1) Solve the following differential equation

$\displaystyle sin x(e^y + 1) = e^y(1 + cos x) \frac{dy}{dx}$

$\displaystyle y(0)=0$

My steps

-I rearranged the formula and got $\displaystyle \frac{sinx}{1+cosx} dx = \frac{e^y}{e^y + 1}dy$

-But Don't know if this is right or how to even take the integral of both sides.

2)A particle moving along the x-axis encounters a resisting force that results in an acceleration of $\displaystyle a=\frac{dv}{dt}=-0.02v^2$. Given that x=0 cm and v=35 cm/s at t=0.Find the velocity and the position function for t as a function of t for t is greater than or equal to 0.

My Steps

$\displaystyle -a=-.02v^2$

$\displaystyle -.02\int v^2 dv= -.02 v^3/3$

Is this irhgt. This doesn't seem right.

Thanks for the help in advance.