Solve the given initial value problem.
y double prime plus 6 y prime plus 34 y equals 0​;   ​y(0)equals2​,   y prime left parenthesis 0 right parenthesisequalsminus2

To solve the given initial value problem:

\[
y'' + 6y' + 34y = 0, \quad y(0) = 2, \quad y'(0) = -2
\]

Let's solve this second-order linear homogeneous differential equation step-by-step.

### Step 1: Find the Characteristic Equation
The given differential equation is:

\[
y'' + 6y' + 34y = 0
\]

To solve this, we first find the characteristic equation by replacing \( y'' \) with \( r^2 \), \( y' \) with \( r \), and \( y \) with 1:

\[
r^2 + 6r + 34 = 0
\]

### Step 2: Solve the Characteristic Equation
To solve the quadratic equation, we use the quadratic formula:

\[
r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
\]

where \( a = 1 \), \( b = 6 \), and \( c = 34 \):

\[
r = \frac{-6 \pm \sqrt{6^2 - 4 \cdot 1 \cdot 34}}{2 \cdot 1}
\]

\[
r = \frac{-6 \pm \sqrt{36 - 136}}{2}
\]

\[
r = \frac{-6 \pm \sqrt{-100}}{2}
\]

\[
r = \frac{-6 \pm 10i}{2}
\]

\[
r = -3 \pm 5i
\]

### Step 3: General Solution
The roots are complex: \( r = -3 + 5i \) and \( r = -3 - 5i \). For complex roots of the form \( \alpha \pm \beta i \), the general solution is:

\[
y(t) = e^{\alpha t} \left( C_1 \cos(\beta t) + C_2 \sin(\beta t) \right)
\]

In this case, \( \alpha = -3 \) and \( \beta = 5 \), so the general solution is:

\[
y(t) = e^{-3t} \left( C_1 \cos(5t) + C_2 \sin(5t) \right)
\]

### Step 4: Apply Initial Conditions
We have the initial conditions \( y(0) = 2 \) and \( y'(0) = -2 \). 

1. **Use \( y(0) = 2 \):**

\[
y(0) = e^{-3 \cdot 0} \left( C_1 \cos(5 \cdot 0) + C_2 \sin(5 \cdot 0) \right) = C_1
\]

Since \( y(0) = 2 \):

\[
C_1 = 2
\]

2. **Find \( y'(t) \) and use \( y'(0) = -2 \):**

First, find the derivative of \( y(t) \):

\[
y(t) = e^{-3t} \left( 2 \cos(5t) + C_2 \sin(5t) \right)
\]

Using the product rule:

\[
y'(t) = \frac{d}{dt} \left( e^{-3t} \right) \cdot \left( 2 \cos(5t) + C_2 \sin(5t) \right) + e^{-3t} \cdot \frac{d}{dt} \left( 2 \cos(5t) + C_2 \sin(5t) \right)
\]

\[
y'(t) = (-3e^{-3t}) \left( 2 \cos(5t) + C_2 \sin(5t) \right) + e^{-3t} \left( -10 \sin(5t) + 5 C_2 \cos(5t) \right)
\]

Simplify the expression for \( y'(t) \):

\[
y'(t) = e^{-3t} \left[ -6 \cos(5t) - 3 C_2 \sin(5t) - 10 \sin(5t) + 5 C_2 \cos(5t) \right]
\]

\[
y'(t) = e^{-3t} \left[ (5 C_2 - 6) \cos(5t) + (-3 C_2 - 10) \sin(5t) \right]
\]

Now use the initial condition \( y'(0) = -2 \):

\[
y'(0) = e^{-3 \cdot 0} \left[ (5 C_2 - 6) \cos(5 \cdot 0) + (-3 C_2 - 10) \sin(

Solve the given initial value problem: y'' + 6y' + 34y = 0, y(0) = 2, y'(0) = -2

This page provides a step-by-step solution to the second-order linear homogeneous differential equation y'' + 6y' + 34y = 0 with initial conditions y(0) = 2 and y'(0) = -2. Learn about characteristic equations, solving quadratic equations, and applying initial conditions for complex roots. Suitable for undergraduate-level math students.

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