Answer
$\dfrac{y^3}{3}-e^x+e^{-x}-\dfrac{1}{3}=0$
Work Step by Step
Separate the differentials and integrate.
$\int (\dfrac{e^{2x}+1}{e^x}) dx=\int y^2 dy$
Re-write as: $\int (\dfrac{e^{2x}}{e^x}+\dfrac{1}{e^x}) dx=\int y^2 dy$
This implies that $e^x-e^{-x}+c=(\dfrac{1}{3})y^3+c$
Applying the initial conditions $y(0)=1$, we get
Thus, $e^0-e^{-0}+c=(\dfrac{1}{3}) \implies c=\dfrac{1}{3}$
Hence, $\dfrac{y^3}{3}-e^x+e^{-x}-\dfrac{1}{3}=0$
