Answer
$y=\frac{1}{3}tan(3x)+\frac{7}{3}$
Work Step by Step
$\frac{dy}{dx}=sec^2(3x);$ $y(\frac{\pi}{4})=2$
$dy=sec^2(3x)dx$
$\int dy=\frac{1}{3}\int 3sec^2(3x)dx$
$y=\frac{1}{3}tan(3x)+C$
Substitute in the initial condition to solve for $C$
$2=\frac{1}{3}tan(\frac{3\pi}{4})+C$
$C=\frac{7}{3}$
$y=\frac{1}{3}tan(3x)+\frac{7}{3}$
