https://file.aiquickdraw.com/mathbot/file/1727742487251tn2k9kar.jpg

The problem presents a second-order linear differential equation:

\[
2y'' + 21y' + 54y = 0
\]
with initial conditions \( y(0) = 0 \) and \( y'(0) = 1 \).

### Step-by-step Solution:

1. **Convert the equation into characteristic form:**

   The characteristic equation is obtained by assuming the solution is of the form \( y(t) = e^{rt} \), where \( r \) is a constant. Substituting this into the differential equation gives:
   \[
   2r^2 + 21r + 54 = 0
   \]
   Solve the quadratic equation using the quadratic formula:
   \[
   r = \frac{-21 \pm \sqrt{21^2 - 4(2)(54)}}{2(2)}
   \]
   \[
   r = \frac{-21 \pm \sqrt{441 - 432}}{4}
   \]
   \[
   r = \frac{-21 \pm \sqrt{9}}{4}
   \]
   \[
   r = \frac{-21 \pm 3}{4}
   \]
   So, the roots are:
   \[
   r_1 = \frac{-21 + 3}{4} = -\frac{9}{2}, \quad r_2 = \frac{-21 - 3}{4} = -6
   \]

2. **General Solution:**

   Since the roots are real and distinct, the general solution is:
   \[
   y(t) = C_1 e^{r_1 t} + C_2 e^{r_2 t}
   \]
   Substituting the values of \( r_1 \) and \( r_2 \):
   \[
   y(t) = C_1 e^{-\frac{9}{2} t} + C_2 e^{-6t}
   \]

3. **Apply Initial Conditions:**

   - **Initial Condition 1:** \( y(0) = 0 \)
     \[
     0 = C_1 e^0 + C_2 e^0
     \]
     \[
     C_1 + C_2 = 0 \quad \Rightarrow \quad C_1 = -C_2
     \]

   - **Initial Condition 2:** \( y'(0) = 1 \)
   
     First, differentiate \( y(t) \):
     \[
     y'(t) = -\frac{9}{2} C_1 e^{-\frac{9}{2} t} - 6 C_2 e^{-6t}
     \]
     Substituting \( t = 0 \):
     \[
     y'(0) = -\frac{9}{2} C_1 - 6 C_2 = 1
     \]
     Using \( C_1 = -C_2 \), substitute into the above equation:
     \[
     -\frac{9}{2} (-C_2) - 6 C_2 = 1
     \]
     \[
     \frac{9}{2} C_2 - 6 C_2 = 1
     \]
     Multiply the entire equation by 2 to eliminate fractions:
     \[
     9 C_2 - 12 C_2 = 2
     \]
     \[
     -3 C_2 = 2 \quad \Rightarrow \quad C_2 = -\frac{2}{3}
     \]
     Thus, \( C_1 = \frac{2}{3} \).

4. **Final Solution:**

   Substituting the values of \( C_1 \) and \( C_2 \) into the general solution:
   \[
   y(t) = \frac{2}{3} e^{-\frac{9}{2} t} - \frac{2}{3} e^{-6t}
   \]

This corresponds to the third option in the image:
\[
y(t) = \frac{2}{3} e^{-\frac{9}{2} t} - \frac{2}{3} e^{-6t}
\]

### Would you like more details or have any questions?

Here are 5 related questions to expand on this topic:
1. How would the solution change if the roots of the characteristic equation were complex?
2. What is the significance of the initial conditions in solving differential equations?
3. Can you derive the characteristic equation for higher-order differential equations?
4. How do you handle repeated roots in the characteristic equation?
5. What methods can be used to solve non-homogeneous second-order differential equations?

**Tip:** Always check the signs of the characteristic roots carefully, as they affect the growth or decay behavior of the solution.

Solve 2y'' + 21y' + 54y = 0, y(0) = 0, y'(0) = 1.

This problem involves solving the second-order differential equation 2y'' + 21y' + 54y = 0 with initial conditions y(0) = 0 and y'(0) = 1. Learn how to solve it step-by-step using the characteristic equation method.

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