A) \[y=c{{e}^{x-{{x}^{2}}}}\]
B) \[y=c{{e}^{{{x}^{2}}-x}}\]
C) \[y=c{{e}^{x}}\]
D) \[y=c{{e}^{-{{x}^{2}}}}\]
Correct Answer: A
Solution :
\[\frac{dy}{dx}+2xy=y\] \[\Rightarrow \] \[\frac{dy}{dx}+(2y-1)y=0\] \[\therefore \] \[IF={{e}^{\int_{{}}^{{}}{(2x-1)dx}}}\] \[={{e}^{{{x}^{2}}-x}}\] \[\therefore \]The solution is \[y\,{{e}^{{{x}^{2}}-x}}=c\] \[\Rightarrow \] \[y=c{{e}^{x-{{x}^{2}}}}\] Alternate solution: \[\frac{dy}{dx}+2xy=y\] \[\Rightarrow \] \[dy+y(2x-1)dx=0\] \[\Rightarrow \] \[\frac{1}{y}dx+(2x-1)dx=0\] On integrating both sides, we get \[\log \,y+({{x}^{2}}-x)=\log \,c\] \[\Rightarrow \] \[\log \frac{c}{y}={{x}^{2}}-x\] \[\Rightarrow \] \[c=y{{e}^{x-{{x}^{2}}}}\] \[\Rightarrow \] \[y=c{{e}^{x-{{x}^{2}}}}\]
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