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First Order Non-Linear Ordinary D.E.

dearcomp

New member
Oct 19, 2013
7
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.
 

Attachments

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: First Order Non Linear Ordinary D.E.

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, :)

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

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
Re: First Order Non Linear Ordinary D.E.

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,

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

dearcomp

New member
Oct 19, 2013
7
Re: First Order Non Linear Ordinary D.E.

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
 

Klaas van Aarsen

MHB Seeker
Staff member
Mar 5, 2012
8,851
Re: First Order Non Linear Ordinary D.E.

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
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?
 

dearcomp

New member
Oct 19, 2013
7
Re: First Order Non Linear Ordinary D.E.

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?
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 :(
 

Sudharaka

Well-known member
MHB Math Helper
Feb 5, 2012
1,621
Re: First Order Non Linear Ordinary D.E.

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 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
Yes indeed it has missing parenthesis and I typed it incorrectly in Wolfram. Sorry about that.

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 :(
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. :p
 

Jester

Well-known member
MHB Math Helper
Jan 26, 2012
183
Re: First Order Non Linear Ordinary D.E.

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.