General Solution of 2y'+y-(2y')*ln(y')=0

[nothingkwt]

Homework Statement

Find the general solution of 2y' + y - (2y')*ln(y') = 0

Homework Equations

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.

 

[clamtrox]

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

 

[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')

 

[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.

 

[nothingkwt]

r = y'
dr = y"

r - 2dr * ln(r) = 0

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

 

[Mark44]

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')

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.

 

[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

 

[Mark44]

r = y'
dr = y"

r - 2dr * ln(r) = 0

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

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

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

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.

 

[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?

 

[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.

 

[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 separate the y.

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

 

[clamtrox]

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

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

 

[nothingkwt]

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

This was after the integration.

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

This is what I integrated.