A) \[{{x}^{2}}+{{y}^{2}}+xy-x+y=c\]
B) \[{{x}^{2}}+{{y}^{2}}-xy+x+y=c\]
C) \[{{x}^{2}}-{{y}^{2}}+2xy-x+y=c\]
D) \[{{x}^{2}}-{{y}^{2}}-2xy+x-y=c\]
Correct Answer: B
Solution :
\[(2x-y+1)dx+(2y-x+1)dy=0\] \[\frac{dy}{dx}=\frac{2x-y+1}{x-2y-1}\], put \[x=X+h\], \[y=Y+k\] \[\frac{dY}{dX}=\frac{2X-Y+2h-k+1}{X-2Y+h-2k-1}\] \[2h-k+1=0\] Þ \[h-2k-1=0\] On solving \[h=-1\], \[k=-1\]; \\[\frac{dY}{dX}=\frac{2X-Y}{X-2Y}\] Put \[Y=vX\]; \ \[\frac{dY}{dX}=v+X\frac{dv}{dX}\] \[v+X\frac{dv}{dX}=\frac{2X-vX}{X-2vX}=\frac{2-v}{1-2v}\] \[X\frac{dv}{dX}=\frac{2-2v+2{{v}^{2}}}{1-2v}=\frac{2({{v}^{2}}-v+1)}{1-2v}\] \ \[\frac{dX}{X}=\frac{(1-2v)}{2({{v}^{2}}-v+1)}dv\] Put \[{{v}^{2}}-v+1=t\] Þ \[(2v-1)dv=dt\] \ \[\frac{dX}{X}=-\frac{dt}{2t}\] \ \[\log X=\log {{t}^{-1/2}}+\log c\] \ \[X={{t}^{-1/2}}c\] Þ \[X={{({{v}^{2}}-v+1)}^{-1/2}}.c\] \[{{X}^{2}}({{v}^{2}}-v+1)=\]constant \[{{(x+1)}^{2}}\left( \frac{{{(y+1)}^{2}}}{{{(x+1)}^{2}}}-\frac{(y+1)}{x+1}+1 \right)=\]constant \[{{(y+1)}^{2}}-(y+1)(x+1)+{{(x+1)}^{2}}=c\] \[{{y}^{2}}+{{x}^{2}}-xy+x+y=c\].
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