# Solving ODE by inspection

• Aug 22nd 2010, 07:41 PM
BayernMunich
Solving 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 PM
BayernMunich
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 AM
Jester
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 AM
BayernMunich
Thanks ..

so the final answer will be :

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

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