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How do i solve this differential equation

  • Thread starter seto6
  • Start date
  • #1
251
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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...
 

Answers and Replies

  • #2
1,796
53
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:

[tex]x(y')^2-(2x+3y)y'+6y=0[/tex]
 
Last edited:
  • #3
HallsofIvy
Science Advisor
Homework Helper
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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 "-").
 
  • #4
1,796
53
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 "-").
No, I meant y using that method but your way was much simpler. :)
 
  • #5
251
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i see i thought of quadratic formula them said nah don't know why now i see how it works out!
 

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