Question

Let \(x \frac{dy}{dx} - \sin 2y = x^3(2 - x^3) \cos^2 y ; y(2) = 0\), then find \(\tan(y(1))\) :

• A. 5/4
• B. 3/4
• C. 7/4
• D. 9/4

Hint:

For Bernoulli-type equations or non-linear equations, always look for a substitution like \( v = f(y) \) that makes the equation linear in \( v \).

Answer 1

The Correct Option is C

Step 1: Understanding the Concept:
This is a non-linear differential equation. Dividing by \( \cos^2 y \) transforms it into a linear equation.
Step 2: Key Formula or Approach:
Divide by \( x \cos^2 y \):
\( \frac{1}{\cos^2 y} \frac{dy}{dx} - \frac{2 \sin y \cos y}{x \cos^2 y} = x^2(2 - x^3) \)
\( \sec^2 y \frac{dy}{dx} - \frac{2 \tan y}{x} = 2x^2 - x^5 \).
Let \( v = \tan y \), then \( \frac{dv}{dx} = \sec^2 y \frac{dy}{dx} \).
Step 3: Detailed Explanation:
Equation becomes: \( \frac{dv}{dx} - \frac{2}{x}v = 2x^2 - x^5 \).
Integrating factor \( IF = e^{\int -2/x dx} = x^{-2} \).
Solution: \( v \cdot x^{-2} = \int (2x^2 - x^5)x^{-2} dx = \int (2 - x^3) dx \)
\( \frac{\tan y}{x^2} = 2x - \frac{x^4}{4} + C \).
Using \( y(2) = 0 \): \( \frac{0}{4} = 4 - 4 + C \implies C = 0 \).
So, \( \tan y = x^2(2x - \frac{x^4}{4}) = 2x^3 - \frac{x^6}{4} \).
At \( x=1 \), \( \tan(y(1)) = 2 - 1/4 = 7/4 \).
Step 4: Final Answer:
\(\tan(y(1)) = 7/4\).