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y(y-2x(y'))^3=(y')^2

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- Thread starter math_addict
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- #1

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y(y-2x(y'))^3=(y')^2

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HallsofIvy

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Are numeric or series solutions okay?

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y(y-2x(y'))^3=(y')^2

Hello math_addict,

This equation can be solved by the following method. If a DE of the first order but not first degree can be written as [itex]x-\phi(y,y')=0[/itex] the way to solve it is by taking the derivative of it and consider y and y' as variables. Then you need to substitute dy/y' for dx and you get a first degree DE in y and y'. After integrating this equation you need to eliminate y' between this solution and the original equation to obtain the general solution. In this process you need to make sure that any factors that are eliminated are to be looked upon with care afterwards because they can be singular solutions. Let's try this on your problem.

OK, first rewriting the original DE gives:

[tex]x=\frac{y}{2y'}-\frac{1}{2y^{1/3}y'^{1/3}}[/tex]

The derivative of this is:

[tex]dx=\frac{1}{2}\left[\frac{1}{y'}+\frac{1}{3y'^{1/3}y^{4/3}}\right]dy+

\frac{1}{2}\left[-\frac{y}{y'^2}+\frac{1}{3y^{1/3}y'^{4/3}}\right]dy'[/tex]

Substituting [itex]\frac{dy}{y'}=dx[/itex] gives after rewriting:

[tex]\frac{1}{2y'}\left[1-\frac{y'^{2/3}}{3y^{4/3}}\right]dy=

-\frac{y}{2y'^2}\left[1-\frac{y'^{2/3}}{3y^{4/3}}\right]dy'[/tex]

The factor between brackets can be eliminated but is important for later on to see if it is a singular solution. We have thus so far:

[tex]\frac{dy'}{y'}=-\frac{dy}{y}[/tex]

Which has as solution:

[tex]y'=\frac{C}{y}[/tex]

Substituting this in the original equation gives you:

[tex]y\left(y-2x\frac{C}{y}\right)^3=\frac{C^2}{y^2}[/tex]

Or, when we set C=K^3:

[tex]y^2=K^2+2xK^3[/tex]

This is the general solution to the DE. Now the factor that was eliminated needs to be looked upon. Setting it equal to zero:

[tex]1-\frac{y'^{2/3}}{3y^{4/3}}=0[/tex]

Or, after rewriting:

[tex]y'=3\sqrt{3}y^2[/tex]

Which has the solution:

[tex]-\frac{1}{y}=3\sqrt{3}x[/tex]

Or:

[tex]27x^2y^2=1[/tex]

And it is not possible to get this by setting K to any number and thus is a singular solution because it is also a solution to the original DE.

Hope this helps, coomast

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Thanks, Coomast, for help.

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