Problem with 2 differential equations

[RobinM]

I'm having difficultiesunderstanding a substition when solving 2 different equations
(what my professor did in the solutions)
In the first equation:
Xy'' =y'*ln(y'/x)
He substitutes p=y'
And thus: x*p'=p*ln(p/x)
Which is logic because y''turns into p'

this i get,
But in the second equation hesolves:
y*y''=2yy'
He substitutes p=y'
But then he says thaty''=(dp/dy)*P
And i don't understand how he gets that for y''


Thanks

 

[HallsofIvy]

I'm having difficultiesunderstanding a substition when solving 2 different equations
(what my professor did in the solutions)
In the first equation:
Xy'' =y'*ln(y'/x)
He substitutes p=y'
And thus: x*p'=p*ln(p/x)
Which is logic because y''turns into p'

this i get,
But in the second equation hesolves:
y*y''=2yy'
He substitutes p=y'
But then he says thaty''=(dp/dy)*P
And i don't understand how he gets that for y''


Thanks

It's a method called "quadrature". You will agree that, as in the first equation, replacing y' with p gives
y*p'= 2yp. Now look at the "chain rule": \(\displaystyle \frac{dp}{dx}= \frac{dp}{dy}\frac{dy}{dx}\). That would be true for any x, y, or p but, here, \(\displaystyle \frac{dy}{dx}= p\) by definition so \(\displaystyle y''= \left(\frac{dp}{dy}\right)p\).

That makes the equation py(dp/dy)= 2yp so that dp/dy= 2.

Though, frankly, from yp'= 2yp it would be simpler to just cancel the "y" terms: p'=2p where p', of course, is dp/dx.

From dp/dy= 2 we have p= dy/dx= 2y+ C and that differential equation has general solution \(\displaystyle y= De^{2x}- C/2\).

From p'= dp/dx= 2p we have \(\displaystyle p= dy/dx= Ce^{2x}\) and, integrating, \(\displaystyle y= (C/2)e^{2x}+D\), which is exactly the same as the first equation with different "labels" for the constants.