Answer to Question #205280 in Differential Equations for shreyas

Question

(D³ – 1)y = 3cos2x

Answer 1

$$y'''-y=3\cos(2x)$$

Homogeneous Equation

$$y'''-y=0$$

The characteristic (auxiliary) equation

$$r^3-1=0$$

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

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

$$y_h=c_1e^{x}+e^{-x/2}\bigg(c_2\cos(\dfrac{\sqrt{3}}{2}x)+c_3\sin(\dfrac{\sqrt{3}}{2}x)\bigg)$$

$$y_p=A\sin(2x)+B\cos(2x)$$

$$y_p'=2A\cos(2x)-2B\sin(2x)$$

$$y_p''=-4A\sin(2x)-4B\cos(2x)$$

$$y_p'''=-8A\cos(2x)+8B\sin(2x)$$

Then

$$-8A\cos(2x)+8B\sin(2x)$$

$$-A\sin(2x)-B\cos(2x)=3\cos(2x)$$

$$8B-A=0$$

$$-8A-B=3$$

$$A=-\dfrac{24}{65}$$

$$B=-\dfrac{3}{65}$$

Therefore

$$y(x)=c_1e^{x}+e^{-x/2}\bigg(c_2\cos(\dfrac{\sqrt{3}}{2}x)+c_3\sin(\dfrac{\sqrt{3}}{2}x)\bigg)$$

$$-\dfrac{24}{65}\sin(2x)-\dfrac{8}{65}\cos(2x)$$