Problem Statement
Solve the differential equation $\frac{dy}{dx} + y\tan x = \cos^2 x$, given that $y = 2$ when $x = 0$, for $-\frac{\pi}{2} < x < \frac{\pi}{2}$. [7 Marks]
Verified Solution & Marking Scheme
Compute Integrating Factor I(x)
Equation is in standard form $\frac{dy}{dx} + P(x)y = Q(x)$ with $P(x) = \tan x$ and $Q(x) = \cos^2 x$.
$$I(x) = e^{\int \tan x \, dx} = e^{\ln|\sec x|} = \sec x$$
(since $\sec x > 0$ on $-\frac{\pi}{2} < x < \frac{\pi}{2}$).
Multiply and Integrate
$$\frac{d}{dx}\left(y \sec x\right) = \cos^2 x \cdot \sec x = \cos^2 x \cdot \frac{1}{\cos x} = \cos x$$
Integrate both sides:
$$y \sec x = \int \cos x \, dx = \sin x + C$$
Apply Initial Condition y(0) = 2
$$2 \sec(0) = \sin(0) + C \implies 2(1) = 0 + C \implies C = 2$$
Therefore:
$$y \sec x = \sin x + 2 \implies y = \frac{\sin x + 2}{\sec x} = \sin x\cos x + 2\cos x = \frac{1}{2}\sin 2x + 2\cos x$$