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Homework Statement
x(y')^{2}-(2x+3y)(y')+6y=0
Homework Equations
non
The Attempt at a Solution
i'm kinda lost on how to approch meaning what method to use..
all i know is that its a non linear differential equation...
No, I meant y using that method but your way was much simpler. :)Perhaps jackmell meant "The first step is to solve for y' ". That is what I recommend.
That is a quadratic equation in y'. You start by solving for y' using the quadratic formula:
[tex]y'= \frac{2x+ 3y\pm\sqrt{(2x+3y)^2- 24xy}}{2x}[/tex]
It helps to note that [itex](2x+ 3y)^2- 24y= 4x^2+ 12xy+ 9y^2- 24xy[/itex][itex]= 4x^2- 12xy+ 9y^2= (2x- 3y)^2[/itex]!
The differential equation reduces to [itex]y'= 2[/itex] (using the "+") or [itex]y'= 3y/x[/itex] (using the "-").