Hello llllllllllll,
We are given the second order inhomogeneous ODE:
$$2y''+6y'+9y=\sin^2(x)$$
Using a power reduction formula, we may write:
$$2y′′+6y′+9y=\frac{1}{2}-\frac{1}{2}\cos(2x)$$
Observing that the differential operator:
$$A\equiv D\left(D^2+4 \right)$$
annihilates the RHS of the ODE, and that none of the roots of this operator are the characteristic roots of the associated homogenous ODE, we know the particular solution must take the form:
$$y_p(x)=A+B\cos(2x)+C\sin(2x)$$
Computing the derivatives of this solution, we find:
$$y_p'(x)=-2B\sin(2x)+2C\cos(2x)$$
$$y_p''(x)=-4B\cos(2x)-4C\sin(2x)$$
Substituting into the ODE, we obtain:
$$2\left(-4B\cos(2x)-4C\sin(2x) \right)+6\left(-2B\sin(2x)+2C\cos(2x) \right)+9\left(A+B\cos(2x)+C\sin(2x) \right)=\frac{1}{2}-\frac{1}{2}\cos(2x)$$
$$9A+(B+12C)\cos(2x)+(C-12B)\sin(2x)=\frac{1}{2}+\left(-\frac{1}{2} \right)\cos(2x)+0\cdot\sin(2x)$$
Equating coefficients, we obtain the system:
$$9A=\frac{1}{2}\implies A=\frac{1}{18}$$
$$B+12C=-\frac{1}{2}$$
$$C-12B=0\implies C=12B\implies B=-\frac{1}{290},\,C=-\frac{6}{145}$$
Thus, the particular solution is:
$$y_p(x)=\frac{1}{18}-\frac{1}{290}\cos(2x)-\frac{6}{145}\sin(2x)$$