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

by **Guest** » Fri Oct 09, 2020 12:32 pm

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!

by **HallsofIvy** » Wed Dec 23, 2020 3:59 pm

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.

by **Guest** » Tue Oct 12, 2021 4:43 pm

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

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

Since y(0)= 0, [tex]e^0= 1= 0+ c[/tex] so c=1 so [tex]e^y= x+ 1[/tex].

Then [tex]e^{y(0.5)}= 0.5+ 1= 3/2[/tex].
[tex]y(0.5)= ln(3/2)[/tex].

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

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

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

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

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