7. Solve the initial value problem \( y^{\prime}=-3 y+1, y(0)=1 \). \[ \begin{array}{l}\text { A. } y(t)=\frac{1}{3} e^{-3 t}+\frac{1}{3} \\ \text { B. } y(t)=\frac{3}{2} e^{2 t}-\frac{3}{2} \\ \text { C. } y(t)=\frac{2}{3} e^{-3 t}+\frac{1}{3} \\ \text { D. } y(t)=\frac{5}{2} e^{-3 t}-\frac{3}{2}\end{array} \]

To solve the initial value problem \( y' = -3y + 1 \) with the condition \( y(0) = 1 \), we can use the method of integrating factors. The equation is a first-order linear ordinary differential equation. First, we rewrite the equation in standard form \( y' + 3y = 1 \). The integrating factor is \( e^{\int 3 \, dt} = e^{3t} \). Multiplying both sides by the integrating factor gives: \[ e^{3t} y' + 3e^{3t} y = e^{3t}. \] This simplifies to: \[ \frac{d}{dt}(e^{3t} y) = e^{3t}. \] Integrating both sides gives: \[ e^{3t} y = \frac{1}{3} e^{3t} + C, \] where \( C \) is a constant. Dividing through by \( e^{3t} \): \[ y = \frac{1}{3} + Ce^{-3t}. \] Using the initial condition \( y(0) = 1 \): \[ 1 = \frac{1}{3} + C \Rightarrow C = \frac{2}{3}. \] Thus, the solution is: \[ y(t) = \frac{2}{3} e^{-3t} + \frac{1}{3}. \] So, the answer is: **C. \( y(t) = \frac{2}{3} e^{-3t} + \frac{1}{3}. \)** --- Did you know that the study of differential equations dates back to the days of Isaac Newton and Gottfried Wilhelm Leibniz? These equations are fundamental in modeling real-world phenomena like population growth, cooling of objects, and the motion of pendulums! If you're interested in diving deeper into the world of differential equations, consider reading "Elementary Differential Equations and Boundary Value Problems" by William E. Boyce and Richard C. DiPrima. It’s a classic that balances theory with practical applications, making it a great resource for anyone serious about mastering this topic!