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