Let y=y(x) be the solution of the differential equation [( x)^1/2,dy = (^3 x - ( x)^3/2y)d…

Question

Let $y=y(x)$ be the solution of the differential equation
$$(\tan x)^{1/2}\,dy = \left(\sec^3 x - (\tan x)^{3/2}y\right)dx,\quad 0<x<\frac{\pi}{2},\qquad y\!\left(\frac{\pi}{4}\right)=\frac{6\sqrt2}{5}.$$
If $y\!\left(\frac{\pi}{3}\right)=\dfrac{4}{5}\alpha$, then $\alpha^4$ equals \_\_\_\_\_\_.

Answer: 48

Accepted Answer

Rewrite as
$$\frac{dy}{dx} + y\tan x = \frac{\sec^3 x}{\sqrt{\tan x}}.$$

Integrating factor $= e^{\int \tan x\,dx} = \sec x$. Then
$$y\sec x = \int \frac{\sec^4 x}{\sqrt{\tan x}}\,dx = \int \frac{(1+\tan^2 x)}{\sqrt{\tan x}}\sec^2 x\,dx.$$

Put $\tan x = t$:
$$y\sec x = 2\sqrt{\tan x} + \frac{2}{5}(\tan x)^{5/2} + c.$$

Using $y(\pi/4)=\dfrac{6\sqrt2}{5}$ gives $c=0$. Then
$$y\!\left(\frac{\pi}{3}\right) = \frac{8}{5}\cdot 3^{1/4} \implies \alpha = 2\times 3^{1/4} \implies \alpha^4 = 48.$$

Solutions · 1