How to solve this ODE?

  • Thread starter AdrianZ
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  • #1
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Homework Statement


[tex]2xy'(x-y^2)+y^3=0[/tex]

Homework Equations





The Attempt at a Solution


What kind of an equation is that? I first thought that might be a Bernoulli differential equation with respect to x but I failed to convert it that form. I also checked if the equation could have single variable integrating factors but the answer is negative. How can I solve it?
 

Answers and Replies

  • #2
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I would approach it parametrically:

[tex]\frac{dy}{dx}=\frac{-y^3}{2x(x-y^2)}[/tex]

so:

[tex]\frac{dy}{dt}=-y^3[/tex]

[tex]\frac{dx}{dt}=2x(x-y^2)[/tex]

Note the first is de-coupled so solve that one and substitute it into the second one. Maybe though there is something I'm missing and would allow an easier approach.
 
  • #3
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Thanks for your help jackmell but I already solved it as a Bernoulli equation with respect to x. I'll write down my answer so that it might be useful for someone else.
[tex]2xy^' (x-y^2 )+y^3=0[/tex]
[tex]2x(x-y^2 )+y^3 x^'=0[/tex]
[tex]x^'+(2x^2-2xy^2)/y^3 =0[/tex]
[tex]x^'+(-2/y)x=(-2/y^3 ) x^2[/tex]

so it's a Bernoulli equation and can be solved easily.
[tex]-x^{-2}+(2/y) x^{-1}=(2/y^3 )[/tex]
[tex]u=x^{-1}:du/dx=-x^{-2}[/tex]
[tex]{du/dx} {dx/dy}+(2/y)u=(2/y^3 )[/tex]
[tex]du/dy+(2/y)u=(2/y^3 )[/tex]
[tex]μ=e^{∫2dy/y}=y^2[/tex]
[tex]uy^2=∫y^2 (2/y^3 )dy[/tex]
[tex]Ce^{y^2/x}=y^2[/tex]
 

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