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The problem you're trying to solve is an initial value problem (IVP) of a second-order linear differential equation. The equation is:

\[
16y'' - 24y' + 9y = 0
\]
with the initial conditions:
\[
y(0) = 3, \quad y'(0) = 5.
\]

### Step-by-Step Solution:

1. **Characteristic Equation:**
   The differential equation is linear with constant coefficients. First, we solve the characteristic equation by assuming \( y = e^{rt} \), which gives:
   \[
   16r^2 - 24r + 9 = 0.
   \]

2. **Solve the Quadratic Equation:**
   Solving for \( r \) using the quadratic formula:
   \[
   r = \frac{-(-24) \pm \sqrt{(-24)^2 - 4(16)(9)}}{2(16)} = \frac{24 \pm \sqrt{576 - 576}}{32} = \frac{24 \pm 0}{32} = \frac{24}{32} = \frac{3}{4}.
   \]
   So, the roots are real and repeated: \( r = \frac{3}{4} \).

3. **General Solution:**
   Since the characteristic equation has a repeated root \( r = \frac{3}{4} \), the general solution takes the form:
   \[
   y(t) = (C_1 + C_2t)e^{\frac{3}{4}t}.
   \]

4. **Use Initial Conditions:**
   Now, apply the initial conditions to solve for \( C_1 \) and \( C_2 \).
   
   - First, use \( y(0) = 3 \):
     \[
     y(0) = C_1e^{0} = C_1 = 3.
     \]
     So, \( C_1 = 3 \).
   
   - Next, use \( y'(0) = 5 \). First, compute \( y'(t) \):
     \[
     y'(t) = \left(C_1 + C_2t\right)\frac{3}{4}e^{\frac{3}{4}t} + C_2e^{\frac{3}{4}t} = \left( \frac{3}{4}(C_1 + C_2t) + C_2 \right)e^{\frac{3}{4}t}.
     \]
     Now, substitute \( t = 0 \) and \( C_1 = 3 \) into this:
     \[
     y'(0) = \left( \frac{3}{4}(3) + C_2 \right)e^{0} = \frac{9}{4} + C_2 = 5.
     \]
     Solving for \( C_2 \):
     \[
     C_2 = 5 - \frac{9}{4} = \frac{20}{4} - \frac{9}{4} = \frac{11}{4}.
     \]

5. **Final Solution:**
   The solution to the differential equation is:
   \[
   y(t) = \left(3 + \frac{11}{4}t\right)e^{\frac{3}{4}t}.
   \]

### Behavior of the Solution:
The solution is of the form \( (3 + \frac{11}{4}t)e^{\frac{3}{4}t} \), which involves exponential growth because of the positive exponential term \( e^{\frac{3}{4}t} \). Therefore, the correct option from the list provided would be:

- **Increasing without bounds**.

---

Would you like any further details or have any questions?

Here are 5 related questions to expand your understanding:
1. What is the significance of repeated roots in solving characteristic equations?
2. How does the nature of the roots affect the behavior of the solution (real, complex, or repeated)?
3. What would the solution look like if the characteristic equation had complex roots?
4. How would the solution change if the initial conditions were different?
5. Can you solve a non-homogeneous version of this problem?

**Tip:** When solving differential equations, always check the form of the characteristic equation, as it helps determine the general solution structure.

Solve the initial value problem 16y'' - 24y' + 9y = 0, y(0) = 3, y'(0) = 5.

Learn how to solve the initial value problem 16y'' - 24y' + 9y = 0 with given conditions y(0) = 3 and y'(0) = 5. This detailed solution covers solving the characteristic equation, determining the general solution, and applying initial conditions. Suitable for college-level students studying differential equations.

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