Can someone help me with this equation? y'=e^-y, y(0)=0, y(0.5)=?

teyyyyyi

New member
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!

Staff member

HallsofIvy

Elite Member
This is NOT a "linear equation" so some of the methods I mention in your previous thread will not work. Yes, it is very easy to solve $$\displaystyle \frac{dy}{dx}= e^{-y}$$ by integration by parts. I still don't know why you need two other methods!

teyyyyyi

New member
Separation of variables.
dy/dx = e^-y
e^ydy = dx
integrate both sides u get
e^y = x + c
y = log (x+c)
y(0)=0 so
0 = log (0 + c)
c = 1
y(0.5) = log (1.5) = 0.4054651081

This is how I did it.

Jomo

Elite Member
Separation of variables.
dy/dx = e^-y
e^ydy = dx
integrate both sides u get
e^y = x + c
y = log (x+c)
y(0)=0 so
0 = log (0 + c)
c = 1
y(0.5) = log (1.5) = 0.4054651081

This is how I did it.
Looks good to me.

Now let u = e^-y

teyyyyyi

New member
This is NOT a "linear equation" so some of the methods I mention in your previous thread will not work. Yes, it is very easy to solve $$\displaystyle \frac{dy}{dx}= e^{-y}$$ by integration by parts. I still don't know why you need two other methods!
hahaha it's for my assignment, he asked us to find 3 methods to solve it thank you for your help anyway!

teyyyyyi

New member
Looks good to me.

Now let u = e^-y
I have some difficulty in continuing the equation.
y' = e^-y, let u = e^-y so du/dx = -e^-y
is it right or do I need to make it in terms of y? I am sorry I'm new to this.

Subhotosh Khan

Super Moderator
Staff member
I have some difficulty in continuing the equation.
y' = e^-y, let u = e^-y so du/dx = -e^-y.................................. Incorrect
is it right or do I need to make it in terms of y? I am sorry I'm new to this.
u = e^(-y)

u' = - e^(-y) * dy/dx = - u2

teyyyyyi

New member
u = e^(-y)

u' = - e^(-y) * dy/dx = - u2
i have gotten the answer, thank you so much for your help!