In Problems 1–8, find a general solution to the differential equation using the method of variation of parameters.

3.y''-2y'+y=t-1et

Short Answer

Expert verified

The general solution is y(t)=c1et+c2tet-tet+lnttet.

Step by step solution

01

Find the particular solution.

The homogenous equation is r2-2r+1=0.

Two independent solutions are r=1,1.

Theny1=et,y2=tet

yh(t)=c1et+c2tet

The particular solution isyp=v1(t)et+v2(t)tet

02

Evaluate v1  and  v2.

Hereyp=v1(t)et+v2(t)tet

And referring to (9)y(t)=c1eαtcosβtc2eαtsinβt, and solve the system then

Put the value ofyp=0

v1'et+v2'tet=0v1'et+v2'et+tet=fa-v1'et+v2'et+tet=t-1et-v1'+v2'1+t=t-1

03

Find v'1 and v1.

v1'=-f(t)y2(t)ay1(t)y'2(t)-y'1(t)y2(t)=-t-1et.tet1et.et+tet-et.tet=-e2te2t=-1

Now integrating this;

v1(t)=-1dt=-t+C

04

Determine v'2 and v2.

v2'=f(t)y1(t)ay1(t)y'2(t)-y'1(t)y2(t)=t-1et.et1e2t=t-1

On integrating, we get:

v2(t)=t-1dt=lnt+C

Therefore, the particular solution is:

yp=-t+Cet+(lnt+C)tetyp=-tet+lnttet

Therefore, the general solution is:

y(t)=yh(t)+yp(t)y(t)=c1et+c2tet-tet+lnttet

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