y'=e^-y, y(0)=0, y(0.5)=?

y'=e^-y, y(0)=0, y(0.5)=?

Hi! I have solved the question above using the separation of variables method, can you guys help me to find the other two methods to find the exact answer (not the approximation answer)? Thank you!
Guest

Re: y'=e^-y, y(0)=0, y(0.5)=?

Where did you get this problem? Was it given to you in a class? You say "THE other two methods"! What makes you think there are exactly two other methods of solving this equation.

HallsofIvy

Posts: 341
Joined: Sat Mar 02, 2019 9:45 am
Reputation: 123

Re: y'=e^-y, y(0)=0, y(0.5)=?

Since this has been here awhile, I would solve it by writing the equation as
$$\frac{dy}{dx}= e^{-y}$$.

Separating the values, which the OP did not want to do, $$e^ydy= dx$$. Integrating,
$$e^y= x+ c$$.

Since y(0)= 0, $$e^0= 1= 0+ c$$ so c=1 so $$e^y= x+ 1$$.

Then $$e^{y(0.5)}= 0.5+ 1= 3/2$$.
$$y(0.5)= ln(3/2)$$.
Guest