Answer to Question #142297 in Differential Equations for NikHil

Question

Solve: xp^2- (2x+3y)p + 6y=0 , p= dy/dx

Answer 1

$xp^2-(2x+3y)p+6y=x(p^2-2p)-3y(p-2)=xp(p-2)-3y(p-2)$=

$=(xp-3y)(p-2)$=0

so we have two equations:

$x\frac{dy}{dx}-3y=0, \frac{dy}{dx}-2=0$

from the first equation:

$\frac{dy}{y}=3\frac{dx}{x}$

$\ln{y}=3\ln{x}+C$

$y=C'x^3$ - the first solution

from the second equation:

$\frac{dy}{dx}=2$

$y=2x+C$ - the second solution