Solution of First Order Differential Equation Involving Trigonometric Terms

To solve the given differential equation: $(\tan x)^{1/2} dy = \left( \sec^3 x - (\tan x)^{3/2} y \right) dx, \quad 0 < x < \frac{\pi}{2}$

We can rewrite it in the standard first-order linear differential equation form $\frac{dy}{dx} + P(x)y = Q(x)$:

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

Step 1: Find the Integrating Factor (I.F.)

$\text{I.F.} = e^{\int P(x) dx} = e^{\int \tan x \, dx} = e^{\ln(\sec x)} = \sec x \quad \text{for } x \in \left(0, \frac{\pi}{2}\right)$

Step 2: Solve the Differential Equation

Multiplying the equation by the Integrating Factor: $y \cdot \sec x = \int \left( \frac{\sec^3 x}{\sqrt{\tan x}} \right) \cdot \sec x \, dx + C$ $y \sec x = \int \frac{\sec^4 x}{\sqrt{\tan x}} \, dx + C$

To evaluate the integral $I = \int \frac{\sec^4 x}{\sqrt{\tan x}} \, dx$, let $t = \tan x$, so $dt = \sec^2 x \, dx$ and $\sec^2 x = 1 + t^2$:

$I = \int \frac{1 + t^2}{\sqrt{t}} \, dt = \int \left( t^{-1/2} + t^{3/2} \right) dt$ $I = 2\sqrt{t} + \frac{2}{5}t^{5/2} = 2\sqrt{\tan x} + \frac{2}{5}(\tan x)^{5/2}$

Thus, the general solution is: $y \sec x = 2\sqrt{\tan x} + \frac{2}{5}(\tan x)^{5/2} + C$

Step 3: Determine the Constant of Integration $C$

Using the given condition $y\left(\frac{\pi}{4}\right) = \frac{6\sqrt{2}}{5}$:

$\left( \frac{6\sqrt{2}}{5} \right) \sec\left(\frac{\pi}{4}\right) = 2\sqrt{\tan\left(\frac{\pi}{4}\right)} + \frac{2}{5}\left(\tan\left(\frac{\pi}{4}\right)\right)^{5/2} + C$ $\left( \frac{6\sqrt{2}}{5} \right) (\sqrt{2}) = 2(1) + \frac{2}{5}(1) + C$ $\frac{12}{5} = \frac{12}{5} + C \implies C = 0$

So, the solution to the differential equation is: $y \sec x = 2\sqrt{\tan x} + \frac{2}{5}(\tan x)^{5/2}$

Step 4: Evaluate $y\left(\frac{\pi}{3}\right)$ and Find $\alpha^4$

At $x = \frac{\pi}{3}$: $\sec\left(\frac{\pi}{3}\right) = 2 \quad \text{and} \quad \tan\left(\frac{\pi}{3}\right) = \sqrt{3} = 3^{1/2}$

Substitute these values into the particular solution: $y\left(\frac{\pi}{3}\right) \cdot 2 = 2\left(3^{1/2}\right)^{1/2} + \frac{2}{5}\left(3^{1/2}\right)^{5/2}$ $2 y\left(\frac{\pi}{3}\right) = 2 \cdot 3^{1/4} + \frac{2}{5} \cdot 3^{5/4}$ $y\left(\frac{\pi}{3}\right) = 3^{1/4} + \frac{1}{5} \cdot 3^{5/4} = 3^{1/4} \left( 1 + \frac{3}{5} \right) = \frac{8}{5} \cdot 3^{1/4}$

Given that $y\left(\frac{\pi}{3}\right) = \frac{4}{5}\alpha$: $\frac{4}{5}\alpha = \frac{8}{5} \cdot 3^{1/4}$ $\alpha = 2 \cdot 3^{1/4}$

Now, calculating $\alpha^4$: $\alpha^4 = \left(2 \cdot 3^{1/4}\right)^4 = 2^4 \cdot 3 = 16 \times 3 = 48$