# Reduction of order

Use reduction of order to find a second solution of the given differential equation
t^2y"-4ty'+6y=0 t>0 y_1(t)=t^2

first I put in in the standard form
y"-4/ty'+6/t^2y =0
then y= v(t)t^2
y'=v'(t)t^2+v(t)2t
y"=y"(t)t^2+v'(t)2t+v'(t)2t+v(t)2 = t^2v"(t)+4tv'(t)+2v(t)
putting these into the original equation gives me
t^2v"(t)+4tv'(t)+2v(t)-4tv'(t)-8v(t)+6t^2v(t)
simplifying gives me
t^2v"(t)+6t^2v(t)-6v(t)
I don't know what to do from here. can someone please help?

## Answers and Replies

HallsofIvy
Science Advisor
Homework Helper
Punchlinegirl said:
Use reduction of order to find a second solution of the given differential equation
t^2y"-4ty'+6y=0 t>0 y_1(t)=t^2

first I put in in the standard form
y"-4/ty'+6/t^2y =0
then y= v(t)t^2
y'=v'(t)t^2+v(t)2t
y"=y"(t)t^2+v'(t)2t+v'(t)2t+v(t)2 = t^2v"(t)+4tv'(t)+2v(t)
putting these into the original equation gives me
t^2v"(t)+4tv'(t)+2v(t)-4tv'(t)-8v(t)+6t^2v(t)
The is wrong. You have added 6y but it should be $\frac{6}{t^2}y$, just 6v, not 6t^3 v.
You should have
[tex]t^2v"+ 4tv'+ 2v- 4tv'- 8v+ 6v= t^2v"= 0[/itex]
That should be easy!

Simplifying gives me
t^2v"(t)+6t^2v(t)-6v(t)
I don't know what to do from here. can someone please help?