What method should I use to find the General Solution of this?Code:`(x^2+y^2-5)dx=(x+xy)dy`

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- Apr 28th 2013, 04:24 AMshinmeiDifferential Equation<don't know what method to useCode:
`(x^2+y^2-5)dx=(x+xy)dy`

- Apr 28th 2013, 06:13 AMtopsquarkRe: Differential Equation<don't know what method to use
- May 5th 2013, 04:47 AMQwobRe: Differential Equation<don't know what method to use
$\displaystyle (x^2+y^2-5)dx=(x+xy)dy$

$\displaystyle \frac{dy}{dx}(x+xy)=x^2+y^2-5$

$\displaystyle \frac{dy}{dx}=\frac{x^2+y^2-5}{x+xy}$

$\displaystyle \int \frac{dy}{dx} dx=\int \frac{x^2+y^2-5}{x+xy} dx$

$\displaystyle \int dy=\int \frac{x^2+y^2-5}{x+xy} dx$

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. - May 5th 2013, 11:23 AMQwobRe: Differential Equation<don't know what method to use
^ the problem to me seems to be separating the unknowns $\displaystyle x$ and $\displaystyle y$ onto the different sides of the equation. I can't seem to factorise them myself.