Question

If $y=y(x)$ is the solution curve of the differential equation $\frac{d y}{d x}+y \tan x=x \sec x, 0 \leq x \leq \frac{\pi}{3}, y(0)=1$ then $y\left(\frac{\pi}{6}\right)$ is equal to

• A. $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right)$
• B. $\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _e\left(\frac{2}{e \sqrt{3}}\right)$
• C. $\frac{\pi}{12}+\frac{\sqrt{3}}{2} \log _e\left(\frac{2 \sqrt{3}}{e}\right)$
• D. $\frac{\pi}{12}-\frac{\sqrt{3}}{2} \log _e\left(\frac{2 \sqrt{3}}{e}\right)$

Answer 1

The Correct Option is A

1. We are given the differential equation: \(\frac{dy}{dx} + y \tan x = x \sec x\).
2. This is a linear first-order differential equation of the form: \(\frac{dy}{dx} + P(x) y = Q(x)\), where \(P(x) = \tan x\) and \(Q(x) = x \sec x\).
3. We solve this differential equation using an integrating factor. The integrating factor, \(\mu(x)\), is given by: \(\mu(x) = e^{\int P(x) \, dx} = e^{\int \tan x \, dx}\).
4. Calculate the integral \(\int \tan x \, dx\): \(\int \tan x \, dx = \int \frac{\sin x}{\cos x} \, dx = -\log \cos x\). Thus, the integrating factor is: \(\mu(x) = e^{-\log \cos x} = \frac{1}{\cos x} = \sec x\).
5. Multiply the entire differential equation by our integrating factor \(\sec x\): \(\sec x \frac{dy}{dx} + y \sec x \tan x = x \sec^2 x\).
6. The left side of the equation becomes an exact derivative: \(\frac{d}{dx} (y \sec x) = x \sec^2 x\).
7. Integrate both sides with respect to \(x\): \(y \sec x = \int x \sec^2 x \, dx\).
8. We solve the integral \(\int x \sec^2 x \, dx\) using integration by parts: Set \(u = x\) and \(dv = \sec^2 x \, dx\), such that \(du = dx\) and \(v = \tan x\). Hence, \(\int x \sec^2 x \, dx = x \tan x - \int \tan x \, dx\). We solved \(\int \tan x \, dx = -\log \cos x\) earlier. Therefore, \(\int x \sec^2 x \, dx = x \tan x + \log \cos x\).
9. Substituting back, we have: \(y \sec x = x \tan x + \log \cos x + C\). Solving for \(y\), we get: \(y(x) = \sec x (x \tan x + \log \cos x + C)\).
10. Using the initial condition \(y(0) = 1\): Calculate \(y(0) = \sec(0) (0 \cdot \tan(0) + \log \cos(0) + C) = 1\). \(1 = C\). Thus, \(C = 1\).
11. The solution becomes: \(y(x) = \sec x (x \tan x + \log \cos x + 1)\).
12. Substitute \(x = \frac{\pi}{6}\) to find \(y\left(\frac{\pi}{6}\right)\): \(\sec \frac{\pi}{6} = \frac{2}{\sqrt{3}}, \tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\). Substitute: \(y\left(\frac{\pi}{6}\right) = \frac{2}{\sqrt{3}} \left(\frac{\pi}{6} \cdot \frac{1}{\sqrt{3}} + \log \cos \frac{\pi}{6} + 1\right)\).
13. Since \(\cos \frac{\pi}{6} = \frac{\sqrt{3}}{2}\), the logarithm becomes: \(\log \cos \frac{\pi}{6} = \log\left(\frac{\sqrt{3}}{2}\right) = -\log 2 + \frac{1}{2}\log 3\).
14. After simplifying the expression: \(\frac{2}{\sqrt{3}} \left(\frac{\pi}{6 \sqrt{3}} + (-\log 2 + \frac{1}{2}\log 3) + 1\right) = \frac{\pi}{12} - \frac{\sqrt{3}}{2}\log_e\left(\frac{2}{e\sqrt{3}}\right)\).
15. Hence, the correct answer is: \(\frac{\pi}{12} - \frac{\sqrt{3}}{2} \log_e\left(\frac{2}{e \sqrt{3}}\right)\).