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find a general solution to the given equation.

y'''+4y''+y'-26y=e-3xsin2x+x

Short Answer

Expert verified

y(x)=c3e2x+c4e-3xcos2x+c6e-3xsin2x-1676-x26+129xe-3xcos2x-158xe-3xsin2x

Step by step solution

01

Find the corresponding auxiliary equation

The auxiliary equation ofthe corresponding homogeneous equation:

r2(r-2)r2+6r+132=0

The solutions of the auxiliary equation are:

r1=r2=0,r3=2,r4=r5=-3+2i,r6=r7=-3-2i

Therefore a general solution to the homogeneous equation is:

yh(x)=c3e2xc4e-3xcos2x+c6e-3xsin2x

02

Find the particular solution

Let the particular solution be;

yp(x)=c1+c2x+c5xe-3xcos2x+c7xe-3xsin2x

Then,

yp'(x)=c2+c5e-3xcos2x+c7e-3xsin2x+-3c5+2c7xe-3xcos2x+-2c5-3c7xe-3xsin2x

yp''(x)=-6c5+4c7e-3xcos2x+-4c5-6c7e-3xsin2x+5c5-12c7xe-3xcos2x+12c5+5c7xe-3xsin2x

yp'''(x)=15c5-36c7e-3xcos2x+36c5+15c7e-3xsin2x+9c5+46c7xe-3xcos2x+-46c5+9c7xe-3xsin2x

Then,

yp'''+4yp''+yp'-36yp=-8c5-20c7e-3xcos2x+20c5-8c7e-3xsin2x+c2-26c1-26c2x=e-3xsin2x+x

Then we have:

-8c5-20c7=0c2-26c1=020c5+8c7=1-26c2=1

Therefore,

c1=-1676,c2=-126,c5=129,c7=-158

Hence,

yp(x)=-1676-126x+129xe-3xcos2x-158xe-3xsin2x.

03

Step 3: y(x)=yh+yp

Thus, the required general solution is:

y(x)=c3e2x+c4e-3xcos2x+c6e-3xsin2x-1676-x26+129xe-3xcos2x-158xe-3xsin2x

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