Chapter 9: First-Order Differential Equations - Section 9.2 - First-Order Linear Equations - Exercises 9.2 - Page 536: 4

$y=\sin x \cos x+c$

Re-write the given differential equation as: $y'+(\tan x)y=\cos^2(x)$ The integrating factor is: $v(x)=e^{\tan x dx}=\dfrac{1}{\cos x}$ Now, $\int (\dfrac{1}{\cos x})y'+(\dfrac{\sin x}{\cos^2 x})y=\int \cos x dx$ $\int (\dfrac{1}{\cos x} y)'=\int \cos x$ or, $\dfrac{1}{\cos x}y=\sin x+c$ Hence, the General solution is: $y=\sin x \cos x+c$