Let the solution curve of the differential equation x(dy)/(dx) - y...

Given,

$x\dfrac{dy}{dx} - y = \sqrt{{y}^{2} + 16{x}^{2}}$

$⇒ \dfrac{dy}{dx} = \dfrac{\sqrt{{y}^{2} + 16{x}^{2}} + y}{x}$

Let $y = xt ⇒ \dfrac{dy}{dx} = x\dfrac{dt}{dx} + t$

$⇒ x\dfrac{dt}{dx} + t = t + \sqrt{{t}^{2} + 16}$

$⇒ \dfrac{dt}{\sqrt{{t}^{2} + 16}} = \dfrac{dx}{x}$

Now integrating both side we get,

$⇒ ∫ \dfrac{dt}{\sqrt{{t}^{2} + 16}} = ∫ \dfrac{dx}{x}$

$⇒ \ln \left|t + \sqrt{{t}^{2} + 16}\right| = \ln x + lnc$

$⇒ \dfrac{y}{x} + \dfrac{\sqrt{{y}^{2} + 16{x}^{2}}}{x} = cx$

$∵ y\left(1\right) = 3$ so $⇒ \dfrac{3}{1} + \dfrac{\sqrt{{3}^{2} + 16 \times {1}^{2}}}{1} = c \times 1 ⇒ c = 8$

So solution becomes,

$\dfrac{y}{x} + \dfrac{\sqrt{{y}^{2} + 16{x}^{2}}}{x} = 8x$

Now $y\left(2\right) = \dfrac{y}{2} + \dfrac{\sqrt{{y}^{2} + 16 \times {2}^{2}}}{2} = 8 \times 2$

$⇒ y + \sqrt{{y}^{2} + 64} = 32$

$⇒ y = 15$