1. ## differential equations

i need help with some of these, i kind of know how to do it, but i get stuck...

1. dy/dx = (ycosx)/(1+y^2) y(0)=1
i know how to get to the integral of both sides what to i do becuase i get a y^2 on the left?
we are supposed to solve the equation using the values given..

2. dz/dt + e^(t+z)

im not sure how to separate the equations here

3. dp/dt = sqrt(pt) p(1) = 2
im not sure if i got this right but i got p = sqrt (1/4sqrt(t) + 15/4)

2. $\displaystyle \frac{dy}{dx} = \frac{y\cos{x}}{1+y^2}$

$\displaystyle \frac{1+y^2}{y} \, dy = \cos{x} \, dx$

$\displaystyle \int \frac{1}{y} + y \, dy = \int \cos{x} \, dx$

$\displaystyle \frac{dz}{dt} = e^{t+z}$

$\displaystyle \frac{dz}{dt} = e^t \cdot e^z$

$\displaystyle \int e^{-z} \, dz = \int e^t \, dt$

$\displaystyle \frac{dp}{\sqrt{p}} = \sqrt{t} \, dt$

$\displaystyle 2p^{\frac{1}{2}} = \frac{2}{3} t^{\frac{3}{2}} + C$

$\displaystyle p^{\frac{1}{2}} = \frac{1}{3} t^{\frac{3}{2}} + C$

$\displaystyle p = \left(\frac{1}{3} t^{\frac{3}{2}} + C\right)^2$