1. ## 2 Differential Equations

1) Solve the following differential equation
$sin x(e^y + 1) = e^y(1 + cos x) \frac{dy}{dx}$
$y(0)=0$

My steps
-I rearranged the formula and got $\frac{sinx}{1+cosx} dx = \frac{e^y}{e^y + 1}dy$
-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 $a=\frac{dv}{dt}=-0.02v^2$. 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
$-a=-.02v^2$
$-.02\int v^2 dv= -.02 v^3/3$
Is this irhgt. This doesn't seem right.

Thanks for the help in advance.

2. NUMBER 1.) Well I have no clue if you are supposed to set up that diff. eq like that (I don't know any about them). But those are very easily integrated.

$\int{\frac{\sin{x}}{1+\cos{x}}dx} =$

u = 1 + cos(x)
du = -sin(x)

so:

$- \int{\frac{-\sin{x}}{1+\cos{x}}dx} = -ln(1 + \cos{x}) + c$

AND -

$\int{\frac{e^y}{e^y + 1}dy} =$

u = e^y + 1
du = e^y

so:

$\int{\frac{e^y}{e^y + 1}dy} = ln(e^y + 1) + c$

NUMBER 2.) It seems odd... but appears that it should be correct? The integral of the acceleration function is the velocity function, which is what you did. Integrate that again to get the position function. But it does look odd, especially as the variable inside the function is a v. Shouldn't it be a t?

3. yea i know that is why i am un easy about doing the problems. they just seem obscure.

4. I think you'd rather want to integrate with respect to t than v. Read te problem again.