Solving a DE

**dwsmith** · January 5th 2011, 12:11 PM

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

How can I approach this?

The Cauchy Euler method didn't work.

**Ackbeet** · January 5th 2011, 12:13 PM

It's separable.

**mr fantastic** · January 5th 2011, 12:13 PM

Originally Posted by dwsmith

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

How can I approach this?

The Cauchy Euler method didn't work.

What you have posted can be re-written as $\psi'(y) (1+y^2) - \lambda\psi(y)=0$ and is separable.

**Prove It** · January 5th 2011, 06:56 PM

Originally Posted by mr fantastic

What you have posted can be re-written as $\psi'(y) (1+y^2) - \lambda\psi(y)=0$ and is separable.

It's also first order linear if written as

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

**Chris L T521** · January 5th 2011, 07:54 PM

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.

**Ackbeet** · January 6th 2011, 02:06 AM

...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?

**Chris L T521** · January 6th 2011, 07:41 AM

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.

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