1. ## General solution

find general solution of

xy(1+x^2)dy/dx-(1+y^2)=0

Solution:

this becomes

x(1+x^2) dy/dx=(1+y^2)/y

getting x's on one side and y's on the other
then the difficulties start, i tried to get the dy on the side of the y's to differenciate but when i do this i get 1/dy which seems weird

then tried

x(1+x^2) dy = (1/y+y) dx and differenciated left side(x's) with respect to y and the right side (y's) with respect to x which seems weird...
is this right?

2. Originally Posted by fitzgilbert
find general solution of

xy(1+x^2)dy/dx-(1+y^2)=0

Solution:

this becomes

x(1+x^2) dy/dx=(1+y^2)/y

...
This is only a try because I don't know what you are supposed to do:

$\frac{x(1+x^2)}{dx}=\frac{1+y^2}{y \cdot dy}~\implies~ \frac1{x(1+x^2)} \cdot dx=\frac{y}{1+y^2} \cdot dy$

3. Originally Posted by earboth
This is only a try because I don't know what you are supposed to do:

$\frac{x(1+x^2)}{dx}=\frac{1+y^2}{y \cdot dy}~\implies~ \frac1{x(1+x^2)} \cdot dx=\frac{y}{1+y^2} \cdot dy$
Integrate both sides and then solve for y.