Question

Solve the second-order differential equation: \[ y'' + 0.8 y' + 0.16 y = 0, \quad y(0) = 3, \quad y'(0) = 4.5 \]

Hint:

For solving second-order differential equations with constant coefficients, always find the characteristic equation first. The roots of the characteristic equation give you the form of the general solution.

Answer 1

This is a second-order linear homogeneous differential equation. We first find the characteristic equation: \[ r^2 + 0.8r + 0.16 = 0 \] Solving this quadratic equation using the quadratic formula: \[ r = \frac{-0.8 \pm \sqrt{(0.8)^2 - 4(1)(0.16)}}{2(1)} \] \[ r = \frac{-0.8 \pm \sqrt{0.64 - 0.64}}{2} = \frac{-0.8 \pm 0}{2} \] Thus, we have a repeated root: \[ r = -0.4 \] The general solution to the differential equation is: \[ y(t) = (C_1 + C_2 t) e^{-0.4t} \] Using the initial conditions \( y(0) = 3 \) and \( y'(0) = 4.5 \): 1. \( y(0) = 3 \) gives: \[ C_1 = 3 \] 2. \( y'(t) = C_2 e^{-0.4t} - 0.4(C_1 + C_2 t) e^{-0.4t} \) and \( y'(0) = 4.5 \) gives: \[ C_2 = 4.5 \] Thus, the solution is: \[ y(t) = (3 + 4.5t) e^{-0.4t} \] Now, to find \( y(1) \): \[ y(1) = (3 + 4.5 \times 1) e^{-0.4} = (3 + 4.5) e^{-0.4} \approx 7.5 \times 0.6703 = 5.03 \] Thus, the final answer is: \[ y(1) \approx \boxed{5.03} \]