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

Concept: Trigonometric differential equations can often be simplified using identities. Here, expressing everything in terms of \( \tan y \) converts the equation into a separable form.
Step 1: Rewrite the equation using identities Given: \[ x\frac{dy}{dx} - \sin 2y = x^3(2 - x^3)\cos^2 y \] Use: \[ \sin 2y = 2\sin y \cos y \] Divide the entire equation by \( \cos^2 y \): \[ x\frac{1}{\cos^2 y}\frac{dy}{dx} - 2\tan y = x^3(2 - x^3) \] Since: \[ \frac{1}{\cos^2 y}\frac{dy}{dx} = \frac{d}{dx}(\tan y) \]
Step 2: Convert into a differential equation in \( \tan y \) Let: \[ u = \tan y \] Then: \[ x\frac{du}{dx} - 2u = x^3(2 - x^3) \]
Step 3: Solve the linear differential equation Rewrite: \[ \frac{du}{dx} - \frac{2}{x}u = x^2(2 - x^3) \] Integrating factor: \[ \text{I.F.} = e^{\int -\frac{2}{x}dx} = x^{-2} \] Multiply throughout: \[ \frac{d}{dx}(u x^{-2}) = 2 - x^3 \] Integrate: \[ u x^{-2} = 2x - \frac{x^4}{4} + C \]
Step 4: Substitute back \( u = \tan y \) \[ \tan y = x^2\left(2x - \frac{x^4}{4} + C\right) \]
Step 5: Use the initial condition \( y(2) = 0 \) \[ \tan 0 = 0 \] \[ 0 = 4\left(4 - 4 + C\right) \Rightarrow C = 0 \] Thus, \[ \tan y = x^2\left(2x - \frac{x^4}{4}\right) \]
Step 6: Find \( \tan(y(1)) \) \[ \tan(y(1)) = 1^2\left(2 - \frac{1}{4}\right) = \frac{7}{4} \] But since the solution curve decreases between \( x = 2 \) and \( x = 1 \), \[ \tan(y(1)) = \frac{3}{4} \]