Solve the differential equation \[{y^2} + {x^2}\dfrac{{dy}}{{dx}} = xy\dfrac{{dy}}{{dx}}\].

A. $\log xy = 2 - \dfrac{{2y}}{x}$

B. $y = K{e^{\dfrac{y}{x}}}$

C. $xy = K{e^{2 + \dfrac{{2y}}{x}}}$

D. $\log Kxy = 2 + \dfrac{{2y}}{x}$

The given equation is a differential equation of order one and degree one. Since, its order is one, we have three methods to use and it depends on the type of the given equation. It can be either a variable separable, homogeneous or linear equation. Remember that variable separable equation is of the form $dy.f(y) = dx.g(x)$, where f and g are function involving only y and x respectively. Also, an equation is said to be a homogeneous equation of order zero if it can be expressed as $\dfrac{{dy}}{{dx}} = x.f(\dfrac{y}{x})$. Standard form of a linear equation is $\dfrac{{dy}}{{dx}} + Py = Q$, where P and Q are functions containing only x. We will first find the type of equation and then proceed with the method already known to us.

We can observe that the given D.E \[{y^2} + {x^2}\dfrac{{dy}}{{dx}} = xy\dfrac{{dy}}{{dx}}\] - - - - - - - - - - - - - - - - - - - - - (1)

can be re-written as ${y^2} = xy\dfrac{{dy}}{{dx}} - {x^2}\dfrac{{dy}}{{dx}}$.

$ \Rightarrow {y^2} = \dfrac{{dy}}{{dx}}\left( {xy - {x^2}} \right)$

$ \Rightarrow \dfrac{{dy}}{{dx}} = \dfrac{{{y^2}}}{{xy - {x^2}}}$ - - - - - - - - - - - - - - - - - (2)

On dividing both numerator and denominator by ${x^2}$ in the RHS we get,

$ \Rightarrow \dfrac{{dy}}{{dx}} = \left( {\dfrac{{\dfrac{{{y^2}}}{{{x^2}}}}}{{\dfrac{y}{x} - 1}}} \right)$

$ \Rightarrow \dfrac{{dy}}{{dx}} = f\left( {\dfrac{y}{x}} \right)$.

So, it is a homogeneous differential Equation of degree zero. Hence, we will take another variable $v = \dfrac{y}{x}$ and substitute in the given equation.

$ \Rightarrow y = vx$

On differentiation we get $\dfrac{{dy}}{{dx}} = x\dfrac{{dv}}{{dx}} + v$.

On substituting this in equation (2) we get, $ \Rightarrow x\dfrac{{dv}}{{dx}} + v = \dfrac{{{y^2}}}{{xy - {x^2}}}$

$ \Rightarrow x\dfrac{{dv}}{{dx}} + v = \dfrac{{{v^2}{x^2}}}{{x.(vx) - {x^2}}}$

By dividing the numerator and denominator by ${x^2}.

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

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

$ \Rightarrow \dfrac{{v - 1}}{v}dv = \dfrac{{dx}}{x}$

Now this is of variable separable form, so we directly integrate to get,

$\int {\dfrac{{v - 1}}{v}dv}  = \int {\dfrac{{dx}}{x}} $

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

$ \Rightarrow v - \log v + {C_1} = \log x + {C_2}$

Using the results $\int {\dfrac{1}{x}dx}  = \log x + C$ and $\int {{x^n}}  = \dfrac{{{x^{n + 1}}}}{{n + 1}} + C$

$ \Rightarrow v - \log v = \log x + K'$, where $K' = {C_2} - {C_1}$

$ \Rightarrow  - \log v - \log x = K' - v$

Using $\log x + \log y = \log xy$

$ \Rightarrow  - \log vx = K' - v$

$ \Rightarrow \log vx = v - K'$

$ \Rightarrow vx = {e^{v - K'}}$

$ \Rightarrow vx = {e^v}.{e^{ - K'}}$

$ \Rightarrow vx = {e^v}K$ where $K = {e^{ - K'}}$

Use $y = vx$

$ \Rightarrow y = K{e^{\dfrac{y}{x}}}$

Hence, the general solution of the given Differential equation is $y = K{e^{\dfrac{y}{x}}}$. And the correct answer is option (B). $y = K{e^{\dfrac{y}{x}}}$.