- #1

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**dx/dt = y^2 - x^2**

dy/dt = -2xy

dy/dt = -2xy

?

Sorry about not using LaTex, I know it looks better.

Thanks!

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- Thread starter atomqwerty
- Start date

- #1

- 94

- 0

dy/dt = -2xy

?

Sorry about not using LaTex, I know it looks better.

Thanks!

- #2

AlephZero

Science Advisor

Homework Helper

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y' = -2xy

Differentiate the first equation

x''= 2yy' - 2x

Use the second equation to eliminate y'

Then use the first equation again to eliminate y

- #3

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So I obtain

x'' +4x'x +4x^3 + 2x == 0

and now?

thanks

x'' +4x'x +4x^3 + 2x == 0

and now?

thanks

- #4

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- #5

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perfect! I got it! Thank you very much :D

- #6

AlephZero

Science Advisor

Homework Helper

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So I obtain

x'' +4x'x +4x^3 + 2x == 0

and now?

thanks

The standard method to solve that type of equation is

Let x' = p

Then x'' = dp/dt = dp/dx dx/dt = p dp/dx

So you get

p dp/dx + 4px + 4x^3 + 2x = 0

Integrating with respect to x gives

p^2/2 + I can't see how to integrate the 4px term + x^4 + x^2 = C

But I like the polar coordinates method better.

Now we know the answer, my method seems to be heading in right direction, but that's no use unless we can see how to finish it.

- #7

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- #8

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I agree :

x'' + 6 x x' +4 x^3 = 0

2 y y'' -3 (y')^2 +4 y^4 = 0

OK. far to be easier, but possible, even without knowing the solution.

- #9

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