A) \[\operatorname{x}+y\,-\ell n\,\left| x+y \right|\,\,=\,\,c\]
B) \[3x+y+2\ell n\,\,\left| 1-x-y \right|=c\]
C) \[\operatorname{x}+3y\,-2\ell n\,\,\left| 1-x-y \right|\,\,=\,\,c\]
D) None of these
Correct Answer: B
Solution :
This is the form in which \[\frac{{{a}_{1}}}{{{a}_{2}}}=\frac{{{b}_{1}}}{{{b}_{2}}}\] The given equation can be rewritten as \[\frac{dy}{dx}=\frac{1-3(a+y)}{1+(a+y)}=f(x+y)\] Substitute \[x+y=z\Rightarrow 1+\frac{dy}{dx}=\frac{dz}{dx}\]. The equation then becomes. \[\frac{dz}{dx}-1=\frac{1-3z}{1+z}\Rightarrow \frac{dz}{dx}=\frac{1-3z+1+z}{1+z}=\frac{2-2z}{1+z}\] \[\Rightarrow \,\,\,\frac{1+z}{2(1-z)}dz=dx\] On integrating we get \[\frac{1}{2}\int{\frac{1+z}{1-z}}\,dz=\int{dx+a\Rightarrow \frac{1}{2}\int{\left[ \frac{2}{1-z}-1 \right]}\,dz=}\,x+a\] \[=\,\,-\ell n\,\left| 1-z \right|-\frac{1}{2}z=x+a\] \[=\,\,-\ell n\,\left| 1-x-y \right|-\frac{1}{2}(x+y)=x+a\] \[=\,\,-2\ell n\,\left| 1-x-y \right|-3x-y=2a\] \[\Rightarrow \,\,3x+y+2\ell n\left| 1-x-y \right|=c\] where \[c=-2a\]
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