- #1

itsthewoo

- 2

- 0

2x * (dy/dx) + y = y

^{2}* sqrt(x - (x

^{2}* y

^{2}))

Any ideas how to solve it or begin solving it?

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- Thread starter itsthewoo
- Start date

In summary, the conversation discusses a complicated equation and the process of finding its solution. The solution is found using an advanced technique called Lie point transformations and an integrating factor. The final solution is confirmed to be the same as the one given by other mathematical software. A book recommendation is also provided for further information.

- #1

itsthewoo

- 2

- 0

2x * (dy/dx) + y = y

Any ideas how to solve it or begin solving it?

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- #2

NoMoreExams

- 623

- 0

Hmm maple gives a fairly straightforward answer maybe a subst. u = xy, just a guess

- #3

coomast

- 279

- 1

I finally found the solution. It kept running in my head and I couldn't put it aside, so here it is. First of all I used an advanced technique called Lie point transformations to find the answer. It comes down for this equation to find an integrating factor so the equation becomes exact. Writing the equation as:

[tex]\left(2x\right)\cdot dy-\left(-y+y^2\sqrt{x-x^2y^2}\right)\cdot dx=0[/tex]

The integrating factor is the following:

[tex]M=\frac{1}{xy^2\sqrt{x-x^2y^2}}[/tex]

The new (exact) equation is now:

[tex]\left(\frac{2x}{xy^2\sqrt{x-x^2y^2}}\right)\cdot dy+

\left(\frac{1}{xy\sqrt{x-x^2y^2}}-\frac{1}{x}\right)\cdot dx=0[/tex]

Which has as solution:

[tex]x=Ke^{-\frac{2\sqrt{x-x^2y^2}}{xy}}[/tex]

This is the same solution as maxima (like maple or mathematica) gives. If any more info is necessary, I would recommend the following book:

"Ordinary differential equations, an elementary text-book with an introduction

to Lie's theory of the group of one parameter." written by James Morris Page

Hope this helps,

coomast

- #4

Mathwebster

- 185

- 0

[tex]u = x y^2[/tex] will separate your equation.

- #5

coomast

- 279

- 1

I considered the following one-parameter group:

[tex]Uf=ax\frac{\partial f}{\partial x}+by\frac{\partial f}{\partial y}+(b-a)y'\frac{\partial f}{\partial y'}[/tex]

And found the unknown numbers to bea=-1 and b=1/2. From this it is possible to find the integrating factor, but also the transformation you gave. It works indeed. I did not think of doing it like that although I should have . Many thanks for the hint.

coomast

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