Question #268519

a.) 2y′′′−7y′′+12y′+8y=0

b.) y"-2y'+5y=e^xsinx

c.) y′' + y = 2xsinx


1
Expert's answer
2021-11-22T19:27:19-0500

a)


2y7y+12y+8y=02y'''-7y''+12y'+8y=0

Auxiliary equation (or characteristic equation)


2r37r2+12r+8=02r^3-7r^2+12r+8=0

2r2(r+12)8r(r+12)+16(r+12)=02r^2(r+\dfrac{1}{2})-8r(r+\dfrac{1}{2})+16(r+\dfrac{1}{2})=0

2(r+12)(r24r+8)=02(r+\dfrac{1}{2})(r^2-4r+8)=0

2(r+12)((r2)2+4)=02(r+\dfrac{1}{2})((r-2)^2+4)=0

r1=12,r2=22i,r3=2+2ir_1=-\dfrac{1}{2}, r_2=2-2i, r_3=2+2i

The general solution of the homogeneous differential equation is


y=c1ex/2+e2x(c2cos(2x)+c3sin(2x))y=c_1e^{-x/2}+e^{2x}(c_2\cos(2x)+c_3\sin(2x))

b)


y2y+5y=exsinxy''-2y'+5y=e^x\sin x

The corresponding homogeneous differential equation is


y2y+5y=0y''-2y'+5y=0

Auxiliary equation (or characteristic equation)


r22r+5=0r^2-2r+5=0

(r1)2=4(r-1)^2=-4

r1=12i,r2=1+2ir_1=1-2i, r_2=1+2i

The general solution of the homogeneous differential equation is


yh=ex(c1cos(2x)+c2sin(2x))y_h=e^{x}(c_1\cos(2x)+c_2\sin(2x))

Find the particular solution of the nonhomogeneous differential equation in form


yp=Aexcosx+Bexsinxy_p=Ae^x\cos x+Be^x\sin x

yp=AexcosxAexsinx+Bexsinx+Bexcosxy_p'=Ae^x\cos x-Ae^x\sin x+Be^x\sin x+Be^x\cos x

yp=Aexcosx2AexsinxAexcosxy_p''=Ae^x\cos x-2Ae^x\sin x-Ae^x\cos x

+Bexsinx+2BexcosxBexsinx+Be^x\sin x+2Be^x\cos x-Be^x\sin x

Substitute


2Aexsinx+2Bexcosx2Aexcosx-2Ae^x\sin x+2Be^x\cos x-2Ae^x\cos x

+2Aexsinx2Bexsinx2Bexcosx+2Ae^x\sin x-2Be^x\sin x-2Be^x\cos x

+5Aexcosx+5Bexsinx=exsinx+5Ae^x\cos x+5Be^x\sin x=e^x\sin x

3B=13B=1

A=0A=0


yp=13exsinxy_p=\dfrac{1}{3}e^x\sin x

The general solution of the nonhomogeneous differential equation is


y=ex(c1cos(2x)+c2sin(2x))+13exsinxy=e^{x}(c_1\cos(2x)+c_2\sin(2x))+\dfrac{1}{3}e^x\sin x

c)


y+y=2xsinxy'' + y = 2x\sin x

The corresponding homogeneous differential equation is


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

Auxiliary equation (or characteristic equation)


r2+1=0r^2+1=0

r1=i,r2=ir_1=i, r_2=i

The general solution of the homogeneous differential equation is


yh=c1cosx+c2sinxy_h=c_1\cos x+c_2\sin x

Find the particular solution of the nonhomogeneous differential equation in form


yp=(Ax2+Bx+C)cosx+(Dx2+Ex+F)sinxy_p=(Ax^2+Bx+C)\cos x+(Dx^2+Ex+F)\sin x

yp=(Ax2+Bx+C)sinx+(2Ax+B)cosxy_p'=-(Ax^2+Bx+C)\sin x+(2Ax+B)\cos x

+(Dx2+Ex+F)cosx+(2Dx+E)sinx+(Dx^2+Ex+F)\cos x+(2Dx+E)\sin x

yp=(Ax2+Bx+C)cosx2(2Ax+B)sinxy_p''=-(Ax^2+Bx+C)\cos x-2(2Ax+B)\sin x

+2Acosx(Dx2+Ex+F)sinx+2A\cos x-(Dx^2+Ex+F)\sin x

+2(2Dx+E)cosx+2Dsinx+2(2Dx+E)\cos x+2D\sin x

Substitute


(Ax2+Bx+C)cosx2(2Ax+B)sinx-(Ax^2+Bx+C)\cos x-2(2Ax+B)\sin x

+2Acosx(Dx2+Ex+F)sinx+2A\cos x-(Dx^2+Ex+F)\sin x

+2(2Dx+E)cosx+2Dsinx+2(2Dx+E)\cos x+2D\sin x

+(Ax2+Bx+C)cosx+(Dx2+Ex+F)sinx+(Ax^2+Bx+C)\cos x+(Dx^2+Ex+F)\sin x

=2xsinx=2x\sin x

2(2Ax+B)sinx+2Acosx-2(2Ax+B)\sin x+2A\cos x

+2(2Dx+E)cosx+2Dsinx=2xsinx+2(2Dx+E)\cos x+2D\sin x=2x\sin x

A=12A=-\dfrac{1}{2}

B=D=0B=D=0

E=A=12E=-A=\dfrac{1}{2}

yp=12x2cosx+12xsinxy_p=-\dfrac{1}{2}x^2\cos x+\dfrac{1}{2}x\sin x

The general solution of the nonhomogeneous differential equation is


y=c1cosx+c2sinx12x2cosx+12xsinxy=c_1\cos x+c_2\sin x-\dfrac{1}{2}x^2\cos x+\dfrac{1}{2}x\sin x


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