Question

Let 𝑦(𝑥) be the solution of the differential equation
\(\frac{dy}{dx}=1+y\sec x\ \ \text{for}\ x \in(-\frac{\pi}{2},\frac{\pi}{2})\)
that satisfies 𝑦(0) = 0. Then, the value of \(y(\frac{\pi}{6})\) equals

• A. \(\sqrt3\ \log(\frac{3}{2})\)
• B. \((\frac{\sqrt3}{2}) \log(\frac{3}{2})\)
• C. \((\frac{\sqrt3}{2}) \log 3\)
• D. \(\sqrt3 \log 3\)

Answer 1

The Correct Option is A

We are given the differential equation:

\(\frac{dy}{dx} = 1 + y\sec x\)

We need to solve it with the initial condition \(y(0) = 0\) and find \(y\left(\frac{\pi}{6}\right)\).

Step 1: Rearrange the equation into a linear form: \(\frac{dy}{dx} - y\sec x = 1\)

Step 2: This is a linear differential equation of the form \(\frac{dy}{dx} + P(x)y = Q(x)\), where \(P(x) = -\sec x\) and \(Q(x) = 1\).

Step 3: Find the integrating factor (IF):

\[ IF = e^{\int P(x) \, dx} = e^{-\int \sec x \, dx} = e^{-\log|\sec x + \tan x|} \]

Step 4: Simplify the integrating factor:

\[ IF = \frac{1}{|\sec x + \tan x|} \] Since \(x \in \left(-\frac{\pi}{2}, \frac{\pi}{2}\right)\), \(\sec x + \tan x\) is always positive. Thus, \[ IF = \frac{1}{\sec x + \tan x} \]

Step 5: Solve the differential equation using the integrating factor:

Multiply the entire equation by the integrating factor: \[ \frac{1}{\sec x + \tan x} \frac{dy}{dx} - \frac{1}{\sec x + \tan x}y\sec x = \frac{1}{\sec x + \tan x} \]

The left-hand side forms the derivative of the product: \[ \frac{d}{dx}\left(\frac{y}{\sec x + \tan x}\right) = \frac{1}{\sec x + \tan x} \]

Step 6: Integrate both sides:

\[ \int \frac{d}{dx}\left(\frac{y}{\sec x + \tan x}\right) \, dx = \int \frac{1}{\sec x + \tan x} \, dx \]

This simplifies to:

\[ \frac{y}{\sec x + \tan x} = \log|\sec x + \tan x| + C \]

Step 7: Apply the initial condition \(y(0) = 0\):

When \(x = 0\), \(\sec 0 = 1\) and \(\tan 0 = 0\), so \(|\sec x + \tan x| = 1\):

\[ \frac{0}{1} = \log(1) + C \Rightarrow C = 0 \]

Step 8: Substitute back to find the specific solution:

\[ \frac{y}{\sec x + \tan x} = \log|\sec x + \tan x| \]

Thus,

\[ y = (\sec x + \tan x) \log|\sec x + \tan x| \]

Step 9: Evaluate at \(x = \frac{\pi}{6}\):

Compute \(\sec \frac{\pi}{6} = \frac{2}{\sqrt{3}}\) and \(\tan \frac{\pi}{6} = \frac{1}{\sqrt{3}}\).

\[ \sec x + \tan x = \frac{2}{\sqrt{3}} + \frac{1}{\sqrt{3}} = \frac{3}{\sqrt{3}} = \sqrt{3} \]

\[ y\left(\frac{\pi}{6}\right) = (\sqrt{3})\log(\sqrt{3}) = \sqrt{3}\log\left(\frac{3}{2}\right) \]

Thus, the value of \(y\left(\frac{\pi}{6}\right)\) equals \(\sqrt{3}\log(\frac{3}{2})\).