Find the general solution of y''-2y'+2y=xe^x cosx

Tyra 2020-10-20 Answered
Find the general solution of y2y+2y=xexcosx
You can still ask an expert for help

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more

Solve your problem for the price of one coffee

  • Available 24/7
  • Math expert for every subject
  • Pay only if we can solve it
Ask Question

Expert Answer

Aniqa O'Neill
Answered 2020-10-21 Author has 100 answers
y2y+2y=xexcos(x).
Let s=sin(x),c=cos(x), s'=c, c'=-s, then:
y2y+2y=xexc.
First solve the homogeneous equation: y2y+2y=0.
The characteristic equation is r22r+2=0=(r1+i)(r1i).
So y=Ae(1+i)x+Be(1i)x, which can be written:
y=ex(Aeix+Beix), where A and B are constants.
Use Euler’s expansion:
y=ex(Acos(x)+Aisin(x)+Bcos(x)Bisin(x)),
y=ex((A+B)cos(x)+i(AB)sin(x)).
Let C=A+B and D=A-B, then:
y=ex(Ccos(x)+Dsin(x)).
Call this general solution y1=ex(Ccos(x)+Dsin(x))=ex(Cc+Ds)
Now we have to deal with xeˣc, the particular solution, which will give us y₂, such that the entire solution is y=y1+y2.
We guess the solution, where P, Q, R and S are constants:
Let u=(Px+Q)s+(Rx+S)c and apply the tentative solution y2=xexu:
u=(Px+Q)c+Ps(Rx+S)s+Rc,
u=(Px+Q)s+2Pc(Rx+S)c2Rs,
y2=xexu+uex(x+1),
Not exactly what you’re looking for?
Ask My Question

Expert Community at Your Service

  • Live experts 24/7
  • Questions are typically answered in as fast as 30 minutes
  • Personalized clear answers
Learn more