# 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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Answer

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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.

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.

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