Solve \begin {align*} y^{\prime } & =\frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}\\ y^{\prime } & =\omega \left ( x,y\right ) \end {align*}
The symmetry condition results in the pde\begin {equation} \eta _{x}+\omega \left ( \eta _{y}-\xi _{x}\right ) -\omega ^{2}\xi _{y}-\omega _{x}\xi -\omega _{y}\eta =0 \tag {1} \end {equation} Let Ansatz be\begin {align*} \xi & =0\\ \eta & =f\left ( x\right ) g\left ( y\right ) \end {align*}
Hence (1) becomes\[ g\left ( y\right ) \frac {df}{dx}+\omega f\left ( x\right ) \frac {dg}{dy}-\omega _{y}f\left ( x\right ) g\left ( y\right ) =0 \] But \(\omega _{x}=\frac {d}{dx}\left ( \frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}\right ) =-\frac {\left ( y+1\right ) }{\left ( x+1\right ) ^{2}}\) and \(\omega _{y}=\frac {x+1+2\sqrt {1+y}}{\sqrt {1+y}\left ( 2+2x\right ) }\). Hence the above becomes\begin {equation} g\left ( y\right ) \frac {df}{dx}+\left ( \frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}\right ) f\left ( x\right ) \frac {dg}{dy}-\frac {x+1+2\sqrt {1+y}}{\sqrt {1+y}\left ( 2+2x\right ) }\left ( f\left ( x\right ) g\left ( y\right ) \right ) =0 \tag {2} \end {equation} The numerator of the normal form of the above is\begin {equation} 2\frac {df}{dx}g\sqrt {1+y}x+2y\sqrt {1+y}f\frac {dg}{dy}+2f\frac {dg}{dy}xy+2\frac {df}{dx}g\sqrt {1+y}-2fg\sqrt {1+y}+2f\frac {dg}{dy}\sqrt {1+y}-fgx+2f\frac {dg}{dy}x+2fy\frac {dg}{dy}-fg+2f\frac {dg}{dy}=0 \tag {3} \end {equation} We can now either collect on \(y\) or \(x\) and try. Let us start with collecting on all terms with \(y\). This gives\begin {equation} g\sqrt {1+y}\left ( 2x\frac {df}{dx}+2\frac {df}{dx}-2f\right ) +y\sqrt {1+y}\frac {dg}{dy}\left ( 2f\right ) +\frac {dg}{dy}\sqrt {1+y}\left ( 2f\right ) +g\left ( xf-f\right ) +y\frac {dg}{dy}\left ( 2xf+2f\right ) +\frac {dg}{dy}\left ( 2xf+2f\right ) =0 \tag {3A} \end {equation} The coefficients of all terms with \(g\left ( y\right ) \) or \(y\) in them are from the above are the following, which each must be zero\begin {align*} 2f & =0\\ xf-f & =0\\ 2xf+2f & =0\\ 2x\frac {df}{dx}+2\frac {df}{dx}-2f & =0 \end {align*}
Now we set each to zero and see if this produces \(f\left ( x\right ) \) which can be used. We have 4 choices to try above. Starting from the most simple one. The first one above gives \(2f=0\) or \(f=0\). But this is not function of \(x\). We try the next one \(xf-f=0\). This gives \(f=0\) or \(x=1\). Hence this does not give \(f\) as function of \(x.\) Next we try \(2xf+2f.\) This also does not give \(f\) as function of \(x.\) The last one is \(2x\frac {df}{dx}+2\frac {df}{dx}-2f=0\) or \(\frac {df}{dx}=\frac {2f}{2x+2}\). Solving this gives \(f=c_{1}\left ( x+1\right ) \). This is successful since \(f\) is function of \(x\). Hence\begin {align*} f\left ( x\right ) & =c_{1}\left ( x+1\right ) \\ \frac {df}{dx} & =c_{1} \end {align*}
Now we need to determine \(g\left ( y\right ) \). Substituting the above into (3) gives\[ 2c_{1}g(y)\sqrt {1+y}x+2\sqrt {1+y}c_{1}(x+1)\frac {dg}{dy}y+2c_{1}(x+1)\frac {dg}{dy}xy+2c_{1}g\sqrt {1+y}-2c_{1}(x+1)g\sqrt {1+y}+2c_{1}(x+1)\frac {dg}{dy}\sqrt {1+y}-c_{1}(x+1)g(y)x+2c_{1}(x+1)\frac {dg}{dy}x+2c_{1}(x+1)\frac {dg}{dy}y-c_{1}(x+1)g(y)+2c_{1}(x+1)\frac {dg}{dy}=0 \] Which simplifies to\begin {equation} 2c_{1}\sqrt {1+y}\frac {dg}{dy}yx+2c_{1}\frac {dg}{dy}x^{2}y-c_{1}gx^{2}+2c_{1}\frac {dg}{dy}\sqrt {1+y}x+2\sqrt {1+y}c_{1}\frac {dg}{dy}y+2c_{1}\frac {dg}{dy}x^{2}+4c_{1}\frac {dg}{dy}xy-2c_{1}xg+2c_{1}\frac {dg}{dy}\sqrt {1+y}+4c_{1}\frac {dg}{dy}x+2c_{1}\frac {dg}{dy}y-c_{1}g(y)+2c_{1}\frac {dg}{dy}=0 \tag {4} \end {equation} Now factoring on all terms with \(x\), and these are \(\left \{ x,x^{2}\right \} \) gives\begin {equation} -c_{1}x^{2}\left ( -2\frac {dg}{dy}y+g-2\frac {dg}{dy}\right ) -c_{1}x\left ( -2\sqrt {1+y}\frac {dg}{dy}y-2\sqrt {1+y}\frac {dg}{dy}-2\frac {dg}{dy}y+g-2\frac {dg}{dy}\right ) +T=0 \tag {4A} \end {equation} Where \(T\) are terms that depends on \(y\) only. Each factor of \(x,x^{2}\) must be zero. Hence the first above implies\begin {align*} -2\frac {dg}{dy}y+g-2\frac {dg}{dy} & =0\\ g^{\prime }\left ( y\right ) & =\frac {g}{2\left ( 1+y\right ) } \end {align*}
Solving gives\begin {equation} g=c_{2}\sqrt {1+y} \tag {5} \end {equation} Substituting (5) into (4) gives\[ c_{1}\left ( 1+x\right ) c_{2}\left ( 1+y\right ) =0 \] Which is not zero. Hence this term does not work. Now we try the second term in (4A) which means\begin {align*} -2\sqrt {1+y}\frac {dg}{dy}y-2\sqrt {1+y}\frac {dg}{dy}-2\frac {dg}{dy}y+g-2\frac {dg}{dy} & =0\\ \frac {dg}{dy} & =\frac {-g}{-2\sqrt {1+y}y-2\sqrt {1+y}-2y-2} \end {align*}
Solving gives\[ g\left ( y\right ) =c_{2}\frac {\sqrt {1+y}}{1+\sqrt {1+y}}\] Again, substituting the above back in (4) gives\[ c_{1}\left ( 1+x\right ) c_{2}\frac {\left ( 1+y\right ) x}{\left ( 1+\sqrt {1+y}\right ) ^{2}}=0 \] Which is not zero. Therefore starting with \(f\left ( x\right ) =c_{1}\left ( x+1\right ) \) has failed to produce a valid \(g\left ( y\right ) \) to satisfy the pde. This means we need to start all over again. Going back to (3) and now collecting on all terms with \(x\) instead. Here is (3) again\begin {equation} 2\frac {df}{dx}g\sqrt {1+y}x+2y\sqrt {1+y}f\frac {dg}{dy}+2f\frac {dg}{dy}xy+2\frac {df}{dx}g\sqrt {1+y}-2fg\sqrt {1+y}+2f\frac {dg}{dy}\sqrt {1+y}-fgx+2f\frac {dg}{dy}x+2fy\frac {dg}{dy}-fg+2f\frac {dg}{dy}=0 \tag {3} \end {equation} Collecting on all terms that depend on \(x\) gives\begin {equation} x\frac {df}{dx}\left ( 2g\sqrt {1+y}\right ) +f\left ( 2y\sqrt {1+y}\frac {dg}{dy}-2g\sqrt {1+y}+2\frac {dg}{dy}\sqrt {1+y}+2y\frac {dg}{dy}+2\frac {dg}{dy}-g\right ) +xf\left ( 2\frac {dg}{dy}y-g+2\frac {dg}{dy}y\right ) =0 \tag {3B} \end {equation} Each term must be zero, hence this gives these trials\begin {align*} 2g\sqrt {1+y} & =0\\ 2\frac {dg}{dy}y-g+2\frac {dg}{dy}y & =0\\ 2y\sqrt {1+y}\frac {dg}{dy}-2g\sqrt {1+y}+2\frac {dg}{dy}\sqrt {1+y}+2y\frac {dg}{dy}+2\frac {dg}{dy}-g & =0 \end {align*}
Starting with the first one above \(2g\sqrt {1+y}=0\) which gives \(g=0\) which does not match the ansatz. Now we try the second one above, which gives \[ \frac {dg}{dy}=\frac {g}{2+2y}\] Solving gives\begin {equation} g=c_{1}\sqrt {1+y} \tag {6} \end {equation} Which meets the requirements of the ansatz. Now we need to use the above to generate \(f\left ( x\right ) \). We do not need to try the third one above unless this fails. Substituting (6) into (3) gives\begin {align} c_{2}\left ( 2\frac {df}{dx}xy+2\frac {df}{dx}x+2\frac {df}{dx}y-fy+2\frac {df}{dx}-f\right ) & =0\nonumber \\ 2\frac {df}{dx}xy+2\frac {df}{dx}x+2\frac {df}{dx}y-fy+2\frac {df}{dx}-f & =0 \tag {7} \end {align}
Collecting on \(y\) gives\[ c_{1}\left ( 1+y\right ) \left ( 2\frac {df}{dx}x+2\frac {df}{dx}-f\right ) =0 \] Hence \(2\frac {df}{dx}x+2\frac {df}{dx}-f\) must be zero. This gives as solution \begin {align*} f\left ( x\right ) & =c_{2}\sqrt {1+x}\\ \frac {df}{dx} & =c_{2}\frac {1}{2\sqrt {1+x}} \end {align*}
Substituting the above into (7) to verify gives\begin {align*} 2\left ( c_{2}\frac {1}{2\sqrt {1+x}}\right ) xy+2\left ( c_{2}\frac {1}{2\sqrt {1+x}}\right ) x+2\left ( c_{2}\frac {1}{2\sqrt {1+x}}\right ) y-\left ( c_{2}\sqrt {1+x}\right ) y+2\left ( c_{2}\frac {1}{2\sqrt {1+x}}\right ) -c_{2}\sqrt {1+x} & =0\\ c_{2}\frac {1}{\sqrt {1+x}}xy+c_{2}\frac {1}{\sqrt {1+x}}x+c_{2}\frac {1}{\sqrt {1+x}}y-c_{2}\sqrt {1+x}y+c_{2}\frac {1}{\sqrt {1+x}}-c_{2}\sqrt {1+x} & =0\\ c_{2}\left ( \frac {1}{\sqrt {1+x}}xy+\frac {1}{\sqrt {1+x}}x+\frac {1}{\sqrt {1+x}}y-\sqrt {1+x}y+\frac {1}{\sqrt {1+x}}-\sqrt {1+x}\right ) & =0\\ 0 & =0 \end {align*}
Verified, Hence we have found \(f\left ( x\right ) ,g\left ( y\right ) \). Therefore\begin {align*} \xi & =0\\ \eta & =f\left ( x\right ) g\left ( y\right ) \\ & =\sqrt {1+x}\sqrt {1+y} \end {align*}
Where we set \(c_{1}=c_{2}=1\). The integrating factor is therefore \begin {align*} \mu \left ( x,y\right ) & =\frac {1}{\eta -\xi \omega }\\ & =\frac {1}{\sqrt {1+x}\sqrt {1+y}} \end {align*}
The next step is to determine the canonical coordinates \(R,S\). Where \(R\) is the independent variable and \(S\) is the dependent variable. This is done by using the standard characteristic equation by writing\begin {equation} \frac {dx}{\xi }=\frac {dy}{\eta }=dS\nonumber \end {equation} For the special case \(\xi =0\,\) we have \(R=x\). \(S\left ( x,y\right ) \) is now found from the last two pair of equations which gives\begin {align*} dS & =\frac {dy}{\eta }\\ dS & =\frac {dy}{\sqrt {1+x}\sqrt {1+y}}\\ S & =2\frac {\sqrt {1+y}}{\sqrt {1+x}} \end {align*}
Hence (constant of integration is set to zero)\begin {align} R & =x\tag {2}\\ S & =2\frac {\sqrt {1+y}}{\sqrt {1+x}}\nonumber \end {align}
Now that \(R\left ( x,y\right ) ,S\left ( x,y\right ) \) are found, the ODE \(\frac {dS}{dR}=\Omega \left ( R\right ) \) is setup. The ODE comes out to be function of \(R\) only, so it is quadrature. This is the main idea of this method. By solving for \(R\) we go back to \(x,y\) and solve for \(y\left ( x\right ) \). How to find \(\frac {dS}{dR}\)? There is an equation to determine this given by\begin {align*} \frac {dS}{dR} & =\frac {\frac {dS}{dx}+\omega \left ( x,y\right ) \frac {dS}{dy}}{\frac {dR}{dx}+\omega \left ( x,y\right ) \frac {dR}{dy}}\\ & =\frac {S_{x}+\omega \left ( x,y\right ) S_{y}}{R_{x}+\omega \left ( x,y\right ) R_{y}} \end {align*}
Everything on the RHS is known. \(S_{x}=-\frac {\sqrt {1+y}}{\left ( 1+x\right ) ^{\frac {3}{2}}},R_{x}=1,S_{y}=\frac {1}{\sqrt {1+x}\sqrt {1+y}},R_{y}=0\). Substituting into the above gives\begin {align*} \frac {dS}{dR} & =-\frac {\sqrt {1+y}}{\left ( 1+x\right ) ^{\frac {3}{2}}}+\omega \left ( x,y\right ) \frac {1}{\sqrt {1+x}\sqrt {1+y}}\\ & =-\frac {\sqrt {1+y}}{\left ( 1+x\right ) ^{\frac {3}{2}}}+\left ( \frac {x\sqrt {1+y}+\sqrt {1+y}+1+y}{1+x}\right ) \frac {1}{\sqrt {1+x}\sqrt {1+y}}\\ & =\frac {1}{\sqrt {x+1}}\\ & =\frac {1}{\sqrt {R+1}} \end {align*}
Hence \[ \frac {dS}{dR}=\frac {1}{\sqrt {R+1}}\] This is quadrature. Solving gives \[ S=2\sqrt {R+1}+c_{1}\] Convecting back to \(x,y\,\) gives\[ 2\frac {\sqrt {1+y}}{\sqrt {1+x}}=2\sqrt {x+1}+c_{1}\]