Question

. use the annihilator method to determinethe form of a particular solution for the given equation.

$y\text{'}\text{'}-6y\text{'}+9y=sin2x+x$

Short answer

$y_p\left(x\right)=c_1+c_{2}x+c_5\cos 2x+c_6\sin 2x$

Step-by-step solution

Solve the homogeneous of the given equation

The homogeneous of the given equation is

$(D^2-6D+9)\left[y\right]=0$

The solution of the homogeneous is

$y_h\left(x\right)=c_1{e}^{3x}+c_{2}x{e}^{3x}$ (1)

Now $\sin 2x+x$is annihilated by$\left(D^4+4D^2\right)$

Then, every solution to the given nonhomogeneous equation also satisfies

.$\left(D^4+4D^2\right)(D^2-6D+9)\left[y\right]=0$

Then

$y\left(x\right)=c_1+c_{2}x+c_3{e}^{3x}+c_{4}x{e}^{3x}+$ $c_5\cos 2x+c_6\sin 2x$ (2)

is the general solution to this homogeneous equation

We know$u\left(x\right)=u_h+u_p$

Comparing (1) & (2)

$y_p\left(x\right)=c_1+c_{2}x+c_5\cos 2x+c_6\sin 2x$