After a little manipulation, the given diff'l equation will look like this:
e^y * dy = (2x + 1) * dx.
x^2
Integrating both sides, we get e^y = 2------- + x + c, or e^y = x^2 + x + c
2
Now let x=0 and y = 1, o find c:
e^1 = 0^2 + 0 + c. So, c = e, and the solution is e^y = x^2 + x + e.
