## University Calculus: Early Transcendentals (3rd Edition)

$\dfrac{y^3}{3}-e^x+e^{-x}-\dfrac{1}{3}=0$
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, $\dfrac{y^3}{3}-e^x+e^{-x}-\dfrac{1}{3}=0$