Question

Let g be a differentiable function such that $ \int_0^x g(t) dt = x - \int_0^x tg(t) dt $, $ x \ge 0 $ and let $ y = y(x) $ satisfy the differential equation $ \frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x) $, $ x \in \left[ 0, \frac{\pi}{2} \right) $. If $ y(0) = 0 $, then $ y\left( \frac{\pi}{3} \right) $ is equal to

• A. \( \frac{2\pi}{3\sqrt{3}} \)
• B. \( \frac{4\pi}{3} \)
• C. \( \frac{2\pi}{3} \)
• D. \( \frac{4\pi}{3\sqrt{3}} \)

Hint:

Recognize and solve the linear differential equation using the integrating factor method.

Answer 1

The Correct Option is B

Given:
\[ \int_0^x g(t)\,dt = x - \int_0^x t g(t)\,dt \] Differentiate both sides with respect to \( x \): \[ g(x) = 1 - xg(x) \Rightarrow g(x)(1 + x) = 1 \Rightarrow g(x) = \frac{1}{1 + x} \] Now consider the differential equation: \[ \frac{dy}{dx} - y \tan x = 2(x+1) \sec x \cdot g(x) = 2(x+1) \sec x \cdot \frac{1}{1+x} = 2 \sec x \] So the equation becomes: \[ \frac{dy}{dx} - y \tan x = 2 \sec x \] This is a linear differential equation. The integrating factor is: \[ \text{IF} = e^{\int -\tan x \, dx} = e^{\ln|\cos x|} = \cos x \] Multiplying both sides by the integrating factor: \[ \cos x \cdot \frac{dy}{dx} - y \cos x \tan x = 2 \Rightarrow \frac{d}{dx}(y \cos x) = 2 \] Integrating both sides: \[ y \cos x = \int 2\,dx = 2x + C \] Apply the initial condition \( y(0) = 0 \): \[ 0 \cdot \cos 0 = 2 \cdot 0 + C \Rightarrow C = 0 \] Therefore, the solution is: \[ y \cos x = 2x \Rightarrow y = \frac{2x}{\cos x} = 2x \sec x \] Now, compute \( y\left( \frac{\pi}{3} \right) \): \[ y\left( \frac{\pi}{3} \right) = 2 \cdot \frac{\pi}{3} \cdot \sec\left( \frac{\pi}{3} \right) = 2 \cdot \frac{\pi}{3} \cdot 2 = \frac{4\pi}{3} \]

Answer 2

To solve the given problem, we need to find \( y\left( \frac{\pi}{3} \right) \) for the differential equation:

\(\frac{dy}{dx} - y \tan x = 2(x+1) \sec x g(x),\) where \( y(0) = 0 \).

We are given another condition:

\(\int_0^x g(t) \, dt = x - \int_0^x t g(t) \, dt.\)

1. First, differentiate both sides of the given integral condition with respect to \( x \) to find \( g(x) \).

Using Leibniz's rule, we get:

\(g(x) = 1 - xg(x).\)

Solving for \( g(x) \), we find:

\(g(x) = \frac{1}{x + 1}.\)

1. Substitute \( g(x) \) in the differential equation:

\(\frac{dy}{dx} - y \tan x = 2(x+1) \sec x \cdot \frac{1}{x+1} = 2 \sec x.\)

1. This transforms the differential equation into:

\(\frac{dy}{dx} - y \tan x = 2 \sec x.\)

1. The solution to this linear first-order differential equation uses an integrating factor. The integrating factor \( \mu(x) \) is:

\(\mu(x) = e^{\int -\tan x \, dx} = e^{-\ln |\sec x|} = \cos x.\)

1. Multiply the entire differential equation by the integrating factor:

\(\cos x \frac{dy}{dx} - y \cos x \tan x = 2.\)

1. Recognize the left part as a derivative of a product:

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

1. Integrate both sides with respect to \( x \):

\(y \cos x = 2x + C.\)

1. Apply the initial condition \( y(0) = 0 \) to solve for \( C \):

\(0 \cdot 1 = 0 + C \rightarrow C = 0.\)

The solution is now:

\(y \cos x = 2x.\)

1. Find \( y\left( \frac{\pi}{3} \right) \):

\(y \left( \frac{\pi}{3} \right) \cdot \cos \left( \frac{\pi}{3} \right) = 2 \cdot \frac{\pi}{3}.\)

\(\frac{1}{2} y \left( \frac{\pi}{3} \right) = \frac{2\pi}{3} \rightarrow y \left( \frac{\pi}{3} \right) = \frac{4\pi}{3}.\)

Thus, the value of \( y\left( \frac{\pi}{3} \right) \) is \(\frac{4\pi}{3}\).