I've just started this unit at university.

A question was to solve this ODE:

y' = y

Why is it that the answer is not y= y^2/2 + c,

but instead y= ce^x

also should it actually be written as y(x)=ce^x, or does that not matter?

Is there an algebraic approach to do this question? I put it in wolfram, and the final two steps were:

y(x) = e^(x+c)

Simplify

=> y(x) = ce^x

I don't quite understand how that simplifies.

Any help is appreciated, thanks!