2t^2+8t+9$$
And the homogeneous solution is:
$$y_h=c_0e^x+c_1xe^x+c_3\cos(x)+c_4\sin(x)$$
It's a polynomial of degree two not four as written in your book.
A: $$y_t+4y_{t-2}+16y_{t-4}=b_t$$
You are given the solution $z=\pm 2i$ with multiplicity two. 
The module of z is $|z|=\sqrt {2^2+0^2}=2$. We can deduce the angle since we have $\dfrac {z}{|z|}=i\implies \phi = \dfrac {\pi} 2$. Therefore the solution to the homogeneous difference equation is:
$$y_h=\color{red}{2^t }\left(c_1 \cos(\frac {\pi}2t)+c_2 \sin(\frac {\pi}2t)+c_3 t\cos(\frac {\pi}2t)+c_4 t\sin(\frac {\pi}2t) \right)$$
You can deduce why you have to multiply by $t^2$ in the third option and why you don't have to multiply for the first and second option. Do you get it ?
