For $$k \in \mathbb{R}$$, let the solutions of the equation $$\cos\left(\sin^{-1}\left(x \cot\left(\tan^{-1}\left(\cos(\sin^{-1} x)\right)\right)\right)\right) = k$$, $$0 < |x| < \frac{1}{\sqrt{2}}$$ be $$\alpha$$ and $$\beta$$, where the inverse trigonometric functions take only principal values. If the solutions of the equation $$x^2 - bx - 5 = 0$$ are $$\frac{1}{\alpha^2} + \frac{1}{\beta^2}$$ and $$\frac{\alpha}{\beta}$$, then $$\frac{b}{k^2}$$ is equal to

The differential equation is $$x\dfrac{dy}{dx} - y = \sqrt{y^2 + 16x^2}$$, with $$y(1) = 3$$.

This is a homogeneous equation. Let $$y = vx$$, so $$\dfrac{dy}{dx} = v + x\dfrac{dv}{dx}$$.

Substituting: $$x\left(v + x\dfrac{dv}{dx}\right) - vx = \sqrt{v^2x^2 + 16x^2}$$

$$x^2\dfrac{dv}{dx} = x\sqrt{v^2 + 16}$$

$$x\dfrac{dv}{dx} = \sqrt{v^2 + 16}$$

Separating variables: $$\dfrac{dv}{\sqrt{v^2 + 16}} = \dfrac{dx}{x}$$

Integrating both sides using the formula $$\displaystyle\int \dfrac{dv}{\sqrt{v^2 + a^2}} = \ln\left(v + \sqrt{v^2 + a^2}\right) + C$$:

$$\ln\left(v + \sqrt{v^2 + 16}\right) = \ln|x| + C$$

$$v + \sqrt{v^2 + 16} = kx$$ where $$k = e^C$$.

Substituting back $$v = y/x$$: $$\dfrac{y}{x} + \sqrt{\dfrac{y^2}{x^2} + 16} = kx$$

Multiplying by $$x$$: $$y + \sqrt{y^2 + 16x^2} = kx^2$$ $$-(1)$$

Using the initial condition $$y(1) = 3$$: $$3 + \sqrt{9 + 16} = k \cdot 1$$, so $$k = 3 + 5 = 8$$.

The solution curve is $$y + \sqrt{y^2 + 16x^2} = 8x^2$$.

At $$x = 2$$: $$y + \sqrt{y^2 + 64} = 32$$.

$$\sqrt{y^2 + 64} = 32 - y$$

Squaring: $$y^2 + 64 = 1024 - 64y + y^2$$

$$64y = 960$$, so $$y = 15$$.

Check: $$32 - y = 17 > 0$$, so the square root step is valid.

The answer is $$y(2) = 15$$.