## Thomas' Calculus 13th Edition

As we are given that $y=\int_{0}^{x} (1+2 \sqrt {\sec t}) dt$ Now, $y'=1+2 \sqrt {\sec x}$ and $y''=2 (\dfrac{1}{2}) (\sec x)^{-(1/2)} (\sec x\tan x)=(\sec x)^{-\frac{1}{2}+1} \tan x$ $\implies y''=(\sec x)^{\frac{1}{2}} \tan x=\sqrt {\sec x} \tan x$ Apply initial conditions $y(0)=0$ $\implies y'(0)=1+2 \sqrt {\sec (0)}=1+2(1)=3$ Hence, the differential equation is satisfied.