Question

Solve for y' =e^(x+y)

Answer 1

I separated the variables. I got 1/e^y dy = e^x dx. Then I integrated both sides. On the RHS, I simply got e^x +C. I'm having trouble with the LHS.

Answer 2

If you write the LHS as e^(-y), then integrating it is similar to the RHS. The resulting equation I got after integrating is: -e^(-y)=e^x + C.

Simplified to solve for y: y = -ln(-e^x+C)

Answer 3

thank you!