Question

Let \( y = y(x) \) be the solution of the differential equation \(\sec^2 x \, dx + \left( e^{2y} \tan^2 x + \tan x \right) dy = 0,\)\( 0 < x < \frac{\pi}{2} \), \( y \left( \frac{\pi}{4} \right) = 0 \). If \( y \left( \frac{\pi}{6} \right) = \alpha \), then \( e^{8\alpha} \) is equal to\(\_\_\_\_\_\).

Answer 1

Correct Answer: 9

Given:

\[ \sec^2 x \frac{dx}{dy} + e^{2y} \tan^2 x + \tan x = 0 \]

Step 1: Substitute \( \tan x \) with \( t \).

Let \( t = \tan x \). Then, \( \frac{dt}{dy} = \sec^2 x \frac{dx}{dy} \).

The equation becomes: \[ \frac{dt}{dy} + e^{2y} t^2 + t = 0 \] Rearranging gives: \[ \frac{dt}{dy} + t = -t^2 e^{2y} \]

Step 2: Divide by \( t^2 \).

\[ \frac{1}{t^2} \frac{dt}{dy} + \frac{1}{t} = -e^{2y} \]

Step 3: Substitute \( \frac{1}{t} \) with \( u \).

Let \( u = \frac{1}{t} = t^{-1} \). Then, \( \frac{du}{dy} = -t^{-2} \frac{dt}{dy} = -\frac{1}{t^2} \frac{dt}{dy} \).

The equation transforms to: \[ -\frac{du}{dy} + u = -e^{2y} \] Multiplying by -1 yields: \[ \frac{du}{dy} - u = e^{2y} \]

Step 4: Solve the linear differential equation.

The integrating factor (I.F.) is: \[ e^{\int -1 \, dy} = e^{-y} \]

Multiply the equation by the I.F.: \[ u e^{-y} = \int e^{-y} \times e^{2y} dy \]

Integrating the right side: \[ u e^{-y} = e^{y} + c \]

Step 5: Substitute back for \( u \).

Replace \( u \) with \( \frac{1}{t} \) and \( t \) with \( \tan x \): \[ \frac{1}{\tan x} \times e^{-y} = e^{y} + c \]

Step 6: Apply the given conditions.

Given \( x = \frac{\pi}{4} \) and \( y = 0 \): \[ \frac{1}{\tan(\pi/4)} e^{-0} = e^{0} + c \Rightarrow 1 \times 1 = 1 + c \Rightarrow c = 0 \]

Given \( x = \frac{\pi}{6} \) and \( y = \alpha \). With \( c = 0 \), the equation becomes: \[ \frac{1}{\tan(\pi/6)} e^{-\alpha} = e^{\alpha} \] \[ \frac{1}{1/\sqrt{3}} e^{-\alpha} = e^{\alpha} \Rightarrow \sqrt{3} e^{-\alpha} = e^{\alpha} \] Rearranging the terms: \[ e^{2\alpha} = \sqrt{3} \] Raising both sides to the power of 4: \[ (e^{2\alpha})^4 = (\sqrt{3})^4 \] \[ e^{8\alpha} = 9 \]

Final Answer:

\[ e^{8\alpha} = 9 \]