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

teyyyyyi

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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! :)
 

tkhunny

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Please demonstrate how you accomplished the task. Then we can talk.
 

HallsofIvy

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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

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Please demonstrate how you accomplished the task. Then we can talk.
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

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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

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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

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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

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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
 
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