How do i solve this differential equation

[seto6]

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...

 

[jackmell]

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

 

[HallsofIvy]

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 "-").

 

[jackmell]

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. :)

 

[seto6]

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