Answer:
Let \[x+y=v.\] Then, \[1+\frac{dy}{dx}=\frac{dv}{dx}\] \[\Rightarrow \] \[\frac{dy}{dx}=\frac{dv}{dx}-1\] Putting \[x+y=v\] and \[\frac{dy}{dx}=\frac{dv}{dx}-1\] the given differential equation, we get \[{{v}^{2}}\left( \frac{dv}{dx}-1 \right)={{a}^{2}}\] \[\Rightarrow \] \[{{v}^{2}}\frac{dv}{dx}={{a}^{2}}+{{v}^{2}}\] \[\Rightarrow \] \[{{v}^{2}}dv=({{a}^{2}}+{{v}^{2}})dx\] \[\Rightarrow \] \[\frac{{{v}^{2}}}{{{v}^{2}}+{{a}^{2}}}dv=dx\] [by separating the variables] \[\Rightarrow \] \[\left( 1-\frac{{{a}^{2}}}{{{v}^{2}}+{{a}^{2}}} \right)dv=dx\] \[\Rightarrow \] \[\int{1\cdot dv-{{a}^{2}}\int{\frac{1}{{{v}^{2}}+{{a}^{2}}}}}dv=\int{dx}+C\] [on integration] \[\Rightarrow \] \[v-a\,\,{{\tan }^{-1}}\left( \frac{v}{a} \right)=x+C\] \[\Rightarrow \] \[(x+y)-a\,\,{{\tan }^{-1}}\left( \frac{x+y}{a} \right)=x+C,\] which is the required solution.
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