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