OpenStudy (anonymous):

Use reduction of Order to find second solution. t^2y''+2ty'-2y=0 y1(t)=t

OpenStudy (anonymous):

I think u put y' = x -> y'' = 1 and y = x^2/2 + c Then use the condition to calculate c. etc

OpenStudy (anonymous):

um what? this is Differential Equations

OpenStudy (anonymous):

Yes, I am reducing to two first order de's....

OpenStudy (anonymous):

o well thats wrong y2'= v'(t) +v(t)

OpenStudy (anonymous):

?

OpenStudy (anonymous):

thats now how you solve reduction of order problems, you DONT need to split them up into 2 1st order Diff EQs

OpenStudy (anonymous):

OK

OpenStudy (anonymous):

Now I see, y1(t) = t is a solution, not an initial condition....duh.

OpenStudy (anonymous):

So u have put y2 = v(t)t -> y2'= v'(t) +v(t) Is this where the problem is?

OpenStudy (anonymous):

After subbing in the DE and simplifying I got t^3v'' +4t^2v' = 0 and putting w =v' (so w' =v'') -> t^3w' +4t^2w for a first order de.