Solution of the IVP: $\,y'=\mathrm{e}^{-y^2}-1,\, y(0)=0$

Question

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.

Answer 1

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!