# 2nd order differential equation using reduction of order

1. Oct 9, 2008

### efghi

1. The problem statement, all variables and given/known data

Use the method of reduction of order to find a second solution of the given differential equation

t2y'' - 4ty' + 6y, t>0; y1(t) = t2

3. The attempt at a solution

Here's what I have so far:
y = vt2
y' = 2tv + t2v'
y'' = 2v + 4tv' + t2v''

so

t2 (2v + 4tv' + t2v'') - 4t (2tv + t2v') + 6(vt2) = 0

t4v'' = 0

I'm stuck here. Am I supposed to divide by y, which would make it t2v'' = 0
wouldn't that make v'' = 0 anyways? What do I do now? I know i'm supposed to integrate it some how but I am just utterly lost from here...

1. The problem statement, all variables and given/known data

2. Relevant equations

3. The attempt at a solution

2. Oct 9, 2008

### Hootenanny

Staff Emeritus
You're absolutely spot on here. Let's look at your final line

t4v''=0

Now the boundary conditions state that t is positive, hence this means that t4 must be non-zero. Furthermore, this implies that

v''=0

Now you have a second order differential equation

$$\frac{d^2 v}{dt^2}=0$$

Simply integrate twice with respect to t. Do you follow?

3. Oct 9, 2008

### efghi

So the 2nd general solution should be y2 = t6 / 18 ?

4. Oct 9, 2008

### Hootenanny

Staff Emeritus
How did you get that?

5. Oct 9, 2008

### efghi

integrate twice with respect to t

integral of t4v'' = t5 / 3
integral of t5 / 3 = t6 / 18?

6. Oct 9, 2008

### Hootenanny

Staff Emeritus
That's wrong. As I said in my previous post, you need only solve the ODE

v''=0

And then multiply the solution by y1 to obtain the general solution.

7. Oct 9, 2008

### efghi

How do i solve the ODE?

8. Oct 9, 2008

### Hootenanny

Staff Emeritus
As I said before integrate twice with respect to t.