1. ## Implicit Solution

Show that $xy^3 - xy^3\sin{(x)} = 1$ is an implicit solution to

$\frac{dy}{dx} = \frac{(x\cos{(x)} + \sin{(x)} - 1)y}{3(x - x\sin{(x)})}$

on the interval $\left(0,\frac{\pi}{2}\right)$

2. Originally Posted by Aryth
Show that $xy^3 - xy^3\sin{(x)} = 1$ is an implicit solution to

$\frac{dy}{dx} = \frac{(x\cos{(x)} + \sin{(x)} - 1)y}{3(x - x\sin{(x)})}$

on the interval $\left(0,\frac{\pi}{2}\right)$
Differentiate both sides:

$\frac{d}{dx} (xy^3 - xy^3\sin{(x)}) = \frac{d}{dx}1$

$\frac{d}{dx} (xy^3) - \frac{d}{dx}( xy^3\sin{(x)}) =0$

Product rule on the 2nd term!

$\frac{d}{dx} (xy^3) - sin(x) \frac{d}{dx}( xy^3) - xy^3 \frac{d}{dx}(\sin{(x)}) =0$

Take out a factor of $\frac{d}{dx} (xy^3)$ on LHS:

$(1-\sin{(x)}) \frac{d}{dx} (xy^3) - xy^3 \frac{d}{dx}(\sin{(x)}) =0$

Using implicit differentiation with the product rule: $\frac{d}{dx} (xy^3) = y^3 + 3y^2 x \frac{dy}{dx}$, and also $\frac{d}{dx} \sin{(x)} = cos(x)$ hence:

$(1-\sin{(x)})( y^3 + 3y^2 x \frac{dy}{dx}) - xy^3 \cos{(x)}) =0$

Now rearrange to get into explicit form, and then manipulate to get it into the form posed by the question!