- Thread starter teyyyyyi
- Start date

- Joined
- Jan 27, 2012

- Messages
- 6,704

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

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.

- Joined
- Dec 30, 2014

- Messages
- 8,291

Looks good to me.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.

Now let u = e^-y

hahaha it's for my assignment, he asked us to find 3 methods to solve it thank you for your help anyway!

I have some difficulty in continuing the equation.Looks good to me.

Now let u = e^-y

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.

- Joined
- Jun 18, 2007

- Messages
- 22,405

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

i have gotten the answer, thank you so much for your help!u = e^(-y)

u' = - e^(-y) * dy/dx = - u^{2}