This is the problem:

For the differential equation


find the solution for the initial value problem $y(0)=0$.

I tried to plug in 0 into the equation, which lead me nowhere - $0=exp(x)$. What am I doing wrong, and how do I get the correct answer? Thanks.

  • $\begingroup$ Hint: $y'\, e^y=e^x $. $\endgroup$ – Pocho la pantera Sep 4 '13 at 0:52

The equation can be written as $$ e^y \,dy=e^x \, dx $$ Integrating we obtain $$ e^y=e^x +c $$ So, if $y(0)=0$ then $c=0$. Therefore $y=x$.

  • $\begingroup$ Oh, so you have to mess around with the $e$'s to get to the answer. That definitely helps a lot, thank you. $\endgroup$ – user91971 Sep 4 '13 at 6:16

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.