# Integrating y''/y

1. Aug 29, 2011

### *FaerieLight*

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

How would you find the integral of y''/y (with respect to x)?

2. Relevant equations

3. The attempt at a solution

I have absolutely no idea how to begin. All I know is that if it were y'/y, it would be log(y) + c. Perhaps the integral here has something to do with logs.

Thanks a lot.

2. Aug 29, 2011

### Tomer

Try integration in parts, where f(x) = y''(x), g'(x) = 1/y(x).
You should arrive to a new integral you can solve...

3. Aug 30, 2011

### HallsofIvy

Staff Emeritus
Tomer's suggestion is good but I think a more common notation would be u= 1/y, dv= y'' dy.

4. Aug 30, 2011

### LCKurtz

I don't see how either of these substitutions help to solve this problem. y is a function of x and the variable of integration is x.

5. Aug 30, 2011

### Tomer

I haven't for a second assumed x isn't the variable of integration.
However, trying to solve it now on paper I noticed that what I first though would be a solvable integral isn't really.
Do you have any ideas then?

6. Aug 30, 2011

### jackmell

If:

$$\frac{d}{dx}\left[\frac{y'}{y}+\left(\frac{d}{dx}\log y\right)^2\right]=\frac{y''}{y}$$

then isn't:

$$\int \frac{y''}{y}=\frac{y'}{y}+\left(\frac{d}{dx}\log y\right)^2$$

Not entirely sure guys. Just a start.

7. Aug 30, 2011

### Tomer

What you say is right, but I don't see how the first formula you wrote is correct.

8. Aug 30, 2011

### jackmell

Ok, it's wrong. Sorry.

9. Aug 30, 2011

### Tomer

It's ok, apparently we're all wrong :-)

10. Aug 30, 2011

### PAllen

No, it seems that the starting point in post #6 can't possibly be right. It could only be right if the derivative of the second term in brackets is zero, which can't possibly be true, in general.

11. Aug 30, 2011

### LCKurtz

I will be a bit more emphatic. You aren't going to find a nice closed form general solution with any such techniques.

12. Aug 30, 2011

### PAllen

Yeah, it is equivalent to asking what is a general formula for the solution of the following homogeneous, linear second order equation with non-constant coefficients:

y'' - f(x) y = 0

which is absurd (unless f(x) is special in some way).

13. Aug 30, 2011

### Tomer

Why is this absurd? If I can say that $\int\frac{y'(x)}{y(x)}$ = ln(y(x)), which is a closed form general solution, why should the given integral be absurd?

I don't see any way to find a closed formula, but I also don't see why the question should be meaningless.

Last edited: Aug 30, 2011
14. Aug 30, 2011

### PAllen

It isn't meaningless, it is just well understood - it has been studied (stated as diff.eq.) for centuries. All facts about it are known. It is known that there are solutions for many common f(x), but no formula for a general solution.

15. Aug 31, 2011

### Tomer

I understand :p