How do you solve the initial value problem of #t^2y'' - 4ty' + 4y = -2t^2# given #y(1) = 2# and #y'(1) =0#?

1 Answer
Feb 19, 2015

Answer : #y : t \mapsto 2t - t^4 + t^2#

1) First, solve #t^2y''-4ty' + 4y = 0#.

  • Consider #y(t) = t^k# where #k# is real. It's a solution if and only if
    #t^2k(k-1)t^(k-2) - 4tkt^(k-1) + 4t^k = 0#
    or if and only if #k^2 - 5k + 4 = 0#. The solutions are 1 et 4.

  • Therefore, all the solutions of #t^2y''-4ty' + 4y = 0# are

# t \mapsto \lambda t + \mu t^4 #

where #lambda# and #mu# are arbitrary real constants.

2) Second, find a particular solution of #t^2y''-4ty' + 4y = -2t^2#.

  • Try a function like #y(t) = mt^k#. You find easily that #m=1# and #k=2# are ok.

  • Therefore all the solutions of #t^2y''-4ty' + 4y = -2t^2# are

#y : t \mapsto \lambda t + \mu t^4 + t^2#

where #lambda# and #mu# are arbitrary real constants.

3) Finally, you want to have #y(1)=2# and #y'(1)=0#.

  • Because #y(1) = \lambda + mu + 1#, you have #lambda + mu = & 1#

  • Because #y'(t) = lambda + 4mu t^3 + 2t#, you have #lambda + 4 mu + 2 = 0#.

  • You solve the system above. You get #mu = -1# and #lambda = 2#.

Conclusion The unique solution is #t \mapsto 2t^2 - t^4 + t^2#.