We are given the second-order linear differential equation:

$2 \frac{d^2 s}{dt^2} + 6 \frac{ds}{dt} + 4.5s = 0$

with the initial conditions:

$s(0) = 0, \quad \frac{ds}{dt}(0) = 4.$

Step 1: Convert to a Characteristic Equation

Assume a solution of the form $s(t) = e^{rt}$. Substituting into the equation:

$2r^2 e^{rt} + 6r e^{rt} + 4.5 e^{rt} = 0.$

Dividing by $e^{rt}$ (which is never zero), we get the characteristic equation:

$2r^2 + 6r + 4.5 = 0.$

Step 2: Solve for $r$

Using the quadratic formula:

$r = \frac{-6 \pm \sqrt{6^2 - 4(2)(4.5)}}{2(2)}.$

$r = \frac{-6 \pm \sqrt{36 - 36}}{4} = \frac{-6 \pm 0}{4} = \frac{-6}{4} = -1.5.$

Since both roots are equal ($r_1 = r_2 = -1.5$), the general solution for repeated roots is:

$s(t) = (C_1 + C_2 t)e^{-1.5t}.$

Step 3: Apply Initial Conditions

First Condition: $s(0) = 0$

$0 = (C_1 + C_2(0))e^{0} \Rightarrow C_1 = 0.$

Thus, the solution simplifies to:

$s(t) = C_2 t e^{-1.5t}.$

Second Condition: $\frac{ds}{dt}(0) = 4$

Differentiate $s(t)$:

$\frac{ds}{dt} = C_2 e^{-1.5t} + C_2 t (-1.5 e^{-1.5t}).$

At $t = 0$:

$4 = C_2 e^{0} + C_2(0)(-1.5 e^{0}).$

$4 = C_2.$

Step 4: Final Solution

$s(t) = 4t e^{-1.5t}.$

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Would you like a step-by-step verification, or have any questions?

Related Questions:

1. How do repeated roots affect the general solution of a second-order differential equation?
2. What is the physical interpretation of the damping coefficient in this equation?
3. How would the solution change if the system had an external force?
4. What are the long-term behaviors of the solution?
5. How can Laplace transforms be used to solve this problem?

Tip:

For second-order differential equations with constant coefficients, always check whether the roots are real and distinct, real and repeated, or complex conjugates. This determines the form of the general solution.