Question

Using annihilator method obtain particular solution of y"'+y= cosx+sinx

Answer 1

Corresponding homogeneous differential equation

$$y'''+y=0$$

Characteristic equarion

$$r^3+1=0$$

$$(r+1)(r^2-r+1)=0$$

$$r_1=-1, r_2=\dfrac{1}{2}-\dfrac{\sqrt{3}}{2}i, r_3=\dfrac{1}{2}+\dfrac{\sqrt{3}}{2}i$$

The general solution of the homogeneous differential equation

$$y_h=c_1e^{-x}+c_2e^{x/2}\sin (\dfrac{\sqrt{3}}{2}x)+c_3e^{x/2}\cos (\dfrac{\sqrt{3}}{2}x)$$

$$(D^3+1)y=\cos x+\sin x$$

Find the particular solution of the nonhomogeneous differential equation

$$y_p=A\cos x+B\sin x$$

Then

$$y_p'=-A\sin x+B\cos x$$

$$y_p''=-A\cos x-B\sin x$$

$$y_p'''=A\sin x-B\cos x$$

Substitute

$$A\sin x-B\cos x+A\cos x+B\sin x=$$

$$=\cos x+\sin x$$

$$A+B=1$$

$$-B+A=1$$

$$A=1$$

$$B=0$$

The particular solution of the nonhomogeneous differential equation is

$$y_p=\cos x$$