Answer to Question #278431 in Differential Equations for Audrey

Question #278431

Solve the following differential equation:



1. (D²+D)y=sin x



2. (D²+4D+5) y= 50x + 13e^3x



3. (D³+D²-4D-4)y= 4 sin x



4.(D³-D)y=x



5. (D²-4D+4)y=e^x

1
Expert's answer
2021-12-14T10:03:48-0500

1.

k2+1=0k^2+1=0

k=±ik=\pm i

yh=c1cosx+c2sinxy_h=c_1cosx+c_2sinx


yp=Axcosx+Bxsinxy_p=Axcosx+Bxsinx

yp=AcosxAxsinx+Bsinx+Bxcosx=(A+Bx)cosx+(BAx)sinxy'_p=Acosx-Axsinx+Bsinx+Bxcosx=(A+Bx)cosx+(B-Ax)sinx

yp=Bcosx(A+Bx)sinx+(BAx)cosxAsinxy''_p=Bcosx-(A+Bx)sinx+(B-Ax)cosx-Asinx


Bcosx(A+Bx)sinx+(BAx)cosxAsinx+Axcosx+Bxsinx=sinxBcosx-(A+Bx)sinx+(B-Ax)cosx-Asinx+Axcosx+Bxsinx=sinx

B=0B=0

AA=1-A-A=1

A=1/2A=-1/2


y=yh+yp=c1cosx+c2sinxxcosx/2y=y_h+y_p=c_1cosx+c_2sinx-xcosx/2


2.

k2+4k+5=0k^2+4k+5=0

k=4±16202=2±2ik=\frac{-4\pm\sqrt{16-20}}{2}=-2\pm 2i


yh=e2x(c1cos2x+c2sin2x)y_h=e^{-2x}(c_1cos2x+c_2sin2x)


yp1=Ax+By_{p1}=Ax+B

4A+5Ax+5B=50x4A+5Ax+5B=50x

A=10,B=40/5=8A=10,B=-40/5=-8

yp1=10x8y_{p1}=10x-8


yp2=Ae3xy_{p2}=Ae^{3x}

9Ae3x+12e3x+5e3x=13e3x9Ae^{3x}+12e^{3x}+5e^{3x}=13e^{3x}

A=4/9A=-4/9

yp2=4e3x/9y_{p2}=-4e^{3x}/9


y=e2x(c1cos2x+c2sin2x)+10x84e3x/9y=e^{-2x}(c_1cos2x+c_2sin2x)+10x-8-4e^{3x}/9


3.

k3+k24k4=0k^3+k^2-4k-4=0

k(k24)+k24=0k(k^2-4)+k^2-4=0

k1=1,k2=2,k3=2k_1=-1,k_2=-2,k_3=2


yh=c1ex+c1e2x+c1e2xy_h=c_1e^{-x}+c_1e^{-2x}+c_1e^{2x}


yp=Acosx+Bsinxy_p=Acosx+Bsinx

yp=BcosxAsinxy'_p=Bcosx-Asinx

yp=BsinxAcosxy''_p=-Bsinx-Acosx

yp=Bcosx+Asinxy'''_p=-Bcosx+Asinx


Bcosx+AsinxBsinxAcosx4(BcosxAsinx)4(Acosx+Bsinx)=-Bcosx+Asinx -Bsinx-Acosx-4(Bcosx-Asinx)-4(Acosx+Bsinx)=

=4sinx=4 sin x

5B5A=0-5B-5A=0

5A5B=45A-5B=4

10A=4    A=2/5,B=2/510A=4\implies A=2/5,B=-2/5


y=c1ex+c1e2x+c1e2x+2(cosxsinx)/5y=c_1e^{-x}+c_1e^{-2x}+c_1e^{2x}+2(cosx-sinx)/5


4.

k31=0k^3-1=0

(k1)(k2+k+1)=0(k-1)(k^2+k+1)=0

k1=1k_1=1

k2+k+1=0k^2+k+1=0

k2,3=1±i32k_{2,3}=\frac{-1\pm i\sqrt{3}}{2}


yh=c1ex+ex/2(c2cos(x3/2)+c3sin(x3/2))y_h=c_1e^x+e^{-x/2}(c_2cos(x\sqrt 3/2)+c_3sin(x\sqrt 3/2))


yp=Ax+By_p=Ax+B

AxB=x-Ax-B=x

A=1,B=0A=-1,B=0


y=c1ex+ex/2(c2cos(x3/2)+c3sin(x3/2))xy=c_1e^x+e^{-x/2}(c_2cos(x\sqrt 3/2)+c_3sin(x\sqrt 3/2))-x


5.

k24k+4=0k^2-4k+4=0

k1,2=2k_{1,2}=2

yh=c1e2x+c2xe2xy_h=c_1e^{2x}+c_2xe^{2x}


yp=Aexy_p=Ae^x

A4A+4A=1A-4A+4A=1

A=1A=1


y=c1e2x+c2xe2x+exy=c_1e^{2x}+c_2xe^{2x}+e^x


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