# First order ODE.

1. Nov 25, 2013

### nothingkwt

1. The problem statement, all variables and given/known data
Find the general solution of 2y' + y - (2y')*ln(y') = 0

2. Relevant equations

3. The attempt at a solution
I have no idea how to deal with this i mean none of the first order techniques work and it's mainly because I don't know how to deal with the ln(y').

I tried seperating the ln to lndy - lndx but it didn't do any good.

Last edited: Nov 25, 2013
2. Nov 25, 2013

### clamtrox

I think a reasonable first step is to differentiate the whole equation once. That should make the logarithm term more approachable.

3. Nov 25, 2013

### nothingkwt

Seemed like a good idea but when I differentiated I feel like I complicated it even more by increasing the order

dy/dx (2y' + y - (2y') * ln(y') = 0)

2y" + y' - 2y" * ln(y') - 2y'*y''/y' = 0
= 2y" + y' - 2y" * ln(y') - 2y''
= y' - 2y" * ln(y')

Last edited: Nov 25, 2013
4. Nov 25, 2013

### clamtrox

That's how it may seem, but look closer. It's actually a first order ODE for y', and now the logarithm term makes much more sense. There's an obvious substitution you can make to simplify the equation further.

5. Nov 25, 2013

### nothingkwt

r = y'
dr = y"

r - 2dr * ln(r) = 0

I feel like something is wrong here maybe I differentiated r incorrectly?

6. Nov 25, 2013

### Staff: Mentor

In the first line above, your notation indicates that you are multiplying an equation by dy/dx. What you're actually doing is taking the derivative of each side of an equation.

This notation -- dy/dx(x2) -- means the derivative of some unknown function times x2. Without knowing what dy/dx is, it can't be further simplified.

This notation -- d/dx(x2) means the derivative, with respect to x, of x2, which is 2x.

Your 3rd and 4th lines should be their own equations, with 0 on the right side.

7. Nov 25, 2013

### nothingkwt

Sorry you're right.

2y" + y' - 2y" * ln(y') - 2y'*y''/y' = 0
2y" + y' - 2y" * ln(y') - 2y'' = 0
y' - 2y" * ln(y') = 0

8. Nov 25, 2013

### Staff: Mentor

Yes, instead of dr = y'' you should have dr/dy = y'' or r' = y''.

Using the tip from clamtrox, if you let u = y', so u' = y''. This will give you a first order DE in u that is separable.

9. Nov 26, 2013

### nothingkwt

So it would become . . .

r dy - 2 (ln(r)) dr = 0

since r = y'

int ( y'dy) - 2 int (ln(r)) dr) = 0

Is that right?

Last edited: Nov 26, 2013
10. Nov 26, 2013

### clamtrox

No, that makes no sense. Let's say y is a function of a variable called x. Then y' = dy/dx. This is also a function of x. Now you say r = y'. Also r is a function of x. So your equation is r - 2 dr/dx * ln(r) = 0. Now you can move all r's on one side, and all x's on the other side, and integrate.

11. Nov 27, 2013

### nothingkwt

So after integrating r - 2 dr/dx * ln(r) = 0

I get x - 2ln(ln(r)) = 0

x - 2ln(ln(y')) = 0

x = ln(ln(y')^2)

e^x = (ln(y'))^2

I got stuck here not sure how to seperate the y.

Sorry about replying so late every time it's cause of the time zone.

12. Nov 28, 2013

### clamtrox

Something goes wrong here. The integration is very straightforward if you substitute z = ln(r).

13. Nov 29, 2013

### nothingkwt

This was after the integration.

r - 2 dr/dx * ln(r) = 0

This is what I integrated.