Let the solution curve of the differential equation [xdy dx - y = y^2 + 16x^2 ], y(1) = 3 …

Question

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:

Accepted Answer

Given,

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

$$ \Rightarrow x{{dy} \over {dx}} = y + \sqrt {{y^2} + 16x} $$

$$ \Rightarrow {{dy} \over {dx}} = {y \over x} + \sqrt {{{\left( {{y \over x}} \right)}^2} + 16} $$

This is a homogenous different equation.

Let $${y \over x} = v$$

$$ \Rightarrow y = vx$$

$$ \Rightarrow {{dy} \over {dx}} = v + x{{dv} \over {dx}}$$

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

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

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

Integrating both sides, we get

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

$$ \Rightarrow \ln \left| {v + \sqrt {{v^2} + 16} } \right| = \ln x + \ln c$$

$$ \Rightarrow v + \sqrt {{v^2} + 16}  = cx$$

Now putting, $$v = {y \over x}$$, we get

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

$$ \Rightarrow {y \over x} + \sqrt {{{{y^2} + 16{x^2}} \over {{x^2}}}}  = cx$$

$$ \Rightarrow y + \sqrt {{y^2} + 16{x^2}}  = c{x^2}$$ ...... (1)

Given, $y(1) = 3$

$$\therefore$$ When x = 1 then y = 3.

Putting in equation (1) we get,

$$3 + \sqrt {9 + 16}  = c.\,1$$

$$ \Rightarrow c = 8$$

$$\therefore$$ Solution of equation,

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

Now, y(2) means when x = 2 then y = ?

$$\therefore$$ $$y + \sqrt {{y^2} + 16 \times 4}  = 8 \times 4$$

$$ \Rightarrow y = 15$$

Solutions · 1