A) \[x-y=k{{e}^{x-y}}\]
B) \[x+y=k{{e}^{x+y}}\]
C) \[x+y=k(x-y)\]
D) \[x-y=k{{e}^{x+y}}\]
E) \[x+y=k{{e}^{x-y}}\]
Correct Answer: E
Solution :
\[(x+y)dx-(x+y)dy=dx+dy\] \[\Rightarrow \] \[(x+y-1)dx=(x+y+1)dy\] \[\Rightarrow \] \[\frac{dy}{dx}=\left( \frac{x+y-1}{x+y+1} \right)\] Let\[x+y=v\]and \[\frac{dy}{dx}=\frac{dv}{dx}-1\] \[\therefore \] \[\frac{dv}{dx}-1=\frac{v-1}{v+1}\] \[\Rightarrow \] \[\frac{dv}{dx}=\frac{v-1+v+1}{v+1}\] \[\Rightarrow \] \[\int{\left( \frac{v+1}{2v} \right)}\,dv=\int{dx}\] \[\Rightarrow \] \[\frac{1}{2}\int{1\,dv}+\frac{1}{2}\int{\frac{1}{v}}dv=\int{1\,dx}\] \[\Rightarrow \] \[\frac{1}{2}v+\frac{1}{2}\log v=x+\log c\] \[\Rightarrow \]\[\frac{1}{2}(x+y)+\frac{1}{2}\log (x+y)=x+\log c\] \[\Rightarrow \] \[\log \left( \frac{x+y}{{{c}^{2}}} \right)=x-y\] \[\Rightarrow \] \[x+y=k{{e}^{x-y}}\]
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