Question

Let \( y = y(x) \) be the solution of the differential equation \[ x\frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y,\; x \ne 0. \] If \( y(2) = 0 \), then \( \tan(y(1)) \) is equal to:

• A. \( \dfrac{3}{4} \)
• B. \( -\dfrac{3}{4} \)
• C. \( \dfrac{7}{4} \)
• D. \( -\dfrac{7}{4} \)

Hint:

When trigonometric functions appear with derivatives, try converting the equation into a function of \( \tan y \) or \( \sin y \) to simplify.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
This is a non-linear differential equation. It can be converted into a linear differential equation using a suitable substitution (Bernoulli-style approach).
Step 2: Key Formula or Approach:
Divide the entire equation by \(\cos^2 y\) and use the substitution \(v = \tan y\).
Step 3: Detailed Explanation:
Given: \(x \frac{dy}{dx} - 2 \sin y \cos y = x^3(2 - x^3) \cos^2 y\).
Dividing by \(x \cos^2 y\):
\(\sec^2 y \frac{dy}{dx} - \frac{2 \tan y}{x} = x^2(2 - x^3)\).
Let \(v = \tan y \Rightarrow \frac{dv}{dx} = \sec^2 y \frac{dy}{dx}\).
Substituting this in:
\(\frac{dv}{dx} - \frac{2}{x}v = 2x^2 - x^5\).
This is a linear differential equation.
Integrating factor \(IF = e^{\int -\frac{2}{x} dx} = e^{-2 \ln x} = \frac{1}{x^2}\).
Multiplying by \(IF\):
\(\frac{d}{dx}(v \cdot \frac{1}{x^2}) = \frac{2x^2 - x^5}{x^2} = 2 - x^3\).
Integrating both sides:
\(\frac{v}{x^2} = 2x - \frac{x^4}{4} + C\).
Since \(y(2) = 0\), \(\tan(0) = 0 \Rightarrow v = 0\) when \(x = 2\).
\(0 = 4 - \frac{16}{4} + C \Rightarrow 0 = 4 - 4 + C \Rightarrow C = 0\).
So, \(v = x^2(2x - \frac{x^4}{4}) = 2x^3 - \frac{x^6}{4}\).
For \(x = 1\):
\(\tan(y(1)) = 2(1)^3 - \frac{1^6}{4} = 2 - \frac{1}{4} = \frac{7}{4}\).
Step 4: Final Answer:
The value of \(\tan(y(1))\) is \(\frac{7}{4}\).