# Second order DE

1. Dec 3, 2012

### fluidistic

1. The problem statement, all variables and given/known data
Find the general solution to $A(x)y''+A'(x)y'+\frac{y}{A(x)}=0$ where A(x) is a known function and y(x) is the unknown one.
Hint:Eliminate the term that contains the first derivative.

2. Relevant equations
Not sure.

3. The attempt at a solution
So I don't really know how to tackle it. I guess that the hint suggests a change of variable that would get rid of the y' term. So I tried z=y'A, z=Ay and z=y/A. All failed to express the ODE as a function of z and its derivative(s) and A and its derivative(s).
I am therefore stuck. Any other hint will be welcome!

2. Dec 3, 2012

### micromass

Staff Emeritus
What if you try the following substition $y=Bz$, where B is a known function that is yet to be decided.

What do you get if you perform this substitution?

What should we take for B if we want the z' term to vanish?

3. Dec 3, 2012

### fluidistic

Good idea! $x''A'B'+z'(AB+AB'+A'B)+z (A'B'+B/A)=0$.

$B=Ce^{-x}/A$. I checked out and indeed the terms in front of $z'$ vanishes.
I'm left to solve $z''A'Ce^{-x} \left ( \frac{1}{A}+ \frac{A'}{A^2} \right ) + z \left ( \frac{A'}{A} + \frac{A'^2}{A^2}- \frac{Ce^{-x}}{A^2} \right ) =0$. That really does not look beautiful/easy to me.

4. Dec 3, 2012

### micromass

Staff Emeritus
I get something different here. I get that the term of z' is
$$A^\prime B + 2A B^\prime$$

The form of my B is also much simpler.

5. Dec 3, 2012

### fluidistic

I see, I made a mistake; I don't know why I make so many of them. I reach the same as yours, so $B=kA^{-1/2}$ where k is a constant.

Edit: I might have rushed through the math but I reach that $z''+z \left ( 1-\frac{2}{A'} \right ) =0$. Which has different solutions depending on the sign of the term in front of z. I guess I made another mistakes.

Last edited: Dec 3, 2012
6. Dec 3, 2012

### micromass

Staff Emeritus
OK, I get the same B. But there is no real need of the constant. You just need to find one single B that makes z' disappear, so you can take k=1.

Anyway, with that choice of B, what does your equation become then?

7. Dec 3, 2012

### fluidistic

I now get $B'=-\frac{1}{2}A^{-3/2}$, $B''=\frac{3}{4}A^{-5/2}A'^2-\frac{A^{-3/2}A''}{2}$.
The equation becomes $z''AB+z(AB''+A'B'+B/A)=0$. Now I have to replace the B and its derivatives in that.
Edit: I've just done it and it's not beautiful so far.