A) \[y=1+c\,{{e}^{-x}}\]
B) \[y=1-c\,{{e}^{-x}}\]
C) \[y=x+c\,{{e}^{-x}}\]
D) \[y=x-c\,{{e}^{-x}}\]
Correct Answer: A
Solution :
\[\frac{dy}{dx}+y=1\]; I.F. \[={{e}^{\int{Pdx}}}={{e}^{\int{dx}}}={{e}^{x}}\] Hence solution is \[y.{{e}^{x}}=\int{{{e}^{x}}dx+c}\] \[y{{e}^{x}}={{e}^{x}}+c\] Þ \[y=1+c{{e}^{-x}}\].You need to login to perform this action.
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