What method should I use to find the General Solution of this?Code:(x^2+y^2-5)dx=(x+xy)dy
$\displaystyle (x^2+y^2-5)dx=(x+xy)dy$
$\displaystyle \frac{dy}{dx}(x+xy)=x^2+y^2-5$
$\displaystyle \frac{dy}{dx}=\frac{x^2+y^2-5}{x+xy}$
$\displaystyle \int \frac{dy}{dx} dx=\int \frac{x^2+y^2-5}{x+xy} dx$
$\displaystyle \int dy=\int \frac{x^2+y^2-5}{x+xy} dx$
dy is the same as 1.dy, and so integrating 1 (which is congruent to $\displaystyle y^0$) with respect to y will give you $\displaystyle \frac{y^1}{1} = y$
$\displaystyle y=\int \frac{x^2+y^2-5}{x+xy} dx$
Unfortunately, my high school understanding of calculus doesn't give me a method to integrate the RHS of this equation. If someone here could do that then you should have your general solution. I have no idea if this is correct though, just a long shot.