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(A) Given the differential equation: $(	an x)^{1/2} \frac{dy}{dx} = \sec^3 x - (	an x)^{3/2} y$.
Divide by $(	an x)^{1/2}$: $\frac{dy}{dx} + 	an

Vedclass · Accepted Answer

$48$

Vedclass · Answer

(A) Given the differential equation: $(	an x)^{1/2} \frac{dy}{dx} = \sec^3 x - (	an x)^{3/2} y$.
Divide by $(	an x)^{1/2}$: $\frac{dy}{dx} + 	an x \cdot y = \frac{\sec^3 x}{(	an x)^{1/2}}$.
This is a linear differential equation of the form $\frac{dy}{dx} + P(x)y = Q(x)$, where $P(x) = 	an x$ and $Q(x) = \frac{\sec^3 x}{\sqrt{	an x}}$.
The integrating factor $IF = e^{\int 	an x dx} = e^{\ln|\sec x|} = \sec x$.
The general solution is $y \cdot \sec x = \int Q(x) \cdot IF dx = \int \frac{\sec^3 x}{\sqrt{	an x}} \cdot \sec x dx = \int \frac{\sec^4 x}{\sqrt{	an x}} dx$.
Using $\sec^2 x = 1 + 	an^2 x$, we get $\int \frac{(1 + 	an^2 x) \sec^2 x}{\sqrt{	an x}} dx$.
Let $u = 	an x$, then $du = \sec^2 x dx$. The integral becomes $\int (u^{-1/2} + u^{3/2}) du = 2u^{1/2} + \frac{2}{5}u^{5/2} + C$.
So, $y \sec x = 2\sqrt{	an x} + \frac{2}{5}(	an x)^{5/2} + C$.
Given $y(\frac{\pi}{4}) = \frac{6\sqrt{2}}{5}$, substitute $x = \frac{\pi}{4}$: $y \cdot \sqrt{2} = 2(1) + \frac{2}{5}(1) + C \implies \frac{6\sqrt{2}}{5} \cdot \sqrt{2} = \frac{12}{5} = \frac{12}{5} + C \implies C = 0$.
Thus, $y \sec x = 2\sqrt{	an x} + \frac{2}{5}(	an x)^{5/2}$.
For $x = \frac{\pi}{3}$, $	an(\frac{\pi}{3}) = \sqrt{3}$ and $\sec(\frac{\pi}{3}) = 2$.
$y \cdot 2 = 2\sqrt{\sqrt{3}} + \frac{2}{5}(\sqrt{3})^{5/2} = 2(3)^{1/4} + \frac{2}{5}(3)^{5/4} = 2(3)^{1/4} + \frac{2}{5} \cdot 3 \cdot 3^{1/4} = 2(3)^{1/4} + \frac{6}{5}(3)^{1/4} = \frac{16}{5}(3)^{1/4}$.
$y(\frac{\pi}{3}) = \frac{8}{5}(3)^{1/4} = \frac{4}{5} \cdot 2(3)^{1/4}$.
So, $\alpha = 2(3)^{1/4}$.
$\alpha^4 = (2(3)^{1/4})^4 = 16 \cdot 3 = 48$.

Let $y=y(x)$ be the solution of the differential equation $(\tan x)^{1/2} dy = (\sec^3 x - (\tan x)^{3/2} y) dx$, where $0 < x < \frac{\pi}{2}$ and $y(\frac{\pi}{4}) = \frac{6\sqrt{2}}{5}$. If $y(\frac{\pi}{3}) = \frac{4}{5}\alpha$, then $\alpha^4$ equals . . . . . . .

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