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**Hint:**We see that the given differential equation is a homogeneous differential equation whose standard substitution is $y=vx$.We solve the given homogeneous differential equation by putting $y=vx$ and then using the separation of variables $v,x$ to integrate. We use the standard integration $\int{\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}dx}=\ln f\left( x \right)$ to proceed.

**Complete step-by-step solution:**

We know that a differential equation consists of differentials, functions and variables. We call a first order differential equation homogeneous if $\dfrac{dy}{dx}=f\left( x,y \right)$ if $f\left( x,y \right)=f\left( kx,ky \right)$ for non-zero $k\in R$.We always solve homogeneous differential equation by standard substitution$y=vx$. We are given in the question the following differential equation.

\[\dfrac{dy}{dx}=\dfrac{x-y}{x+y}\]

Let us denote$f\left( x,y \right)=\dfrac{x-y}{x+y}$. For any $k\in R$ we have

\[f\left( kx,ky \right)=\dfrac{kx-ky}{kx+ky}=\dfrac{k\left( x-y \right)}{k\left( x+y \right)}=\dfrac{x-y}{x+y}=f\left( x,y \right)\]

So the given differential equation is a homogeneous differential equation. So let us consider$y=vx$. We differentiate both side with respect to $x$ to have;

\[\begin{align}

& \dfrac{dy}{dx}=v\dfrac{dx}{dx}+x\dfrac{dv}{dx} \\

& \Rightarrow \dfrac{dy}{dx}=v+x\dfrac{dv}{dx}.....\left( 1 \right) \\

\end{align}\]

We put $y=vx$ in the given differential equation to have;

\[\begin{align}

& \dfrac{dy}{dx}=\dfrac{x-vx}{x+vx} \\

& \Rightarrow \dfrac{dy}{dx}=\dfrac{1-v}{1+v}.....\left( 2 \right) \\

\end{align}\]

We equate right hand sides of (1) and (2) to have;

\[v+x\dfrac{dv}{dx}=\dfrac{1-v}{1+v}\]

Now we shall use separation of variables . We have

\[\begin{align}

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

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

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

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

\end{align}\]

We take negative sign both sides and make complete square in the left hand side to have;

\[\begin{align}

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

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

\end{align}\]

We integrate broth dies with the respect to the corresponding variables and have ;

\[\begin{align}

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

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

\end{align}\]

We use the standard integration $\int{\dfrac{{{f}^{'}}\left( x \right)}{f\left( x \right)}dx}=\ln f\left( x \right)$ in both sides f the above step to have;

\[\Rightarrow \dfrac{1}{2}\ln \left| {{\left( v+1 \right)}^{2}}-2 \right|=\ln \left| x \right|+{{c}_{1}}\]

Let us have ${{c}_{2}}=\ln \left| {{c}_{1}} \right|$. We have

\[\begin{align}

& \Rightarrow \dfrac{1}{2}\ln \left| {{\left( v+1 \right)}^{2}}-2 \right|=-\ln \left| x \right|+\ln \left| {{c}_{2}} \right| \\

& \Rightarrow \ln \left| {{\left( v+1 \right)}^{2}}-2 \right|=2\ln \left| \dfrac{{{c}_{2}}}{x} \right| \\

& \Rightarrow \ln \left| {{\left( v+1 \right)}^{2}}-2 \right|=\ln {{\left| \dfrac{{{c}_{2}}}{x} \right|}^{2}} \\

\end{align}\]

We equate the arguments of respective sides to have;

\[\begin{align}

& \Rightarrow {{\left( v+1 \right)}^{2}}-2=\dfrac{c_{2}^{2}}{{{x}^{2}}} \\

& \Rightarrow {{x}^{2}}\left( {{v}^{2}}-2v-1 \right)=c_{2}^{2} \\

& \Rightarrow {{x}^{2}}\left( {{v}^{2}}-2v-1 \right)=c_{2}^{2} \\

\end{align}\]

We put back $v=\dfrac{y}{x}$ in the above step to have;

\[\begin{align}

& \Rightarrow {{x}^{2}}\left( {{\left( \dfrac{y}{x} \right)}^{2}}+2\left( \dfrac{y}{x} \right)-1 \right)=c_{2}^{2} \\

& \Rightarrow {{x}^{2}}\left( \dfrac{{{y}^{2}}+2xy-{{x}^{2}}}{{{x}^{2}}} \right)=c_{2}^{2} \\

& \Rightarrow {{y}^{2}}+2xy-{{x}^{2}}=c_{2}^{2} \\

& \Rightarrow {{x}^{2}}-{{y}^{2}}-2xy=c\left( \text{where }c=-c_{2}^{2} \right) \\

\end{align}\]

**Here ${{c}_{1}},{{c}_{2}},c$ are real constants of integration. So the correct option is D.**

**Note:**We note that the highest differential coefficient of a differential is called order and the highest power on the derivative when expressed in polynomial form. The given differential equation is linear because degree is 1 which makes it a linear homogeneous differential equation. We should remember the logarithmic identities $\ln \left( ab \right)=\ln a+ \ln b$ and $\ln \left( \dfrac{a}{b} \right)=\ln a-\ln b$ while solving homogeneous differential equations.

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