Finding the Exact Solution of the Initial Value Problem for y'=e^(x+y), y(0)=0

[peace-Econ]

Homework Statement

Find the exact solution of the initial value problem. Indicate the interval of existence.

Homework Equations

y'=e^(x+y), i.v.p:y(0)=0

The Attempt at a Solution

this is my attempt:

dy/dx=e^x+y=(e^x)(e^y)

--> dy/e^y=(e^x)dx

Integrating, -e^-y=e^x+C (C is constant) --> e^y=-e^x-C

--> ln(e^-y)=ln(-e^x-C) --> y=-ln(-e^x-C)

Because we have y(0)=0, 0=-ln(-1-C), so C=-2
Therefore, y(x)=-ln(2-e^x) (=ln(1/(2-e^x)))
Then, the interval of existence is (0, ln2).

This is what i did, but I'm not confident for my work. So I want someone to look at it and help me if you find any mistake. Thanks!

 

[SammyS]

Of course, x can be zero. y(0)= 0 . Right?

In fact what makes you say that x can't be negative?

 

[peace-Econ]

you're right actually...

so is the interval of existence actually (-infinity, ln2)?