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

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:

$e^{8\alpha} = 9$