4. (2 points) Solve the following initial value problem: $$ \frac{d^{2} y}{d t^{2}}+10 \frac{d y}{d t}+25 y=0, \quad y(0)=2, \quad y^{\prime}(0)=-4 $$

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We start with the differential equation

$$
\frac{d^2 y}{d t^2} + 10 \frac{d y}{d t} + 25 y = 0.
$$

**Step 1.** Write the characteristic equation by substituting $$ y = e^{rt} $$:

$$
r^2 + 10r + 25 = 0.
$$

**Step 2.** Factor the quadratic:

$$
(r + 5)^2 = 0.
$$

Thus, we have a repeated root

$$
r = -5.
$$

**Step 3.** Write the general solution when there is a double root:

$$
y(t) = C_1 e^{-5t} + C_2 t e^{-5t}.
$$

**Step 4.** Apply the initial condition $$ y(0)=2 $$:

$$
y(0) = C_1 e^0 + C_2 \cdot 0 \cdot e^0 = C_1 = 2.
$$

So,

$$
C_1 = 2.
$$

**Step 5.** Differentiate $$ y(t) $$ to find $$ y'(t) $$:

$$
y(t) = C_1 e^{-5t} + C_2 t e^{-5t}.
$$

Differentiate using the product rule on the second term:

$$
y'(t) = -5C_1 e^{-5t} + C_2 e^{-5t} - 5C_2 t e^{-5t}.
$$

Apply the condition $$ y'(0) = -4 $$:

$$
y'(0) = -5C_1 + C_2 = -4.
$$

Substitute $$ C_1 = 2 $$:

$$
-5(2) + C_2 = -4 \quad \Rightarrow \quad -10 + C_2 = -4.
$$

Solve for $$ C_2 $$:

$$
C_2 = 6.
$$

**Final Answer:** The solution to the initial value problem is

$$
y(t) = 2 e^{-5t} + 6t e^{-5t} = e^{-5t} (2 + 6t).
$$
The solution to the initial value problem is $$ y(t) = e^{-5t} (2 + 6t) $$.

(2 points) Solve the following initial value problem:

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