Differential equation problem

mooshupork34

Junior Member
Joined
Oct 29, 2006
Messages
72
This was confusing me so any help or explanation of how this type of problem is solved would be appreciated.

Find the function y(t), given that:

y' = e^-y * cos(t)

and y(0) = 0.

Also, verify that the solution satisfies both the differential equation adn the initial condition.
 
\(\displaystyle \L\\\frac{dy}{dt}=e^{-y}cos(t)\)

Separate variables:

\(\displaystyle \L\\\frac{y'}{e^{-y}}=cos(t)dt\)

\(\displaystyle \L\\e^{y}=sin(t)+C\)

\(\displaystyle \L\\y=ln(sin(t)+C)\)

Use initial conditions and get C=1

\(\displaystyle \L\\y=ln(sin(t)+1)\)
 
galactus said:
\(\displaystyle \L\\\frac{dy}{dt}=e^{-y}cos(t)\)

Separate variables:

\(\displaystyle \L\\\frac{y'}{e^{-y}}=cos(t)dt\)

\(\displaystyle \L\\e^{y}=sin(t)+C\)

\(\displaystyle \L\\y=ln(sin(t)+C)\)

Use initial conditions and get C=1

\(\displaystyle \L\\y=ln(sin(t)+1)\)

Thanks!

Just to make sure that I'm understanding this...
Would it be correct if I were to rewrite that first part such that...

dy/dt = e^-y cos(t)
dy/dt = 1/e^y * cos(t)
dy/dt = cos(t)/e^y
integral of e^y dy = integral of cos(t) dt
e^y = sin(t) + C
taking the natural log of both sides, we get ln(e^y) = ln(sin(t)+C)
 
Top