Solution of $\frac{\mathrm{d}y}{\mathrm{d}x}=y\mathrm{e}^x$ given $x=0$, $y=\mathrm{e}$

Question

$\dfrac{\mathrm{d}y}{\mathrm{d}x}=y\mathrm{e}^x$, $x=0$ and $y=\mathrm{e}$. Find the particular solution.

Attempt 1 $$ \dfrac{\mathrm{d}y}{\mathrm{d}x}=y\mathrm{e}^x\implies\dfrac{\mathrm{d}y}{y}=\mathrm{e}^x\,\mathrm{d}x\implies \log|y|=\mathrm{e}^x+C\\ x=0,\,y=\mathrm{e}\implies\log|\mathrm{e}|=\log\mathrm{e}=1=1+C\implies C=0\\ \log|y|=\mathrm{e}^x\implies|y|=\mathrm{e}^{\mathrm{e}^x}\implies \color{red}{y=\pm\mathrm{e}^{\mathrm{e}^x}} $$ Attempt 2 $$ \dfrac{\mathrm{d}y}{\mathrm{d}x}=y\mathrm{e}^x\implies\dfrac{\mathrm{d}y}{y}=\mathrm{e}^x\,\mathrm{d}x\implies \mathrm{e}^x=\log|y|+\log|C_1|=\log|C_1y|\\ \mathrm{e}^{\mathrm{e}^x}=|C_1y|=\pm C_1y=Cy\\ x=0,\,y=\mathrm{e}\implies\mathrm{e}=C\mathrm{e}\implies C=1\\ \implies \color{red}{y=\mathrm{e}^{\mathrm{e}^x}} $$

My reference also gives the solution $\log y=\mathrm{e}^x$ as in attempt 2. Why do I seem to get positive and negatve solutions in attempt 1 ?

How do I eliminate the solution $y=-\mathrm{e}^{\mathrm{e}^x}$ in attempt 1 ?

Note: I am not quite familiar with the idea of singularity or intermediate value theorem, as i have only done preliminary maths on first order differential equations.

Answer 1

Your first attempt is incomplete. You got to the point where two solutions $y(x)=e^{e^x}$ and $y(x)=-e^{e^x}$ are possible. The sign can not change inside a solution as they never take the value zero but have to be continuous. A sign change would thus contradict the intermediate value theorem.

As $y(0)=e>0$, you can reduce the number of possible solutions to one.

Answer 2

When dealing with the integral

$$\int \frac{dy}y=\log|y|+C$$ you are not allowed to cross the singularity at $y=0$ so that the solution is more precisely one of

$$\log y=e^x+C\text{ or }\log(-y)=e^x+C$$ (for all $x$).

With the given initial condition, the second option is excluded.