A) \[y={{e}^{x-{{\tan }^{-1}}x}}\]
B) \[y={{e}^{x}}.{{\tan }^{-1}}x+1\]
C) \[y={{\tan }^{-1}}\,x+1\]
D) \[y={{e}^{x+{{\tan }^{-1}}x}}\]
Correct Answer: A
Solution :
Given differential equation is \[\frac{dy}{dx}+\frac{y}{1+{{x}^{2}}}=\frac{{{e}^{x}}}{{{e}^{\tan -1}}x},\,y(0)=1\] \[\therefore \] \[IF={{e}^{\int{\frac{1}{1+{{x}^{2}}}\,dx}}}={{e}^{{{\tan }^{-1}}}}^{x}\] \[\therefore \] The solution is \[y\times {{e}^{{{\tan }^{-1}}x}}=\int{\frac{{{e}^{x}}}{{{e}^{{{\tan }^{-1}}x}}}}\times {{e}^{{{\tan }^{-1}}x}}dx\] \[=\int{{{e}^{x}}\,dx={{e}^{x}}+C}\] \[\Rightarrow \] \[y={{e}^{x-{{\tan }^{-1}}x}}+C{{e}^{-{{\tan }^{-1}}x}}\] \[\because \] \[y(0)=1\] \[\therefore \] \[1={{e}^{0-{{\tan }^{-1}}0}}+C{{e}^{-{{\tan }^{-1}}0}}\] \[\Rightarrow \] \[1=1+C\,\,\Rightarrow \,\,C=0\] \[\therefore \] Required equation is \[y={{e}^{x-{{\tan }^{-1}}x}}.\]
You need to login to perform this action.
You will be redirected in
3 sec
