1. ## Solving a DE

$\displaystyle \psi'(y)+y^2\psi'(y)-\lambda\psi(y)=0$

How can I approach this?

The Cauchy Euler method didn't work.

2. It's separable.

3. Originally Posted by dwsmith
$\displaystyle \psi'(y)+y^2\psi'(y)-\lambda\psi(y)=0$

How can I approach this?

The Cauchy Euler method didn't work.
What you have posted can be re-written as $\displaystyle \psi'(y) (1+y^2) - \lambda\psi(y)=0$ and is separable.

4. Originally Posted by mr fantastic
What you have posted can be re-written as $\displaystyle \psi'(y) (1+y^2) - \lambda\psi(y)=0$ and is separable.
It's also first order linear if written as

$\displaystyle \displaystyle \psi'(y) - \frac{\lambda}{1+y^2}\psi(y) = 0$.

5. Originally Posted by dwsmith
[snip]

The Cauchy Euler method didn't work.

[/snip]
FYI: Cauchy Euler Equations are second order equations, not first order equations. Thus, the technique involved in solving C-E equations is not applicable.

6. ...Cauchy Euler Equations are second order equations...
Really? I thought you could have a CE equation of any order. You just needed to have the appropriate power of x multiplying the corresponding order of differentiation. The wiki certainly seems to indicate that. There's nothing preventing a CE equation from being third order, or fourth. Why not first-order?

7. Originally Posted by Ackbeet
Really? I thought you could have a CE equation of any order. You just needed to have the appropriate power of x multiplying the corresponding order of differentiation. The wiki certainly seems to indicate that. There's nothing preventing a CE equation from being third order, or fourth. Why not first-order?
Ah..I never really thought of that. I was only familiar with them as second order equations.