Take \(x\) = X + l; y = Y + m (\(\because\) l and m are constants)

So, \(\frac{dy}{dx}\) = \(\frac{dY}{dX}\)

Therefore, the equations becomes (in X, Y)

\(\frac{dY}{dX}\) = \(\frac{X+2Y+ \ell+2m-3}{2X+Y+2\ell+m-3}\) = \(\frac{X+2Y}{2X+y}\)

If l, m are chosen to satisfy

\(\ell\) + 2m - 3 = 0

2\(\ell\) + m - 3 = 0

(where \(\ell\) = 1 and m = 1)

In X, Y the equation is homogeneous and of the first degree. Put Y = VX. Then

V + X\(\frac{dV}{dX}\) = \(\frac{X+2VX}{2X+VX}\) = \(\frac{1+2V}{2+V}\)

X\(\frac{dV}{dX}\) = \(\frac{1+2V-(2+V)V}{2+V}\) = \(\frac{1-V^2}{2+V}\)

Separating the variables (X,V) and integrating,

\(\int\frac{2+V}{1-V^2}dV\) = \(\int{\frac{dX}{X}}\)+A

where A is an arbitrary constant.

\(\int\frac{2+V}{1-V^2}dV\) = \(\int{\frac{dX}{X}}\) + A

\(\Rightarrow\) \(\int\frac{2}{1-V^2}dV\) + \(\frac{1}{-2}\int\frac{-2V}{1-V^2}dV\) = \(\int{\frac{dX}{X}}\) + A

\(\Rightarrow\) 2\(\big(\frac{1}{2}log|\frac{1+V}{1-V}|\big)\) - \(\frac{1}{2}\)log(1-V^{2}) = logX + logC

\(\big(\because\int\frac{1}{a^2-x^2}dx\) = \(\frac{1}{2a}log|\frac{a+x}{a-x}|\) and here a = 1 and putting A = logC \(\big)\)

\(\Rightarrow\) log\(|\frac{1+V}{1-V}|\) - log\(\sqrt{1-V^2}\) = logXC (\(\because\) loga^{n} = nloga)

\(\Rightarrow\) log \(\Big(\big(\frac{1+V}{1-V}\big)\frac{1}{\sqrt{1-v^2}}\Big)\) = logXC \(\big(log\frac{A}{B}= logA-logB\big)\)

\(\Rightarrow\) \(\frac{1+V}{1-V}\) \(\frac{1}{\sqrt{1-V^2}}\) = XC

Now putting V = \(\frac{Y}{X}\) (\(\because\) Y = VX)

\(\frac{1+\frac{Y}{X}}{1-\frac{Y}{X}}\) \(\frac{1}{\sqrt{1-\frac{Y^2}{X^2}}}\) = XC

\(\Rightarrow\) \(\frac{X+Y}{X-Y}\) \(\frac{X}{\sqrt{X^2-Y^2}}\) = XC

\(\Rightarrow\)

\(\frac{X+Y}{X-Y}\) \(\frac{1}{\sqrt{X^2-Y^2}}\) = C |

where C is constant

\(\Rightarrow\) X+Y = C (X-Y) \(\sqrt{X^2-Y^2}\)

Now

\(x= X+\ell \) = X + 1 \(\Rightarrow\) X = \(x-1\)

y = Y + m = Y + 1 \(\Rightarrow\) Y = y -1

then, \(x-1\) + y - 1 = C ( \(x-1\) - y + 1) \(\sqrt{(x-1)^2-(y-1)^2}\)

\(\Rightarrow\)

\(x+y-2 \) = C (\(x-y\)) \(\sqrt{(x-1)^2-(y-1)^2}\) |