Calculus: Early Transcendentals (2nd Edition)

$$y\left( \theta \right) = \sqrt 2 \sin \theta + \tan \theta + 1$$
\eqalign{ & y'\left( \theta \right) = \frac{{\sqrt 2 {{\cos }^3}\theta + 1}}{{{{\cos }^2}\theta }} \cr & y\left( \theta \right) = \int {y'\left( \theta \right)} d\theta \cr & then \cr & y\left( \theta \right) = \int {\left( {\frac{{\sqrt 2 {{\cos }^3}\theta + 1}}{{{{\cos }^2}\theta }}} \right)} d\theta \cr & y\left( \theta \right) = \int {\left( {\frac{{\sqrt 2 {{\cos }^3}\theta }}{{{{\cos }^2}\theta }} + \frac{1}{{{{\cos }^2}\theta }}} \right)} d\theta \cr & y\left( \theta \right) = \int {\left( {\sqrt 2 \cos \theta + {{\sec }^2}\theta } \right)} d\theta \cr & y\left( \theta \right) = \int {\sqrt 2 \cos \theta } d\theta + \int {{{\sec }^2}\theta } d\theta \cr & find{\text{ the general solution}} \cr & y\left( \theta \right) = \sqrt 2 \sin \theta + \tan \theta + C \cr & {\text{using the initial condition }}y\left( {\frac{\pi }{4}} \right) = 3 \cr & 3 = \sqrt 2 \sin \left( {\frac{\pi }{4}} \right) + \tan \left( {\frac{\pi }{4}} \right) + C \cr & 3 = \sqrt 2 \left( {\frac{{\sqrt 2 }}{2}} \right) + 1 + C \cr & 3 = 2 + C \cr & C = 1 \cr & {\text{the solution to the initial value problem is}} \cr & y\left( \theta \right) = \sqrt 2 \sin \theta + \tan \theta + 1 \cr}