Reducing Order of Differential Equation: t^2y'' - 4ty' + 6y = 0 [SOLVED]

[jesuslovesu]

[SOLVED] Reduce Order (diff eq)

nevermind i got it, cliffsnotes ftw

Homework Statement

Solve the differential equation using the reduction of order method.
$$t^2 y'' - 4ty' + 6y = 0$$
$$t > 0$$
$$y_1 (t) = t^2$$

Homework Equations

The Attempt at a Solution

Well The first thing I do is
$$y(t) = v(t) t^2$$
Then I find y' and y'' and plug into the original diff eq and get
$$t^4 v'' = 0$$
Which I'll assume is correct.

But now I'm really not sure what to do with that. I could do integration by parts? but that doesn't seem to lead anywhere. How do I get from there to t^3 (the other solution)?

 

[HallsofIvy]

I'm glad you solved it yourself- you would have felt embarassed if someone else had pointed out the obvious!

You divide both sides by t⁴ to get v"= 0. Since v" is 0, v'= C, a constant. Then v= Ct+ D, another constant. Since y= vt², y= Ct³+ Dt² is the general solution. By the way, we can divide by t⁴ only if t is not 0. The equation is singular at t= 0- the existence and uniqueness theorem does not apply if we are given intial conditions at t= 0. For example, there is no solution if we are given y(0)= 1, y'(0)= 0.