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: Substitution

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

\[ \frac{dt}{dy} + e^{2y} t^2 + t = 0 \] \[ \frac{dt}{dy} + t = -t^2 e^{2y} \]

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

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

Step 3: Substitution

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

\[ -\frac{du}{dy} + u = -e^{2y} \] \[ \frac{du}{dy} - u = e^{2y} \]

Step 4: Solving the Linear Differential Equation

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

\[ u e^{-y} = \int e^{-y} \times e^{2y} dy \]

\[ u e^{-y} = e^{y} + c \]

Step 5: Back Substitution

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

Step 6: Applying Conditions

For \( x = \frac{\pi}{4}, y = 0 \Rightarrow c = 0 \)

For \( x = \frac{\pi}{6}, y = \alpha \): \[ \sqrt{3} e^{-\alpha} = e^{\alpha} \] \[ e^{2\alpha} = \sqrt{3} \] \[ e^{8\alpha} = 9 \]

Final Answer:

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

Answer 2

Given the differential equation:

\[ \sec^2 x \, dx + \left( e^{2y} \tan^2 x + \tan x \right) dy = 0 \]

Rearranging terms:

\[ \sec^2 x \, dx = - \left( e^{2y} \tan^2 x + \tan x \right) dy \]

Let:

\[ t = \tan x \implies dt = \sec^2 x \, dx \]

Substituting:

\[ dt = - \left( e^{2y} t^2 + t \right) dy \]

Rearranging:

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

Let:

\[ u = \frac{1}{t} \implies \frac{dt}{dy} = -\frac{1}{u^2} \frac{du}{dy} \]

Substituting:

\[ -\frac{1}{u^2} \frac{du}{dy} + \frac{1}{u} = -e^{2y} \]

Multiplying through by \( -u^2 \):

\[ \frac{du}{dy} - u = e^{2y} u^2 \]

The equation is nonlinear, but we can solve it using separation of variables. Rearranging:

\[ \frac{du}{dy} = u + e^{2y} u^2 \]

Separating variables:

\[ \int \frac{du}{u + e^{2y} u^2} = \int dy \]

Given that \( y \left( \frac{\pi}{4} \right) = 0 \), we substitute the value and integrate to find the general solution. When we evaluate \( y \left( \frac{\pi}{6} \right) = \alpha \), we find:

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