Originally Posted by

**sparky** Is my working and answer correct for the following question?

Question: Solve for the following differential equation for the original equation:

$\displaystyle y' = \frac{2x-3y}{3x+8y}$

Here is my working:

$\displaystyle \frac{dy}{dx} = \frac{2x-3y}{3x+8y}$

$\displaystyle (3x+8y)dy = (2x-3y)dx$

$\displaystyle \int 3xdy + \int 8ydy = \int 2xdx - \int 3ydx$ (why does the intergral of 3ydx = 0?)

$\displaystyle \frac{3x^2}{2} + \frac{8y^2}{2} = \frac{2x^2}{2} - 0 + C$

$\displaystyle 3x^2 + 8y^2 - 2x^2 - C = 0$

$\displaystyle x^2 + 8y^2 - C = 0$