I'm having some trouble with these, could you lend me some help?

This one for example:

y = x.y' + y'.ln(y')

I tried p = y' but all i got was this

dx + (x+ln(p))dp = 0

and then I finish with an impossible integration

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- Sep 11th 2011, 03:37 PMmafraFirst-order nonlinear ordinary differential equations
I'm having some trouble with these, could you lend me some help?

This one for example:

y = x.y' + y'.ln(y')

I tried p = y' but all i got was this

dx + (x+ln(p))dp = 0

and then I finish with an impossible integration - Sep 11th 2011, 04:18 PMchisigmaRe: First-order nonlinear ordinary differential equations
The DE requires a quite particular approach. We have...

$\displaystyle y=x\ y^{'} + y^{'}\ \ln y^{'} $ (1)

Now if we differentiate both terms of (1) we obtain...

$\displaystyle y^{'}= y^{'} + (x+1+\ln y^{'})\ y^{''}$ (2)

... and the (2) is satisfied for...

$\displaystyle y^{''}=0$ (3)

...or...

$\displaystyle x+\ln y^{'}=-1$ (4)

Thesystem of (1) and (3) has solution...

$\displaystyle y=c\ x + c\ \ln c$ (5)

... and that is the general solution of (1). The system of (1) and (4) has as solution a 'particular solution' of (1) and it requires an easy procedure...

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$ - Sep 11th 2011, 04:46 PMmafraRe: First-order nonlinear ordinary differential equations
Thanks, but where's the term for y'(lny')'?

I think I got the singular solution too; -e^-x - Sep 11th 2011, 05:01 PMchisigmaRe: First-order nonlinear ordinary differential equations
Your replay has been useful because I discovered an error. It is...

$\displaystyle \frac{d}{dx} (y^{'}\ \ln y^{'}) = (1+\ln y^{'})\ y^{''}$ (1)

... so that the second equation is...

$\displaystyle x+\ln y^{'}=-1$ (2)

Kind regards

$\displaystyle \chi$ $\displaystyle \sigma$