TITLE: Solution of the IVP: $\,y'=\mathrm{e}^{-y^2}-1,\, y(0)=0$
QUESTION [6 upvotes]: Consider the initial value problem
$$ 
\frac{dy}{dx} = \mathrm{e}^{-y^2} - 1,\quad y(0)=0.
$$
The Method of Separation of Variables provides that:
$$
\int \frac{dy}{e^{-y^2} - 1}  = x+c.
$$
I would be thankful who somone can give me hint.

REPLY [10 votes]: Unfortunately, this Initial Value Problem (IVP) can not be solved using Separation of Variables, since the initial value, $y(0)=0$, kills the flux function: $\,f(y)=\mathrm{e}^{-y^2}-1$ (i.e., $\,\,f(0)=0$), and hence we can not divide by $f(y)$ in order to apply Separation of Variables.
In such case we can only EVOKE UNIQUENESS! 
Our IVP enjoys uniqueness (since that flux $f$ is $C^\infty$ and hence locally Lipschitz continuous), and therefore, any solution of this IVP, regardless how we came up to it, is THE solution $-$ For example, try for a constant one.
So, the one and only solution of this IVP is the constant one: $y(t)\equiv 0$
Note. I have taught ODEs several times, and I always give examples similar to the above, in order to explain that the method of Separation of Variables needs to be treated with care!