First-order nonlinear ordinary differential equations

**mafra** · September 11th 2011, 03:37 PM

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

**chisigma** · September 11th 2011, 04:18 PM

Originally Posted by mafra

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

The DE requires a quite particular approach. We have...

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

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

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

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

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

...or...

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

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

$y=c\ x + c\ \ln c$ (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

$\chi$ $\sigma$

**mafra** · September 11th 2011, 04:46 PM

Thanks, but where's the term for y'(lny')'?

I think I got the singular solution too; -e^-x

**chisigma** · September 11th 2011, 05:01 PM

Originally Posted by mafra

Thanks, but where's the term for y'(lny')'?

I think I got the singular solution too; -e^-x

Your replay has been useful because I discovered an error. It is...

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

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

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

Kind regards

$\chi$ $\sigma$

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