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The problem presented is a second-order linear homogeneous differential equation with constant coefficients:

\[
y'' + 2y' + 17y = 0
\]

with the initial conditions:
- \( y(0) = 6 \)
- \( y'(0) = 5 \)

### Steps to solve:

1. **Solve the characteristic equation**:
   The corresponding characteristic equation for the differential equation is obtained by assuming \( y = e^{rt} \):
   \[
   r^2 + 2r + 17 = 0
   \]

2. **Find the roots**:
   Solve this quadratic equation using the quadratic formula:
   \[
   r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
   \]
   where \( a = 1 \), \( b = 2 \), and \( c = 17 \):
   \[
   r = \frac{-2 \pm \sqrt{2^2 - 4(1)(17)}}{2(1)} = \frac{-2 \pm \sqrt{4 - 68}}{2} = \frac{-2 \pm \sqrt{-64}}{2}
   \]
   \[
   r = \frac{-2 \pm 8i}{2} = -1 \pm 4i
   \]
   The roots are \( r_1 = -1 + 4i \) and \( r_2 = -1 - 4i \).

3. **Form the general solution**:
   Since the roots are complex, the general solution to the differential equation is:
   \[
   y(t) = e^{-t}(C_1 \cos(4t) + C_2 \sin(4t))
   \]
   where \( C_1 \) and \( C_2 \) are constants to be determined using the initial conditions.

4. **Apply initial conditions**:
   - Use \( y(0) = 6 \):
     \[
     y(0) = e^{0}(C_1 \cos(0) + C_2 \sin(0)) = C_1 = 6
     \]
     So, \( C_1 = 6 \).

   - Now, differentiate \( y(t) \) to find \( y'(t) \):
     \[
     y'(t) = e^{-t}(-C_1 \cos(4t) - C_2 \sin(4t)) + e^{-t}(4C_2 \cos(4t) - 4C_1 \sin(4t))
     \]
     Simplify this expression:
     \[
     y'(t) = e^{-t} \left[ (-C_1 + 4C_2) \cos(4t) + (-C_2 - 4C_1) \sin(4t) \right]
     \]

     Apply the initial condition \( y'(0) = 5 \):
     \[
     y'(0) = e^{0}((-6 + 4C_2) \cos(0) + (-C_2 - 4(6)) \sin(0)) = -6 + 4C_2 = 5
     \]
     Solving for \( C_2 \):
     \[
     -6 + 4C_2 = 5
     \]
     \[
     4C_2 = 11 \quad \Rightarrow \quad C_2 = \frac{11}{4}
     \]

5. **Final solution**:
   The solution for \( y(t) \) is:
   \[
   y(t) = e^{-t} \left( 6 \cos(4t) + \frac{11}{4} \sin(4t) \right)
   \]

This is the required solution for \( y \) as a function of \( t \).

---

Would you like further clarification on any of the steps or details?

Here are 5 related questions to deepen your understanding:

1. How would the solution change if the roots of the characteristic equation were real?
2. How does the method of undetermined coefficients differ from the one used here?
3. What would happen to the solution if the initial conditions were different?
4. Can this type of differential equation be solved using a numerical approach, and if so, how?
5. What is the physical interpretation of complex roots in differential equations in terms of oscillations?

**Tip**: When solving differential equations with complex roots, the real part of the exponent describes the damping (decay) effect, while the imaginary part is related to oscillations.

Find y as a function of t if y'' + 2y' + 17y = 0, y(0) = 6, y'(0) = 5.

Learn how to solve the second-order homogeneous differential equation y'' + 2y' + 17y = 0 with given initial conditions y(0) = 6 and y'(0) = 5. Explore the use of the characteristic equation and complex roots to find the solution.

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