A) \[y={{e}^{kt}}+c,\] for some constant \[c\le 0\] and \[k\ge 0\]
B) \[y=c{{e}^{kt}},\] for some constants \[c\ge 0\] and \[k\le 0\]
C) \[y={{e}^{ct}}+k,\] for some constants \[c\le 0\] and \[k\ge 0\]
D) \[y=k{{e}^{ct}},\] for some constants \[c\ge 0\] and \[k\le 0\]
Correct Answer: B
Solution :
Given that, \[\frac{dy}{dt}\propto y\] \[\Rightarrow \] \[\frac{dy}{dt}=ky\] \[\Rightarrow \] \[\frac{1}{y}dy=k\,dt\] On integrating, we get \[\log y=\log c+kt\] \[\Rightarrow \] \[\log y-\log c=kt\] \[\Rightarrow \] \[\log \frac{y}{c}=kt\] \[\Rightarrow \] \[\frac{y}{c}={{e}^{kt}}\] \[\Rightarrow \] \[y=c{{e}^{kt}}\]
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