The problem requires solving a first-order differential equation with an initial condition and then evaluating the solution at a specific point to find a parameter.

Differential Equation Rearrangement

The given differential equation is:

$(\tan x)^{\frac{1}{2}} \text{ dy} = \left(\sec^{3}x - (\tan x)^{\frac{3}{2}}y\right) \text{ dx}$

Rearranging terms to fit the standard linear form $\frac{dy}{dx} + P(x)y = Q(x)$:

$(\tan x)^{\frac{1}{2}} \text{ dy} + (\tan x)^{\frac{3}{2}} y \text{ dx} = \sec^{3}x \text{ dx}$

Divide by $(\tan x)^{\frac{1}{2}} \text{ dx}$:

$\frac{dy}{dx} + \frac{(\tan x)^{\frac{3}{2}}}{(\tan x)^{\frac{1}{2}}} y = \frac{\sec^{3}x}{(\tan x)^{\frac{1}{2}}}$

$\frac{dy}{dx} + (\tan x) y = \sec^{3}x (\cot x)^{\frac{1}{2}}$

Identify DE Components P(x) and Q(x)

From the standard form, we identify:

• $P(x) = \tan x$
• $Q(x) = \sec^{3}x (\cot x)^{\frac{1}{2}}$

Integrating Factor Calculation

The integrating factor (IF) is calculated as $e^{\int P(x) dx}$:

$\int P(x) dx = \int \tan x dx = \ln|\sec x|$

Since $0 < x < \frac{\pi}{2}$, $\sec x$ is positive, so $\int \tan x dx = \ln(\sec x)$.

$IF = e^{\ln(\sec x)} = \sec x$

Integrating the Equation

Multiply the standard linear form by the integrating factor $\sec x$:

$\sec x \frac{dy}{dx} + (\sec x \tan x) y = \sec x \cdot \sec^{3}x (\cot x)^{\frac{1}{2}}$

The left side is the derivative of $(y \cdot IF)$:

$\frac{d}{dx}(y \sec x) = \sec^4 x (\cot x)^{\frac{1}{2}}$

To integrate the right side, let $u = \tan x$. Then $du = \sec^2 x dx$, and $\cot x = 1/u$.

$\int \sec^4 x (\cot x)^{\frac{1}{2}} dx = \int \sec^2 x \sec^2 x (\cot x)^{\frac{1}{2}} dx$

$= \int (1 + \tan^2 x) \sec^2 x (\cot x)^{\frac{1}{2}} dx = \int (1 + u^2) u^{-1/2} du$

$= \int (u^{-1/2} + u^{3/2}) du = \frac{u^{1/2}}{1/2} + \frac{u^{5/2}}{5/2} + C$

$= 2u^{1/2} + \frac{2}{5}u^{5/2} + C = 2(\tan x)^{1/2} + \frac{2}{5}(\tan x)^{5/2} + C$

Thus, the integrated equation is:

$y \sec x = 2(\tan x)^{1/2} + \frac{2}{5}(\tan x)^{5/2} + C$

Apply Initial Condition

Use the initial condition $y\left(\frac{\pi}{4}\right) = \frac{6\sqrt{2}}{5}$. At $x = \frac{\pi}{4}$, $\tan x = 1$ and $\sec x = \sqrt{2}$.

$y\left(\frac{\pi}{4}\right) \sec\left(\frac{\pi}{4}\right) = \left(\frac{6\sqrt{2}}{5}\right) (\sqrt{2}) = \frac{12}{5}$

Substitute into the integrated equation:

$\frac{12}{5} = 2(1)^{1/2} + \frac{2}{5}(1)^{5/2} + C = 2 + \frac{2}{5} + C = \frac{12}{5} + C$

This implies $C = 0$.

Particular Solution Form

The particular solution is:

$y \sec x = 2(\tan x)^{1/2} + \frac{2}{5}(\tan x)^{5/2}$

Or, solving for $y(x)$:

$y(x) = \cos x \left( 2(\tan x)^{1/2} + \frac{2}{5}(\tan x)^{5/2} \right)$

Evaluate Solution at x=pi/3

We need to find $y\left(\frac{\pi}{3}\right)$. At $x = \frac{\pi}{3}$, $\tan x = \sqrt{3}$ and $\sec x = 2$.

Substitute these values into the equation $y \sec x = 2(\tan x)^{1/2} + \frac{2}{5}(\tan x)^{5/2}$:

$y\left(\frac{\pi}{3}\right) (2) = 2(\sqrt{3})^{1/2} + \frac{2}{5}(\sqrt{3})^{5/2}$

$2 y\left(\frac{\pi}{3}\right) = 2 \cdot 3^{1/4} + \frac{2}{5} (3^{1/2})^{5/2} = 2 \cdot 3^{1/4} + \frac{2}{5} \cdot 3^{5/4}$

Divide by 2:

$y\left(\frac{\pi}{3}\right) = 3^{1/4} + \frac{1}{5} \cdot 3^{5/4} = 3^{1/4} + \frac{3}{5} \cdot 3^{1/4}$

$y\left(\frac{\pi}{3}\right) = 3^{1/4} \left(1 + \frac{3}{5}\right) = 3^{1/4} \left(\frac{8}{5}\right) = \frac{8}{5} 3^{1/4}$

Determine Alpha Value

We are given that $y\left(\frac{\pi}{3}\right) = \frac{4}{5} \alpha$. Equating the two expressions for $y\left(\frac{\pi}{3}\right)$:

$\frac{4}{5} \alpha = \frac{8}{5} 3^{1/4}$

Multiply both sides by $\frac{5}{4}$:

$\alpha = \frac{5}{4} \cdot \frac{8}{5} 3^{1/4} = 2 \cdot 3^{1/4}$

Calculate Final Alpha^4

Finally, calculate $\alpha^4$:

$\alpha^4 = (2 \cdot 3^{1/4})^4 = 2^4 \cdot (3^{1/4})^4 = 16 \cdot 3$

$\alpha^4 = 48$