2nd Order Linear DE (homogeneous/wronskian/euler-cauchy)

[anthony:)]

Homework Statement

The equation t^2y'' + 4ty' -4y = 0

has a solution of y(t) = t.

Find a second, linearly independent solution.

2. The attempt at a solution

Ok so I just applied the Cauchy Euler equation method to find a general solution of

y = c1*t^-4 + c2*t

Where c1 and c2 are constants.

The problem stated that t was already a solution so I'm assuming that t^-4 is the other solution.

To determine whether they were linearly independent, I found the Wronkian which was

-5t^-4

I know that two solutions of a DE are linearly independent if the wronkian is non-zero for all points where the solution space is defined, but in this case, the wronskian is only zero if t> 0 or t < 0. But it will be zero if t=0.

The problem does not include a range for t...

So basically, my question is : Could t^-4 be a second linearly independent solution of the DE?

Thank you.

 

[tiny-tim]

welcome to pf!

hi anthony:)! welcome to pf!

(use {} round an exponent or subscript if it's more than one character )

To determine whether they were linearly independent, I found the Wronkian which was

-5t^{-4}

I know that two solutions of a DE are linearly independent if the wronkian is non-zero for all points where the solution space is defined, but in this case, the wronskian is only zero if t> 0 or t < 0. But it will be zero if t=0. …

but y = t and y = t^-4 are obviously independent!

i think you only need to look at the wronskian to check independence if you don't already know what the solutions are

 

[anthony:)]

Sorry I'm kind of slow...

Thanks

 

[LCKurtz]

To determine whether they were linearly independent, I found the Wronkian which was

-5t^-4

I know that two solutions of a DE are linearly independent if the wronkian is non-zero for all points where the solution space is defined, but in this case, the wronskian is only zero if t> 0 or t < 0. But it will be zero if t=0.

The problem does not include a range for t...

You mean the Wronskian is non-zero if t > 0 or t < 0. And it is not zero when t = 0; it is undefined.

Also note that the leading coefficient of the DE is zero when t = 0. That gives a singular point at t = 0 which is why the domain for the solution doesn't include t = 0.