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- Thread starter dearcomp
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- #1

- Feb 5, 2012

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Hi dearcomp,Hello people,

I couldn't solve the given D.E by using exact d.e & substitution method

Thanks in advance.

(x*y*sqrt(x^2-y^2) + x)*y' = (y - x^2*(sqrt(x^2-y^2) )

gif file of d.e can be found in the attachments part.

Use the substitution, \(v=\frac{y}{x}\). The final answer is given >>here<<.

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- Mar 5, 2012

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Hello people,

I couldn't solve the given D.E by using exact d.e & substitution method

Thanks in advance.

(x*y*sqrt(x^2-y^2) + x)*y' = (y - x^2*(sqrt(x^2-y^2) )

gif file of d.e can be found in the attachments part.

Hi dearcomp, welcome to MHB!

Can you verify that you have given the proper DE?

There is at least 1 typo with a missing parenthesis.

In Sudharaka's solution there is another missing parenthesis, meaning a different DE was solved.

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- #5

- Mar 5, 2012

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In that case I'm afraid I can't help you.Yes Sudharaka solved a different D.E

This is the exact problem with the corrected parantheses..

(x*y*sqrt(x^2-y^2) + x)*y' = (y - x^2*(sqrt(x^2-y^2)))

Thank you in advance

It looks like the type of DE that monster Wolfram would be able to solve if humanly possible.

But Wolfram can't, or at least not in closed form.

Of course a numerical approximation is still possible.

Do you have a context for the problem?

Are there perhaps hints, suggestions, or is there specific course material that relates to the DE?

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- #6

Indeed, even wolfram can't handle this problem.In that case I'm afraid I can't help you.

It looks like the type of DE that monster Wolfram would be able to solve if humanly possible.

But Wolfram can't, or at least not in closed form.

Of course a numerical approximation is still possible.

Do you have a context for the problem?

Are there perhaps hints, suggestions, or is there specific course material that relates to the DE?

I'm trying to use WolframMathematica but I can not make it understand the question because I'm not familiar with it...

There are no hints, suggestions related to the question

- Feb 5, 2012

- 1,621

Hi dearcomp, welcome to MHB!

Can you verify that you have given the proper DE?

There is at least 1 typo with a missing parenthesis.

In Sudharaka's solution there is another missing parenthesis, meaning a different DE was solved.

Yes indeed it has missing parenthesis and I typed it incorrectly in Wolfram. Sorry about that.Yes Sudharaka solved a different D.E

This is the exact problem with the corrected parantheses..

(x*y*sqrt(x^2-y^2) + x)*y' = (y - x^2*(sqrt(x^2-y^2)))

Thank you in advance

I am not sure how we can solve this in the general case, but notice that \(y=x\) is a particular solution of this equation. Now at least you have one solution.Indeed, even wolfram can't handle this problem.

I'm trying to use WolframMathematica but I can not make it understand the question because I'm not familiar with it...

There are no hints, suggestions related to the question

- Jan 26, 2012

- 183

Maple solved the problem giving

$\arctan \left( {\dfrac {y }{\sqrt {{x}^{2}-

y ^{2}}}} \right) +\dfrac{1}{2}\left(y^2+{x}^{2}\right)=c_1$

The way I would go about trying to solve this is the following. Expand your ODE and regroup according to the $\sqrt{x^2-y^2}$ So

$x \sqrt{x^2-y^2} (y y' + x) + xy' - y = 0$

or

$y y' + x + \dfrac{xy'-y}{x \sqrt{x^2-y^2}} = 0$

$y y' + x + \dfrac{xy'-y}{x^2 \sqrt{1-\left(\dfrac{y}{x}\right)^2}} = 0$

$\dfrac{1}{2} \dfrac{d}{dx} \left( x^2+y^2 \right) + \dfrac{\dfrac{d}{dx} \left(\dfrac{y}{x}\right)}{ \sqrt{1-\left(\dfrac{y}{x}\right)^2}} = 0$.

Now each piece can be integrated separately.