Solving First Order Linear Differential Equation

To solve the given first-order linear differential equation:

$\frac{dy}{dx} - y\tan x = 2x\sec x$

The equation is in standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = -\tan x$ and $Q(x) = 2x\sec x$.

Step 1: Find the Integrating Factor (I.F.)

$\text{I.F.} = e^{\int P(x)\,dx} = e^{\int -\tan x\,dx} = e^{-\ln|\sec x|} = e^{\ln|\cos x|} = \cos x$

Step 2: General Solution

Multiplying both sides of the differential equation by the integrating factor: $\cos x \cdot \frac{dy}{dx} - y\sin x = 2x$

$\frac{d}{dx}(y\cos x) = 2x$

Integrating both sides with respect to $x$: $y\cos x = \int 2x\,dx$ $y\cos x = x^2 + C$

Step 3: Apply the Initial Condition

Given that $y(0) = 0$: $0\cdot\cos(0) = 0^2 + C \implies C = 0$

Thus, the solution is: $y(x) = x^2\sec x$

Step 4: Derivative of $y(x)$

Differentiating $y(x)$ with respect to $x$ (or using the original differential equation): $y'(x) = 2x\sec x + x^2\sec x\tan x$

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Step 5: Evaluate the Given Options

1. For Option (A): $y\left(\frac{\pi}{4}\right) = \left(\frac{\pi}{4}\right)^2 \sec\left(\frac{\pi}{4}\right) = \frac{\pi^2}{16} \cdot \sqrt{2} = \frac{\pi^2}{8\sqrt{2}}$ Hence, Option (A) is correct.

2. For Option (B): $y'\left(\frac{\pi}{4}\right) = 2\left(\frac{\pi}{4}\right)\sec\left(\frac{\pi}{4}\right) + \left(\frac{\pi}{4}\right)^2\sec\left(\frac{\pi}{4}\right)\tan\left(\frac{\pi}{4}\right) = \frac{\pi}{\sqrt{2}} + \frac{\pi^2\sqrt{2}}{16} \neq \frac{\pi^2}{18}$ Hence, Option (B) is incorrect.

3. For Option (C): $y\left(\frac{\pi}{3}\right) = \left(\frac{\pi}{3}\right)^2 \sec\left(\frac{\pi}{3}\right) = \frac{\pi^2}{9} \cdot 2 = \frac{2\pi^2}{9} \neq \frac{\pi^2}{9}$ Hence, Option (C) is incorrect.

4. For Option (D): $y'\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)\sec\left(\frac{\pi}{3}\right) + \left(\frac{\pi}{3}\right)^2\sec\left(\frac{\pi}{3}\right)\tan\left(\frac{\pi}{3}\right)$ $y'\left(\frac{\pi}{3}\right) = 2\left(\frac{\pi}{3}\right)(2) + \frac{\pi^2}{9}(2)(\sqrt{3}) = \frac{4\pi}{3} + \frac{2\sqrt{3}\pi^2}{9} = \frac{4\pi}{3} + \frac{2\pi^2}{3\sqrt{3}}$ Hence, Option (D) is correct.

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Correct Answer: A, D