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.