A) \[y=\frac{1-{{e}^{-2x}}}{2}\]
B) \[y=\frac{1+{{e}^{-2x}}}{2}\]
C) \[y=1+{{e}^{x}}\]
D) \[y=\frac{1+{{e}^{x}}}{2}\]
Correct Answer: A
Solution :
[a] \[\frac{dy}{dx}+2y=1;\frac{dy}{dx}=1-2y\] \[\int{\frac{dy}{1-2y}=\int{dx}}\] \[-\frac{1}{2}\log \left| 1-2y \right|=x+C\] at \[x=0,\text{ }y=0\] \[-\frac{1}{2}\log 1=0+C\Rightarrow C=0\] \[1-2y={{e}^{-2x}};y=\frac{1-{{e}^{-2x}}}{2}\]You need to login to perform this action.
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