Differential Eqns: dy/d0 = (e^y(sin0)^2) / (y sec 0), etc

[paulxzt]

0 is theta

1) dy/d0 = (e^y(sin0)^2) / (y sec 0)

ysec0*dy = e^y (sin0)^2 d0

(y/e^y)dy=(sin0)^2/(sec0) d0

am i doing this right? how do i proceed from here? do i use u sub.?

2)dy/dx = (ycosx)/(1+y^2)

(1+y^2)/y dy = cosx dx

am i doing this right? I integrated the right side to be sin x + c but cannot seem to do the left side.. should i be using integration by parts?
thanks for any help

 

[galactus]

\(\displaystyle \L\\\frac{dy}{d{\theta}}=\frac{e^{y}sin^{2}{\theta}}{ysec{\theta}}\)

\(\displaystyle \L\\\frac{dy}{d{\theta}}=\frac{e^{y}sin^{2}{\theta}cos{\theta}}{y}\)


First, seperate variables. Which you seem have done OK.

\(\displaystyle \L\\\frac{y}{e^{y}}dy=sin^{2}{\theta}cos{\theta}d{\theta}\)

Integrate:

\(\displaystyle \L\\\int{ye^{-y}}dy=\int{sin^{2}{\theta}cos{\theta}}d{\theta}\)

You can use parts on the left, as you asked.

Let \(\displaystyle \L\\u=y, \;\ dv=e^{-y}dy, \;\ du=dy, \;\ v=-e^{-y}\)

\(\displaystyle \L\\(-y-1)e^{-y}=\frac{sin^{3}{\theta}}{3}+C\)