Question

Given \(\frac{dy}{dx} + 2y\tan x = \sin x\), \(y=0\) at \(x=\frac{\pi}{3}\). If maximum value of \(y\) is \(1/k\), find \(k\).

Hint:

For linear DE, find IF first. For maximum of trigonometric expression, substitute \(t = \cos x\) and use quadratic optimization.

Answer 1

Correct Answer: 8

Concept: Linear differential equation: \(\frac{dy}{dx} + P(x)y = Q(x)\). Integrating factor: \(IF = e^{\int P\,dx}\).

Step 1: Identify \(P(x)\) and \(Q(x)\). \[ P(x) = 2\tan x, \quad Q(x) = \sin x \]

Step 2: Find integrating factor. \[ IF = e^{\int 2\tan x \, dx} = e^{2\ln|\sec x|} = e^{\ln(\sec^2 x)} = \sec^2 x \]

Step 3: Multiply the DE by IF. \[ \sec^2 x \frac{dy}{dx} + 2y\sec^2 x \tan x = \sin x \sec^2 x \] \[ \frac{d}{dx}(y \sec^2 x) = \sin x \cdot \frac{1}{\cos^2 x} = \frac{\sin x}{\cos^2 x} \]

Step 4: Integrate both sides. \[ y \sec^2 x = \int \frac{\sin x}{\cos^2 x} \, dx \] Let \(u = \cos x\), \(du = -\sin x \, dx\): \[ \int \frac{\sin x}{\cos^2 x} \, dx = -\int u^{-2} \, du = -(-u^{-1}) + C = \frac{1}{\cos x} + C = \sec x + C \] Thus: \[ y \sec^2 x = \sec x + C \]

Step 5: Apply initial condition \(y=0\) at \(x = \frac{\pi}{3}\). \[ \sec\frac{\pi}{3} = 2 \] \[ 0 \cdot \sec^2\frac{\pi}{3} = 2 + C \Rightarrow 0 = 2 + C \Rightarrow C = -2 \]

Step 6: General solution. \[ y \sec^2 x = \sec x - 2 \] \[ y = \cos^2 x (\sec x - 2) = \cos x - 2\cos^2 x \]

Step 7: Find maximum value of \(y\). Let \(t = \cos x\), \(t \in [-1, 1]\): \[ y = t - 2t^2 \] \[ \frac{dy}{dt} = 1 - 4t = 0 \Rightarrow t = \frac{1}{4} \] \[ \frac{d^2y}{dt^2} = -4<0 \Rightarrow \text{maximum at } t = \frac{1}{4} \] \[ y_{\text{max}} = \frac{1}{4} - 2\left(\frac{1}{16}\right) = \frac{1}{4} - \frac{1}{8} = \frac{1}{8} \] Given that maximum value of \(y\) is \(1/k\): \[ \frac{1}{k} = \frac{1}{8} \Rightarrow k = 8 \]