Non-Homogenous differential equation with complex roots and duality I have the following problem given:
$y^{(4)}+4y''+16y=b(x)$ and the solutions to the homogenous characteristic polynomial is $z_{1,2}=2i$ and $z_{3,4}=-2i$.
So, the first thing what I had to clarify in my head was that $z_{1,2}=2i$ and $z_{3,4}=-2i$ just means the solution $z=±2i$ appears twice.
$y_h$ hence should be $(1)1(c_cos(2x)+c_2sin(2x))+(1)(xc_3cos(2x)+c_4xsin(2x))$. We multiplied by x in the second half of the term because of duality.
So, back to the problem:
$b(x)=sin(2x)$: the normal trial solution should be $y(x)=acos(2x)+bcos(2x)$, however, since this solution is similar to the homogenous one we multiply both terms by x, but this is also similar to the second term of the homogenous solution. Therefore we multiply by x again leading to the solution: $y(x)=x^2(acos(2x)+bcos(2x))$. Is this alright until now?
$b(x)=e^xsin(2x)$: the trial solution should be $y(x)=de^x(acos(2x)+bcos(2x))$ and we can reformulate the constants $da$ and $db$ as $d_a$, $d_b$ leading to $y(x)=e^x(d_acos(2x)+d_bcos(2x))$. Right? Here we do not need to multiply by x again because we already multiplied by $e^x$ that aready contains an x.
$b(x)=xsin(2x)$: now, I have a problem with this one.The solution is given as $y(x)=x^2((a_0+a_1)sin(2x)+(b_0+b_1)cos(x))$ and I assume that there isa typo and it should be $y(x)=x^2((a_0+a_1)sin(2x)+(b_0+b_1)cos(*2*x))$. So, what I do not get is why we have to multiply by $x^2$ again? And not by x only, since we start with $xsin(2x)$ what appears once in the homogenous solution.
Can someone help?
 A: The ODE does not match the roots, it should likely be $y^{(4)} +8 y'' + 16 y = b(x)$, however we can proceed with the information given.
We are given repeated roots of $\pm 2 i$, thus the homogeneous solution can be written as
$$y_h = c_1 \sin(2x) + c_2 \cos(2x) + x(c_3 \sin(2x) + c_4 \cos(2x))$$
Case 1: We have $b(x) = \sin(2x)$, and because we have that in the homogeneous solution, we multiply by $x$, so we choose
$$y_p = x(x(a \sin(2x) + b  \cos(2x)))$$
Note: we did not need to include the $\sin(2x)$ and $\cos(2x)$ terms because that would just give us the same $x\sin(2x)$ and $x\cos(2x)$ terms we already have in the homogeneous solution.
Case 2: We have $b(x) = x \sin(2x)$, and because we have that in the homogeneous solution, we multiply by $x^2$, so we choose
$$y_p = x^2(a \sin(2x) + b \cos(2x) + x(c \sin(2x) + d \cos(2x)))$$
Note: in this case we do need to include the $\sin(2x)$ and $\cos(2x)$ terms because we do not have $x^2 \sin(2x)$ and $x^2 \cos(2x)$ in the homogeneous solution.
Case 3: We have $b(x) = e^x \sin(2x)$, and because we do not have that in the homogeneous solution, we choose
$$y_p = e^x(a \sin(2x) + b \cos(2x))$$
More details can be found regarding Undetermined Coefficients.
