A) \[(x-2)=k{{e}^{{{\tan }^{-1}}y}}\]
B) \[2x{{e}^{{{\tan }^{-1}}y}}={{e}^{2{{\tan }^{-1}}y}}+k\]
C) \[x.{{e}^{-{{\tan }^{-1}}y}}=ta{{n}^{-1y}}+k\]
D) \[x.{{e}^{2{{\tan }^{-1}}y}}=ta{{n}^{-1y}}+k\]
Correct Answer: C
Solution :
[c] \[(1+{{y}^{2}})-(x+{{e}^{{{\tan }^{-1}}y}})\frac{dy}{dx}=0\] \[\Rightarrow (1+{{y}^{2}})\frac{dy}{dx}-x={{e}^{{{\tan }^{-1}}y}}\] \[\Rightarrow \frac{dy}{dx}-\frac{1}{1+{{y}^{2}}}x=\frac{{{e}^{{{\tan }^{-1}}y}}}{1+{{y}^{2}}}\] \[IF={{e}^{\int{pdy}}}={{e}^{\int{\frac{-1}{1+{{y}^{2}}}dy}}}={{e}^{-{{\tan }^{-1}}y}}\] \[\therefore \]The solution is \[x.{{e}^{-{{\tan }^{-1}}y}}=\int{{{e}^{-{{\tan }^{-1}}y}}}.\frac{{{e}^{-{{\tan }^{-1}}y}}}{1+{{y}^{2}}}.dy+k\] Or \[x.{{e}^{-{{\tan }^{-1}}y}}=\int{\frac{1}{1+{{y}^{2}}}}.dy+k\] Or \[x.{{e}^{-{{\tan }^{-1}}y}}={{\tan }^{-1}}y+k\]
You need to login to perform this action.
You will be redirected in
3 sec
