3. (2 points) Solve the following initial value problem: $$ \frac{d^{2} y}{d t^{2}}+10 \frac{d y}{d t}+24 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-4 $$

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We start by writing the differential equation and its initial conditions:

$$
\frac{d^{2} y}{d t^{2}} + 10 \frac{d y}{d t} + 24 y = 0, \quad y(0)=2,\quad y'(0)=-4.
$$

**Step 1. Find the characteristic equation.**

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

$$
r^2 e^{rt} + 10r e^{rt} + 24 e^{rt} = 0.
$$

Since $$ e^{rt} \neq 0 $$, we have:

$$
r^2 + 10r + 24 = 0.
$$

**Step 2. Solve the characteristic equation.**

Factor the quadratic equation:

$$
r^2 + 10r + 24 = (r+4)(r+6) = 0.
$$

Thus, the roots are:

$$
r = -4 \quad \text{and} \quad r = -6.
$$

**Step 3. Write the general solution.**

Since the characteristic roots are real and distinct, the general solution to the differential equation is:

$$
y(t) = C_1 e^{-4t} + C_2 e^{-6t},
$$

where $$ C_1 $$ and $$ C_2 $$ are constants to be determined by the initial conditions.

**Step 4. Apply the initial conditions.**

1. Using $$ y(0)=2 $$:

$$
y(0) = C_1 e^{0} + C_2 e^{0} = C_1 + C_2 = 2.
$$

2. Find the expression for $$ y'(t) $$ and apply $$ y'(0)=-4 $$:

Differentiate $$ y(t) $$:

$$
y'(t) = -4C_1 e^{-4t} -6C_2 e^{-6t}.
$$

Evaluate at $$ t=0 $$:

$$
y'(0) = -4C_1 - 6C_2 = -4.
$$

Now we have the system of equations:

$$
\begin{cases}
C_1 + C_2 = 2, \
-4C_1 - 6C_2 = -4.
\end{cases}
$$

**Step 5. Solve for $$ C_1 $$ and $$ C_2 $$.**

Solve the first equation for $$ C_1 $$:

$$
C_1 = 2 - C_2.
$$

Substitute into the second equation:

$$
-4(2 - C_2) - 6C_2 = -4.
$$

Simplify:

$$
-8 + 4C_2 - 6C_2 = -4 \quad \Longrightarrow \quad -8 - 2C_2 = -4.
$$

Add 8 to both sides:

$$
-2C_2 = 4 \quad \Longrightarrow \quad C_2 = -2.
$$

Then, substitute back to find $$ C_1 $$:

$$
C_1 = 2 - (-2) = 4.
$$

**Step 6. Write the final solution.**

Substitute $$ C_1 $$ and $$ C_2 $$ into the general solution:

$$
y(t) = 4 e^{-4t} - 2 e^{-6t}.
$$

This is the solution to the initial value problem.
The solution to the initial value problem is: $$ y(t) = 4 e^{-4t} - 2 e^{-6t}. $$

(2 points) Solve the following initial value problem:

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