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

$\displaystyle y' = \frac{7-4x}{4}$

Is my working correct?

$\displaystyle \frac{dy}{dx} = \frac{7-4x}{4}$

$\displaystyle 4dy = (7-4x)dx$

$\displaystyle \int 4dy = \int 7dx - \int 4xdx$

$\displaystyle 4y = 7x - \frac{4x^2}{2} + C$

$\displaystyle 8y = 14x - 4x^2 + 2C$ (multiply both sides by 2 to get rid of the half)

$\displaystyle 8y - 14x + 4x^2 - 2C = 0$