Let $$y = y(x)$$ be the solution of $$(\tan x)^{1/2}\,dy = (\sec^3 x - (\tan x)^{3/2}\,y)\,dx$$, $$0 < x < \frac{\pi}{2}$$. If $$y\left(\frac{\pi}{4}\right) = \frac{6\sqrt{2}}{5}$$, and $$y\left(\frac{\pi}{3}\right) = \frac{4}{5}\alpha$$, then $$\alpha^4$$ is equal to :

The given differential equation is

$$(\tan x)^{1/2}\,dy=(\sec^3 x-(\tan x)^{3/2}y)\,dx,\qquad 0\lt x\lt\frac{\pi}{2}.$$

Divide by $$(\tan x)^{1/2}$$ to write it in the standard linear form $$\dfrac{dy}{dx}+P(x)\,y=Q(x):$$

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

Here $$P(x)=\tan x$$ and $$Q(x)=\dfrac{\sec^3 x}{\sqrt{\tan x}}.$$ For a linear ODE the integrating factor (I.F.) is $$e^{\int P(x)\,dx}.$$

$$\int \tan x\,dx=-\ln\cos x \quad\Rightarrow\quad\text{I.F.}=e^{-\ln\cos x}=\sec x.$$

Multiply equation $$-(1)$$ by $$\sec x:$$

$$\sec x\,\frac{dy}{dx}+\sec x\tan x\,y=\frac{\sec^4 x}{\sqrt{\tan x}}.$$

The left‐hand side is the derivative of $$(\sec x\,y):$$

$$\frac{d}{dx}\bigl(y\sec x\bigr)=\frac{\sec^4 x}{\sqrt{\tan x}}.\qquad -(2)$$

Integrate both sides.

To evaluate the integral on the right, put $$t=\sqrt{\tan x}\;\;(t^2=\tan x).$$
Then $$\sec^2 x\,dx=2t\,dt\;\Rightarrow\;dx=\dfrac{2t\,dt}{\sec^2 x}.$$

$$\int\frac{\sec^4 x}{\sqrt{\tan x}}\,dx =\int\frac{\sec^4 x}{t}\,\frac{2t\,dt}{\sec^2 x} =\int 2\sec^2 x\,dt =\int 2(1+t^4)\,dt =2t+\frac{2}{5}t^5 +C.$$

Resubstitute $$t=\sqrt{\tan x}:$$

$$y\sec x=2\sqrt{\tan x}+\frac{2}{5}(\tan x)^{5/2}+C.\qquad -(3)$$

Therefore

$$y(x)=\cos x\Bigl[2\sqrt{\tan x}+\frac{2}{5}(\tan x)^{5/2}+C\Bigr].\qquad -(4)$$

Using the initial condition $$y\!\left(\frac{\pi}{4}\right)=\frac{6\sqrt2}{5}:$$
At $$x=\frac{\pi}{4},\; \cos x=\frac{\sqrt2}{2},\; \tan x=1.$$ Substitute into $$-(4)$$:

$$\frac{6\sqrt2}{5}=\frac{\sqrt2}{2}\Bigl[2(1)+\frac{2}{5}(1)+C\Bigr] =\frac{\sqrt2}{2}\Bigl[\frac{12}{5}+C\Bigr].$$

Divide by $$\sqrt2$$ and multiply by $$2:$$ $$\frac{12}{5}+C=\frac{12}{5}\;\;\Rightarrow\;\;C=0.$$

Hence the particular solution is

$$y(x)=\cos x\Bigl[2\sqrt{\tan x}+\frac{2}{5}(\tan x)^{5/2}\Bigr].\qquad -(5)$$

Evaluate at $$x=\frac{\pi}{3}:$$
$$\cos\frac{\pi}{3}=\frac12,\quad \tan\frac{\pi}{3}=\sqrt3.$$

Inside the bracket:

$$2\sqrt{\tan x}=2\sqrt{\sqrt3}=2\cdot3^{1/4},$$ $$\frac{2}{5}(\tan x)^{5/2}=\frac{2}{5}(\sqrt3)^{5/2} =\frac{2}{5}\cdot3^{5/4}.$$ Pulling out the common factor $$2\cdot3^{1/4}:$$

$$2\cdot3^{1/4}\Bigl[1+\frac35\Bigr]=2\cdot3^{1/4}\cdot\frac85=\frac{16}{5}3^{1/4}.$$

Therefore

$$y\!\left(\frac{\pi}{3}\right)=\frac12\cdot\frac{16}{5}3^{1/4}=\frac{8}{5}3^{1/4}.$$

The question states $$y\!\left(\frac{\pi}{3}\right)=\frac{4}{5}\alpha,$$ hence

$$\frac{4}{5}\alpha=\frac{8}{5}3^{1/4}\;\;\Rightarrow\;\;\alpha=2\cdot3^{1/4}.$$

Finally,

$$\alpha^4=(2)^4\!\left(3^{1/4}\right)^{4}=16\cdot3=48.$$

So the required value is 48.