Separable ODE, should I take the log of both sides?

1. Jan 13, 2010

JFonseka

1. The problem statement, all variables and given/known data
Solve dy/dx = exp(x-y) given that y = ln 2 at x = 0

2. Relevant equations
None.

3. The attempt at a solution

Firstly let's get the equation in to a form so we can re-arrange the x's and y's, and then re-arrange.

dy/dx = exp(x)/exp(y)

exp(y)*dy = exp(x)*dx

Integrate:

exp(y) = exp(x) + C

Substitute:

exp(ln(2)) = exp(0) + C

2 = 1 + C, therefore C = 1

Hence, exp(y) = exp(x) + 1, and take log to get an equation in the form of y = ...

ln(exp(y)) = ln(exp(x)) + ln(1) =>

y = x

Doesn't really look right though, am I right or wrong?

Edit: Solved-> ln(exp(y)) = ln(exp(x) + 1)
=> y(x) = ln(exp(x) + 1)

Last edited: Jan 13, 2010
2. Jan 13, 2010

vela

Staff Emeritus
Looked OK until the last step when you took the logs. log(a+b) isn't equal to log(a)+log(b).

Note also that the initial condition no longer holds if y=x was the solution.

3. Jan 13, 2010

JFonseka

I solved the problem using Maple; it gives the answer as y = ln(exp x + 1).

So I guess I'm meant to take the log of the entire expression on the r.h.s.