\( y ^ { \prime \prime } - y ^ { \prime } - 6 y = 0 \quad y ( 0 ) = 1 \quad y ^ { \prime } ( 0 ) = 2 \)

To solve the second-order linear homogeneous differential equation \( y'' - y' - 6y = 0 \), we first find the characteristic equation, which is \( r^2 - r - 6 = 0 \). Factoring this gives us \( (r - 3)(r + 2) = 0 \), yielding the roots \( r_1 = 3 \) and \( r_2 = -2 \). Thus, the general solution is \( y(t) = C_1 e^{3t} + C_2 e^{-2t} \). Next, we use the initial conditions \( y(0) = 1 \) and \( y'(0) = 2 \) to find the constants \( C_1 \) and \( C_2 \). Substituting \( t = 0 \) gives us \( C_1 + C_2 = 1 \). Taking the derivative, \( y'(t) = 3C_1 e^{3t} - 2C_2 e^{-2t} \), and substituting \( t = 0 \) gives \( 3C_1 - 2C_2 = 2 \). Solving this system leads to \( C_1 = \frac{4}{5} \) and \( C_2 = \frac{1}{5} \). Therefore, the final solution is \( y(t) = \frac{4}{5} e^{3t} + \frac{1}{5} e^{-2t} \). When analyzing differential equations, it’s useful to remember that initial conditions help tailor the general solution to a specific problem. A common pitfall is overlooking the steps involved in solving for constants—make sure to substitute both initial conditions carefully to avoid miscalculations! If you want to dive deeper into this topic, consider reading about the applications of differential equations in real-world scenarios like population modeling, electrical circuits, or mechanical systems. You'll see how these equations not only describe motion but also impact various fields like engineering and biology!