Hi again. I was having some trouble with this implicit differentiation problem and was wondering if I could get some help.
$\displaystyle \frac{dy}{dx} - (3 + 2x)(1 + y^2) = 0$
Thanks
You use the lazy notation :P.
To be COMPLETELY correct...
$\displaystyle \frac{dy}{dx} = (3 + 2x)(1 + y^2)$
$\displaystyle \frac{1}{1 + y^2}\,\frac{dy}{dx} = 3 + 2x$
$\displaystyle \int{\frac{1}{1 + y^2}\,\frac{dy}{dx}\,dx} = \int{3 + 2x\,dx}$
$\displaystyle \int{\frac{1}{1 + y^2}\,dy} = 3x + x^2 + C_1$
$\displaystyle \arctan{y} + C_2 = 3x + x^2 + C_1$
$\displaystyle \arctan{y} = 3x + x^2 + C$, where $\displaystyle C = C_1 - C_2$
$\displaystyle y = \tan{(3x + x^2 + C)}$.