A) \[\log ({{(y+3)}^{2}}+{{(x+2)}^{2}})+ta{{n}^{-1}}\frac{y+3}{y+2}+C\]
B) \[\log ({{(y+3)}^{2}}+{{(x-2)}^{2}})+ta{{n}^{-1}}\frac{y-3}{x-2}=C\]
C) \[\log ({{(y+3)}^{2}}+{{(x+2)}^{2}})+2ta{{n}^{-1}}\frac{y+3}{x+2}=C\]
D) \[\log ({{(y+3)}^{2}}+{{(x+2)}^{2}})-2ta{{n}^{-1}}\frac{y+3}{x+2}=C\]
Correct Answer: C
Solution :
[c] The intersection of \[y-x+1=0\] and \[y+x+5=0\] is \[\left( -\,2,-\,3 \right)\]. Put \[x=X-2,y=Y-3.\] The given equation reduces to\[\frac{dy}{dx}=\frac{Y-X}{Y+X}\]. Putting \[Y=vX,\]we get \[X\frac{dv}{dX}=-\frac{{{v}^{2}}+1}{v+1}\] Or \[\left( -\frac{v}{{{v}^{2}}+1}-\frac{1}{{{v}^{2}}+1} \right)dv=\frac{dx}{X}\] Or \[-\frac{1}{2}\log \left( {{v}^{2}}+1 \right)-{{\tan }^{-1}}v=\log \left| X \right|+\]constant Or \[\log ({{Y}^{2}}+{{X}^{2}})+2ta{{n}^{-1}}\frac{Y}{X}=cons\operatorname{tant}\] Or \[\log ({{(y+3)}^{3}}+{{(x+2)}^{2}})+2ta{{n}^{-1}}\frac{y+3}{x+2}=C\]
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