A) \[y={{e}^{-x}}(x-1)\]
B) \[y=x{{e}^{x}}\]
C) \[y=x{{e}^{-x}}+1\]
D) \[y=x{{e}^{-x}}\]
Correct Answer: D
Solution :
\[\frac{dy}{dx}+y={{e}^{-x}}\]; I.F. \[={{e}^{\int{dx}}}={{e}^{x}}\] \\[y{{e}^{x}}=\int{{{e}^{-x}}.{{e}^{x}}dx+c}\] Þ \[y{{e}^{x}}=x+c\] Since \[y(0)=0\], \ \[c=0\] Hence, the required solution is \[y{{e}^{x}}=x\] Þ \[y=x{{e}^{-x}}\].You need to login to perform this action.
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