Chapter 9: First-Order Differential Equations - Section 9.2 - First-Order Linear Equations - Exercises 9.2 - Page 537: 21

$y=y_0e^{kt}$

Here, we have $y'-ky=0$ The integrating factor is: $I=e^{\int -k dt}=e^{-kt}$ Now, $e^{-kt}[y'-ky]=0$ This gives: $\int [ye^{-kt}]'=\int e^{-kt} dy-\int e^{-kt} ky dt $ $\implies y=ce^{(kt)}$ After applying the initial conditions, we have $ c=y_0$ So, $y=ce^{(kt)} \implies y=y_0e^{kt}$