How to show that thw ODE $y''+9y=\cos(\alpha t)$ has a solution in a specific form? I am supposed to show that the ODE $y''+9y=\cos(\alpha t), \alpha \in \mathbb{R}$ has a solution of the form
$$(a+bt)\cos(\alpha t)+(c+dt)\sin(\alpha t),$$
with $a,b,c,d \in \mathbb{R}$. In a rather arduous process (involving the matrix exponential), I managed to show that
$$\frac{\cos(\alpha t)}{9-\alpha^2}+K_1\cos(3t)+K_2\sin(3t)$$
solves the equation above, but I don't know how to transform this solution into the specific form.
I tried taking the derivative of the proposed solution and plugging it in the differential equation, but I am not sure how to proceed since the result is a system of two equations with four variables.
I haven't learnt about the Laplace transform yet. Any help would be appreciated!

$$y′′+9y=\cos(αt)$$
Solve the homogeneous equation
$$r^2+9=0 \implies (r-3i)(r+3i)=0 \implies r=\pm3i$$
$$y(t)=c_1\cos(3t)+c_2\sin(3t)$$
For the particular solution as mentionned in the comments you have to treat two cases $\alpha= \pm 3 $ and $\alpha \ne \pm 3$
For the case $\alpha \ne  \pm 3$ The particular solution is simply:
$$y_p=B\cos(\alpha t)$$
$$-B\alpha^2\cos(\alpha t)+9B\cos(\alpha t)=\cos(\alpha t)$$
$$-B\alpha^2+9B=1 \implies B=\frac 1 {9-\alpha^2}$$
And the particular is
$$y_p=\frac {\cos(\alpha t)} {9-\alpha^2}$$
$$y(t)=c_1\cos(3t)+c_2\sin(3t)+\frac {\cos(\alpha t)} {9-\alpha^2}$$
You still need to treat the other case $\alpha=3$.
