Question

y'''' + 3y''' + 3y' + y = x*e^(-x) + x*cosx - 7 + x^(2) *e^(-x) *sinx

Answer 1

The corresponding homogeneous equation

$$y''' + 3y'' + 3y' + y =0$$

Characteristic (auxiliary) equation

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

$$(r+1)^3=0$$

$$r_1=r_2=r_3=-1$$

The general solution of the homogeneous differential equation is

$$y_h=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}$$

Find the partial solution of the nonhomogeneous differential equation

$$y''' + 3y'' + 3y' + y =xe^{-x} - 7$$ $$y_1(x)=x^3(Ax+B)e^{-x}+C$$

$$y_1'(x)=(-Ax^4-Bx^3+4Ax^3+3Bx^2)e^{-x}$$

$$y_1''(x)=(Ax^4+Bx^3-4Ax^3-3Bx^2)e^{-x}$$

$$+(-4Ax^3-3Bx^2+12Ax^2+6Bx)e^{-x}$$

$$y_1'''(x)=(-Ax^4-Bx^3+8Ax^3)e^{-x}$$

$$+(6Bx^2-12Ax^2-6Bx)e^{-x}$$

$$+(4Ax^3+3Bx^2-24Ax^2)e^{-x}$$

$$+(-12Bx+24Ax+6B)e^{-x}$$

Substitute

$$(-Ax^4-Bx^3+8Ax^3)e^{-x}$$

$$+(6Bx^2-12Ax^2-6Bx)e^{-x}$$

$$+(4Ax^3+3Bx^2-24Ax^2)e^{-x}$$

$$+(-12Bx+24Ax+6B)e^{-x}$$

$$+3(Ax^4+Bx^3-4Ax^3-3Bx^2)e^{-x}$$

$$+3(-4Ax^3-3Bx^2+12Ax^2+6Bx)e^{-x}$$

$$+3(-Ax^4-Bx^3+4Ax^3+3Bx^2)e^{-x}$$

$$+(Ax^4+Bx^3)e^{-x}+C =xe^{-x} - 7$$

$$x^4e^{-x}:0=0$$

$$x^3e^{-x}:0=0$$

$$x^2e^{-x}:0=0$$

$$x^1e^{-x}:24A=1$$

$$x^0e^{-x}:B=0$$

$$x^0:C=-7$$

The partial solution of the nonhomogeneous differential equation

$$y''' + 3y'' + 3y' + y =xe^{-x} - 7$$

is

$$y_1(x)=\dfrac{x^4e^{-x}}{24}$$

Find the partial solution of the nonhomogeneous differential equation

$$y''' + 3y'' + 3y' + y =x\cos x$$

$$y_2(x)=Ax\cos x+Bx\sin x+C\cos x+D\sin x$$

$$y_2'(x)=A\cos x-Ax\sin x+B\sin x+Bx\cos x$$

$$-C\sin x+D\cos x$$

$$y_2''(x)=-2A\sin x-Ax\cos x+2B\cos x$$

$$-Bx\sin x-C\cos x-D\sin x$$

$$y_2'''(x)=-3A\cos x+Ax\sin x-3B\sin x$$

$$-Bx\cos x+C\sin x-D\cos x$$

Substitute

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

$$-Bx\cos x+C\sin x-D\cos x$$

$$-6A\sin x-3Ax\cos x+6B\cos x$$

$$-3Bx\sin x-3C\cos x-3D\sin x$$

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

$$-3C\sin x+3D\cos x$$

$$+Ax\cos x+Bx\sin x+C\cos x+D\sin x$$

$$=x\cos x$$

$$x\cos x: 2B-2A=1$$

$$x\sin x: 2B+2A=0$$

$$\cos x: 2D+6B-2C=0$$

$$\sin x: -2C-6A-2D=0$$

$$A=-\dfrac{1}{4}, B=\dfrac{1}{4}, C=\dfrac{3}{4}, D=0$$

The partial solution of the nonhomogeneous differential equation

$$y''' + 3y'' + 3y' + y =x\cos x$$

is

$$y_2(x)=-\dfrac{1}{4}x\cos x+\dfrac{1}{4}x\sin x+\dfrac{3}{4}\cos x$$

Find the partial solution of the nonhomogeneous differential equation

$$y''' + 3y'' + 3y' + y =x^2 e^{-x}\sin x$$

$$y_3(x)=(Ax^2+Bx+C)e^{-x}(D\cos x+E\sin x)$$

$$y_3'(x)=(2Ax+B)e^{-x}(D\cos x+E\sin x)$$

$$-(Ax^2+Bx+C)e^{-x}(D\cos x+E\sin x)$$

$$+(Ax^2+Bx+C)e^{-x}(-D\sin x+E\cos x)$$

$$y_3''(x)=2Ae^{-x}(D\cos x+E\sin x)$$

$$-(4Ax+2B)e^{-x}(D\cos x+E\sin x)$$

$$+(4Ax+2B)e^{-x}(-D\sin x+E\cos x)$$

$$+2(Ax^2+Bx+C)e^{-x}(D\sin x-E\cos x)$$

$$y_3'''(x)=-6Ae^{-x}(D\cos x+E\sin x)$$

$$+6Ae^{-x}(-D\sin x+E\cos x)$$

$$+(12Ax+6B)e^{-x}(D\sin x-E\cos x)$$

$$-2(Ax^2+Bx+C)e^{-x}(D\sin x-E\cos x)$$

$$+2(Ax^2+Bx+C)e^{-x}(D\cos x+E\sin x)$$

After substitution we have

$$y_3(x)=(x^2\cos x-6x\sin x-12\cos x)e^{-x}$$

The general solution of the nonhomogeneous differential equation is

$$y=c_1e^{-x}+c_2xe^{-x}+c_3x^2e^{-x}$$ $$+\dfrac{x^4e^{-x}}{24}-\dfrac{1}{4}x\cos x+\dfrac{1}{4}x\sin x+\dfrac{3}{4}\cos x$$

$$+x^2e^{-x}\cos x-6xe^{-x}\sin x-12e^{-x}\cos x$$