Solution
= Solution
The <initial condition> gives $e^{y(0)}=2=1+C$, so $C=1$. Consequently
$$
\boxed{y(x)=\ln(1+e^x),\qquad x\in\mathbb R.}
$$
Its logarithm has positive argument for every real $x$.
= Solution
The <initial condition> gives $e^{y(0)}=2=1+C$, so $C=1$. Consequently
$$
\boxed{y(x)=\ln(1+e^x),\qquad x\in\mathbb R.}
$$
Its logarithm has positive argument for every real $x$.