Answer:
Given differential equation is. \[(1+{{e}^{2x}})\,dy+(1+{{y}^{2}}){{e}^{x}}dx=0\] \[\Rightarrow \] \[(1+{{e}^{2x}})\,dy=-\,(1+{{y}^{2}}){{e}^{x}}dx\] \[\Rightarrow \] \[\frac{dy}{1+{{y}^{2}}}=-\frac{{{e}^{x}}}{1+{{e}^{2x}}}\,dx\] On integrating both sides, we get \[\int{\frac{dy}{1+{{y}^{2}}}}=-\int{\frac{{{e}^{x}}}{1+{{e}^{2x}}}\,dx}\] Put \[{{e}^{x}}=t\] \[\Rightarrow \] \[{{e}^{x}}dx=dt\] \[\therefore \] We have, \[\int{\frac{dy}{1+{{y}^{2}}}}=-\int{\frac{dt}{1+{{t}^{2}}}}\] \[\Rightarrow \] \[{{\tan }^{-1}}y=-{{\tan }^{-1}}t+C\] \[\Rightarrow \] \[{{\tan }^{-1}}y+{{\tan }^{-1}}{{e}^{x}}=C\] \[[\because \,\,\,t={{e}^{x}}]\] Now, it is given that, y = 1 when x = 0. \[\therefore \] \[{{\tan }^{-1}}(1)+{{\tan }^{-1}}(1)=C\] \[\Rightarrow \] \[\frac{\pi }{4}+\frac{\pi }{4}=C\] \[\Rightarrow \] \[C=2\times \frac{\pi }{4}=\frac{\pi }{2}\] Hence, the required particular solution is \[{{\tan }^{-1}}y+{{\tan }^{-1}}{{e}^{x}}=\frac{\pi }{2}\] \[\Rightarrow \] \[{{\tan }^{-1}}y=\frac{\pi }{2}-{{\tan }^{-1}}{{e}^{x}}\] \[\Rightarrow \] \[{{\tan }^{-1}}y={{\cot }^{-1}}{{e}^{x}}\] \[\Rightarrow \] \[y=\tan ({{\cot }^{-1}}{{e}^{x}})\] \[\Rightarrow \] \[y=\tan \left( {{\tan }^{-1}}\frac{1}{{{e}^{x}}} \right)\] \[\therefore \] \[y=\frac{1}{{{e}^{x}}}\]
You need to login to perform this action.
You will be redirected in
3 sec
