Hello

I stopped at this one:

$\displaystyle y(x^2-y^2+1)dx-x(x^2-y^2-1)dy=0$

I opened the brackets:

$\displaystyle yx^2dx - y^3dx + ydx-x^3dy+xy^2dy+xdy=0$

It will be:

$\displaystyle yx^2dx - y^3dx-x^3dy+xy^2dy+d(xy)=0$

then ?!!!!

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- Aug 22nd 2010, 07:41 PMBayernMunichSolving ODE by inspection
Hello

I stopped at this one:

$\displaystyle y(x^2-y^2+1)dx-x(x^2-y^2-1)dy=0$

I opened the brackets:

$\displaystyle yx^2dx - y^3dx + ydx-x^3dy+xy^2dy+xdy=0$

It will be:

$\displaystyle yx^2dx - y^3dx-x^3dy+xy^2dy+d(xy)=0$

then ?!!!! - Aug 22nd 2010, 07:52 PMBayernMunich
I multiplied the last equation by 3, to get:

$\displaystyle yd(x^3)-3y^3dx-3x^3dy+xd(y^3)+3d(xy)=0$

Now, I stopped. - Aug 23rd 2010, 06:31 AMJester
Re-writing gives

$\displaystyle -(x^2-y^2)(xdy-ydx) + xdy+ydx = 0$

$\displaystyle -(x^2-y^2)x^2 d\left(\dfrac{y}{x}\right) + d(xy) = 0$

$\displaystyle (1-\frac{x^2}{y^2}) x^2y^2d\left(\dfrac{y}{x}\right) + d(xy) = 0$

$\displaystyle \left(1-\dfrac{1}{\left(\frac{y}{x}\right)^2}\right) d\left(\dfrac{y}{x}\right) + \dfrac{d(xy)}{(xy)^2} = 0$ - Aug 24th 2010, 10:01 AMBayernMunich
Thanks ..

so the final answer will be :

$\displaystyle \dfrac{y}{x}+\dfrac{x}{y}-\dfrac{1}{xy}=0$

Right? - Aug 24th 2010, 03:52 PMJester
+ c but that's what I got.