sec(x) dy/dx = e^(y + sin(x))
So far I have :
ln (secx) dy/dx = y + sin x
ln (secx) dy = y + sinx dx
Not sure what to do now... I don't think I can just subtract the y?
you should be able to solve this using implicit differentiation.
so you have $\displaystyle \frac{dy}{dx} = -\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}} = -\frac{F_x}{F_y} $
so then
$\displaystyle \sec(x) dy/dx = e^{y + \sin(x)}$
$\displaystyle = \frac{1}{\cos(x)} \frac{dy}{dx} = e^{y + \sin(x)} $
$\displaystyle \frac{dy}{dx}= e^{y + \sin(x)} \times \cos(x)$
$\displaystyle \frac{\partial F}{\partial y} = e^{y + \sin(x)} \times \cos(x)$
$\displaystyle \frac{\partial F}{\partial x} = \cos(x) \times e^{y + \sin(x)} \times \cos(x) -sin(x) \times e^{y + \sin(x)} = e^{y + \sin(x)}(\cos^2(x) -\sin(x))$
now just combine both function...