Answer to Question #294208 in Differential Equations for Khan

Question

Solve the differential equation (D^3+D^2+D+1)y=Sinx

Answer 1

Corresponding homogeneous equation

$$(D^3+D^2+D+1)y=0$$

Characteristic (auxiliary)equation

$$r^3+r^2+r+1=0$$

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

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

$$r_1=-1, r_2=i, r_3=-i$$

The general solution of the homogeneous differential equation is

$$y_h=c_1e^x+c_2\sin x+c_3\cos x$$

The particular solution of the non homogeneous equation is

$$y_p=Ax\sin x+Bx\cos x$$

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

$$y_p''=2A\cos x-Ax\sin x-2B\sin x-Bx\cos x$$

$$y_p'''=-3A\sin x-Ax\cos x-3B\cos x+Bx\sin x$$

Substitute

$$-3A\sin x-Ax\cos x-3B\cos x+Bx\sin x$$

$$+2A\cos x-Ax\sin x-2B\sin x-Bx\cos x$$

$$+A\sin x+Ax\cos x+B\cos x-Bx\sin x$$

$$+Ax\sin x+Bx\cos x=\sin x$$

$$x\sin x:B-A-B+A=0$$

$$x\cos x:-A-B+A+B=0$$

$$\sin x:-3A-2B+A=1$$

$$\cos x:-3B+2A+B=0$$

$$A=B=-\dfrac{1}{4}$$

The general solution of the given non homogeneous differential equation is

$$y=c_1e^x+c_2\sin x+c_3\cos x-\dfrac{x}{4}\sin x-\dfrac{x}{4}\cos x$$