Find the solution to \[ \frac{d y}{d t}=9 y \] satisfying \[ y(3)=9 \]

The equation you provided is a first-order linear differential equation. To solve it, we can separate variables: \[ \frac{1}{y} \frac{d y}{d t} = 9. \] Integrating both sides gives us: \[ \int \frac{1}{y} \, d y = \int 9 \, d t, \] which results in: \[ \ln |y| = 9t + C, \] where \( C \) is the constant of integration. Exponentiating both sides leads to: \[ |y| = e^{9t+C} = e^{C} e^{9t}. \] Let \( k = e^{C} \), which can be any positive constant. Thus, we have: \[ y = k e^{9t}. \] To find the constant \( k \), we use the initial condition \( y(3) = 9 \): \[ 9 = k e^{9 \cdot 3} \quad \Rightarrow \quad 9 = k e^{27}. \] So, we can solve for \( k \): \[ k = \frac{9}{e^{27}}. \] Thus, the particular solution is: \[ y(t) = \frac{9}{e^{27}} e^{9t} = 9 e^{9t - 27}. \] In conclusion, the solution satisfying the initial condition is: \[ \boxed{y(t) = 9 e^{9t - 27}}. \]