How do you solve the initial value problem of #t^2y''  4ty' + 4y = 2t^2# given #y(1) = 2# andÂ #y'(1) =0#?
1 Answer
Answer :
1) First, solve

Consider
#y(t) = t^k# where#k# is real. It's a solution if and only if
#t^2k(k1)t^(k2)  4tkt^(k1) + 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
where
2) Second, find a particular solution of

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
where
3) Finally, you want to have

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