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# The solution of the differential equation $xdy - ydx = \left( {\sqrt {{x^2} + {y^2}} } \right)dx$ is?A. $y - \sqrt {{x^2} + {y^2}} = C{x^2}$B. $y + \sqrt {{x^2} + {y^2}} = C{x^2}$C. $y + \sqrt {{x^2} + {y^2}} + C{x^2} = 0$D. None of these

Last updated date: 27th Mar 2023
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Hint: Identify the type of differential equation & chose the method to solve it.

Start by solving the given equation.
$xdy - ydx = \sqrt {{x^2} + {y^2}} dx$
$xdy = \left( {y + \sqrt {{x^2} + {y^2}} } \right)dx$
$\dfrac{{dy}}{{dx}} = \dfrac{{y + \sqrt {{x^2} + {y^2}} }}{x}$
Since it is homogeneous equation,
We will equate $y = vx$ will simplify the solution.
Therefore,
Put $y = vx \Rightarrow \dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$
$v + x\dfrac{{dv}}{{dx}} = \dfrac{{vx + x\sqrt {1 + {v^2}} }}{x} = \dfrac{{v + \sqrt {1 + {v^2}} }}{1}$
$x\dfrac{{dv}}{{dx}} = \sqrt {1 + {v^2}}$
$\int {\dfrac{{dv}}{{\sqrt {1 + {v^2}} }} = \int {\dfrac{{dx}}{x}} }$
$\log \left( {v + \sqrt {1 + {v^2}} } \right) = \log x + \log C$

The next step is to re-equate $vx = y$,
$\log \left( {\dfrac{y}{x} + \dfrac{{\sqrt {{x^2} + {y^2}} }}{x}} \right) = \log x + \log C$
$\log \left( {y + \sqrt {{x^2} + {y^2}} } \right) - \log x = \log x + \log C$
$\log \left( {y + \sqrt {{x^2} + {y^2}} } \right) = \log \left( {{x^2}C} \right)$
Cancelling log on both sides, we get,
$\left( {y + \sqrt {{x^2} + {y^2}} } \right) = \left( {{x^2}C} \right)$
Option B is the correct option.

Note: It is important to have a knowledge of the various methods to solve a differential equation.