undetermined coefficient method question for second-order (particular solution) Here are two second-order differential equations.
$$ y''+9y=\sin(2t) \tag 1 $$
$$ y'' +4y =\sin(2t)  \tag 2 $$
I am told to use undetermine coefficients method to solve.
For 1), I use $y_p=A \cos(2t)+B \sin(2t)$ to get $A=0$ and B=$\frac{1}{5}$ and get $y_p=\frac{1}{5} \sin(2t)$
For 2), I realize that that method doesn't work and told to do $y_p=t(A \cos(2t)+B \sin(2t)$ Why does it work then?
 A: Realize that the right-hand side is a constant multiple of one of the solutions of the homogeneous equation
$$y_1=\cos(2t), \hspace{1.3cm} y_2=\sin(2t)$$
A: $$Y''+9Y=\sin 2t=f(t)~~~~~~(1)$$
We first solve $y''+9y=0$ to get $y_1,y_2=\sin 3t, \cos 3t$ and general solution as $y=C_1 \sin 3t+ C_2 \cos 3t$
But the solution of (1) is given by varying the parameters $C_1$ and $C_2$ as functions of $t$. Then
$$Y(t)=C_1(t) \sin 3t + C_2(t) \cos 3t$$
Where $$C_1(t)=-\int \frac{f(t)~ y_2(t) dt}{W(t)} +D_1.~~~ C_2(t)=\int \frac{f(t) ~y_1(t)dt}{W(t)}+D_2$$
Here $W(t)=[y_1 y_2'-y_1' y_2]$
Similarly, the solution of second inhomogeneous ODE can be obtained. $D_1,D_2$ are constant of integration as the ODE is second order. TYhis method is called variation of parameters or method of undetermined co-rfficients: $C_1(t), C_2(t).$
A: $$2)y'' +4y =\sin(2t)$$
in this case the characteristic polynomial is
$$r^2+4=0\implies r=\pm 2i$$
And the solution of the Homogeneous equation is:
$$y_p=c_1 \sin(2t)+c_2 \cos(2t)$$
Thats why, since $\sin(2t)$ is already part of the solution of the homogeneous equation, you need to choose for the particular solution:
$$y_p=t(A\sin(2t)+B \cos(2t))$$
In the first case, the solution of the homogeneous solution is:
$$y''+9y=\sin(2t)$$
$$ r^2+9=0 \implies r=\pm3i$$
$$y_h=c_1\cos(3t)+c_2\sin(3t)$$
No $\sin(2t)$ in the homogeneous solution.
So that the particular solution is simply of the form
$$y_p =A\sin(2t)$$
