# How do i solve this differential equation

## Homework Statement

x(y')2-(2x+3y)(y')+6y=0

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...

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Whenever you want to solve a non-linear one, try all the simple approaches in the chapter on non-linear equations in any DE textbook. One of those will work here. Get a DE textbook, find that chapter, then go over the example on eliminating the dependent variable. The first step is to solve for y in the equation:

$$x(y')^2-(2x+3y)y'+6y=0$$

Last edited:
HallsofIvy
Homework Helper
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:
$$y'= \frac{2x+ 3y\pm\sqrt{(2x+3y)^2- 24xy}}{2x}$$

It helps to note that $(2x+ 3y)^2- 24y= 4x^2+ 12xy+ 9y^2- 24xy$$= 4x^2- 12xy+ 9y^2= (2x- 3y)^2$!

The differential equation reduces to $y'= 2$ (using the "+") or $y'= 3y/x$ (using the "-").

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:
$$y'= \frac{2x+ 3y\pm\sqrt{(2x+3y)^2- 24xy}}{2x}$$

It helps to note that $(2x+ 3y)^2- 24y= 4x^2+ 12xy+ 9y^2- 24xy$$= 4x^2- 12xy+ 9y^2= (2x- 3y)^2$!

The differential equation reduces to $y'= 2$ (using the "+") or $y'= 3y/x$ (using the "-").
No, I meant y using that method but your way was much simpler. :)

i see i thought of quadratic formula them said nah don't know why now i see how it works out!