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**Hint:**Since we have to solve this and by looking at the question we can say that it is a homogeneous differential equation. So we will put $y = vx$and then after equating we will integrate it and by using the partial fraction we will get the values and on further integration, we will get the solution.

**Complete step-by-step answer:**Given, we have the function

$\dfrac{{dy}}{{dx}} = \dfrac{{{x^2} + xy}}{{{x^2} + {y^2}}}$

Since this is a homogeneous differential equation.

So for solving this first of all we will put $y = vx$

On differentiating the above term we get

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

So the above equation will become

$ \Rightarrow v + x\dfrac{{dv}}{{dx}} = \dfrac{{1 + v}}{{1 + {v^2}}}$

And on solving the equation, we get

$ \Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{1 - {v^3}}}{{1 + {v^2}}}$

And it can also be written as

$ \Rightarrow \dfrac{{1 + {v^2}}}{{1 - {v^3}}}dv = \dfrac{{dx}}{x}$

Now, on integrating both the sides we get

$ \Rightarrow \int {\dfrac{{1 + {v^2}}}{{\left( {1 - v} \right)\left( {1 + {v^2} + v} \right)}}} dv = \int {\dfrac{{dx}}{x}} $, let’s name it equation $1$

Now on resolving \[ \Rightarrow \dfrac{{1 + {v^2}}}{{\left( {1 - v} \right)\left( {1 + {v^2} + v} \right)}}\] into the partial fractions, we get

$ \Rightarrow \dfrac{{1 + {v^2}}}{{\left( {1 - v} \right)\left( {1 + {v^2} + v} \right)}} = \dfrac{A}{{\left( {1 - v} \right)}} + \dfrac{{Bv + C}}{{\left( {1 + {v^2} + v} \right)}}$, and we will name its equation$2$.

So on solving the partial fractions, we get

$ \Rightarrow 1 + {v^2} = A\left( {1 + {v^2} + v} \right) + \left( {Bv + c} \right)\left( {1 - v} \right)$

And on comparing the above equation, we get

$ \Rightarrow A - B = 1$ ; $A + C = 1$; $A + B - C = 0$

So now on solving these equations, we get

$ \Rightarrow A = \dfrac{2}{3}{\text{ ; B = }}\dfrac{{ - 1}}{3}{\text{ ; C = }}\dfrac{1}{3}$

So now we will substitute these values in the equation$2$, we get

$ \Rightarrow \dfrac{{1 + {v^2}}}{{\left( {1 - v} \right)\left( {1 + {v^2} + v} \right)}} = \dfrac{2}{{3\left( {1 - v} \right)}} - \dfrac{1}{3}\dfrac{{\left( {v - 1} \right)}}{{\left( {1 + {v^2} + v} \right)}}$

Therefore the equation $1$ will become

$ \Rightarrow \int {\dfrac{2}{{3\left( {1 - v} \right)}}dv - \int {\dfrac{1}{3}\dfrac{{\left( {v - 1} \right)}}{{\left( {1 + {v^2} + v} \right)}}} } dv = \int {\dfrac{{dx}}{x}} $

Now on applying the integral formula, we get

\[ \Rightarrow \dfrac{2}{3}\log \left| {1 - v} \right| - \dfrac{1}{6}\int {\dfrac{{\left( {2v + 1 - 3} \right)}}{{\left( {1 + {v^2} + v} \right)}}} dv = \log x + \log c\]

Now, on further integrating more, we get

\[ \Rightarrow \dfrac{2}{3}\log \left| {1 - v} \right| - \dfrac{1}{6}\int {\dfrac{{\left( {2v + 1} \right)}}{{\left( {1 + {v^2} + v} \right)}}} dv + \dfrac{1}{2}\int {\dfrac{1}{{{{\left( {v + \dfrac{1}{2}} \right)}^2} + {{\left( {\dfrac{{\sqrt 3 }}{2}} \right)}^2}}}dv} = \log x + \log c\]

Now on solving this, we get

\[ \Rightarrow \dfrac{2}{3}\log \left| {1 - \dfrac{y}{x}} \right| - \dfrac{1}{6}\log \left| {{v^2} + v + 1} \right| + \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2v + 1}}{{\sqrt 3 }}} \right) = \log x + \log c\]

Combining and solving the same term together, we get

\[ \Rightarrow \dfrac{2}{3}\log \left| {x - y} \right| - \dfrac{1}{6}\log \left| {{y^2} + xy + {x^2}} \right| + \dfrac{2}{3}\log x + \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\left( {\dfrac{{2y + x}}{{x\sqrt 3 }}} \right) = \dfrac{5}{3}\log x + \log c\]

Now solving for the log by using the log property, we get

$ \Rightarrow \log \dfrac{{c{{\left( {x - y} \right)}^{2/3}}}}{{x{{\left( {{x^2} + xy + {y^2}} \right)}^{1/6}}}} = - \dfrac{1}{{\sqrt 3 }}{\tan ^{ - 1}}\dfrac{{2y + x}}{{x\sqrt 3 }}$

On removing the log, we get

\[ \Rightarrow \dfrac{{c{{\left( {x - y} \right)}^{2/3}}}}{{x{{\left( {{x^2} + xy + {y^2}} \right)}^{1/6}}}} = \exp \left[ { - \dfrac{1}{{\sqrt 3 }}{{\tan }^{ - 1}}\dfrac{{2y + x}}{{x\sqrt 3 }}} \right]\]

**Therefore, the option $\left( a \right)$ will be correct.**

**Note:**So for solving such type of problem we should have to remember the formulas which are going to be used and with practice, we can develop the skills of using the methods for solving this type of problem easily.

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