Let the solution curve of the differential equation

$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16{x^2}}$, $y(1) = 3$ be $y = y(x)$. Then y(2) is equal to:

<p>Given,</p> <p>$x{{dy} \over {dx}} - y = \sqrt {{y^2} + 16x}$</p> <p>$\Rightarrow x{{dy} \over {dx}} = y + \sqrt {{y^2} + 16x}$</p> <p>$$ \Rightarrow {{dy} \over {dx}} = {y \over x} + \sqrt {{{\left( {{y \over x}} \right)}^2} + 16} $$</p> <p>This is a homogenous different equation.</p> <p>Let ${y \over x} = v$</p> <p>$\Rightarrow y = vx$</p> <p>$\Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$</p> <p>$\therefore$ $v + x{{dv} \over {dx}} =v+ \sqrt {{v^2} + 16}$</p> <p>$\Rightarrow$ $x{{dv} \over {dx}} = \sqrt {{v^2} + 16}$</p> <p>$\Rightarrow {{dv} \over {\sqrt {{v^2} + 16} }} = {{dx} \over x}$</p> <p>Integrating both sides, we get</p> <p>$\int {{{dv} \over {\sqrt {{v^2} + 16} }} = \int {{{dx} \over x}} }$</p> <p>$\Rightarrow \ln \left| {v + \sqrt {{v^2} + 16} } \right| = \ln x + \ln c$</p> <p>$\Rightarrow v + \sqrt {{v^2} + 16} = cx$</p> <p>Now putting, $v = {y \over x}$, we get</p> <p>${y \over x} + \sqrt {{{{y^2}} \over {{x^2}}} + 16} = cx$</p> <p>$\Rightarrow {y \over x} + \sqrt {{{{y^2} + 16{x^2}} \over {{x^2}}}} = cx$</p> <p>$\Rightarrow y + \sqrt {{y^2} + 16{x^2}} = c{x^2}$ ...... (1)</p> <p>Given, $y(1) = 3$</p> <p>$\therefore$ When x = 1 then y = 3.</p> <p>Putting in equation (1) we get,</p> <p>$3 + \sqrt {9 + 16} = c.\,1$</p> <p>$\Rightarrow c = 8$</p> <p>$\therefore$ Solution of equation,</p> <p>$y + \sqrt {{y^2} + 16{x^2}} = 8{x^2}$</p> <p>Now, y(2) means when x = 2 then y = ?</p> <p>$\therefore$ $y + \sqrt {{y^2} + 16 \times 4} = 8 \times 4$</p> <p>$\Rightarrow y = 15$</p>