Author Topic: TT2 #1  (Read 826 times)

Victor Ivrii

  • Administrator
  • Elder Member
  • *****
  • Posts: 1332
  • Karma: 0
    • View Profile
    • Personal website of Victor Ivrii
TT2 #1
« on: November 19, 2014, 08:43:59 PM »
Solve the following initial value problem
\begin{gather*}
x^3 y''' - 3 x^2 y'' + 6x y'- 6y = - 24 x^{-1} + 6 \ln x \ , \\[4pt]
y(1)= -5/6 \ ,\ y'(1)=-2 \ ,\ y''(1)=2 \ .
\end{gather*}

Yuan Bian

  • Jr. Member
  • **
  • Posts: 14
  • Karma: 4
    • View Profile
Re: TT2 #1
« Reply #1 on: November 19, 2014, 10:53:37 PM »
homo:  r(r-1)(r-2)-3r(r-1)+6r-6=0
           (r-1)(r-2)(r-3)=0
r1=1,r2=2,r3=3
l(r)=r3-6r2+11r-6
let t=lnx
l(t)=t'''-6t''+11t'-6t=6t-24e-t
let yp1=At+B, yp2=Ce-t
so Ce-t(l(-1))=-24e-t, (l(0)(At+B)+l'(0)A)=6t
C=1, B=-11/6, A=-1
y=c1x+c2x2+c33+1/x-lnx-11/6
y(1)=−5/6 , y′(1)=−2 , y′′(1)=2
s0 c1+c2+c3+1-11/6=-5/6
    c1+2c2+3c3-2=-2
    2c2+6c3+3=2
get c1=-1/2, c2=1, c3=-1/2
y=-x3/2+x2-x/2+1/x-lnx-11/6
« Last Edit: November 19, 2014, 11:21:28 PM by Yuan Bian »