https://file.aiquickdraw.com/mathbot/file/17285171542109iuumapd.jpg

It seems that the problem you're solving is a second-order linear differential equation, given as:

\[
250y'' - 7y' + 14y = 0
\]

with initial conditions:

\[
y(0) = 7, \quad y'(0) = 8.
\]

The solution you entered, as shown in the image, involves a combination of exponential, trigonometric functions, and constants derived from solving the characteristic equation.

Let’s go step by step to identify where the issue might lie.

### Step 1: Find the characteristic equation
The characteristic equation for a differential equation of the form \( ay'' + by' + cy = 0 \) is given by:

\[
250r^2 - 7r + 14 = 0.
\]

This is a quadratic equation, which we can solve using the quadratic formula:

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

where \(a = 250\), \(b = -7\), and \(c = 14\). Substituting the values:

\[
r = \frac{-(-7) \pm \sqrt{(-7)^2 - 4(250)(14)}}{2(250)}
\]
\[
r = \frac{7 \pm \sqrt{49 - 14000}}{500}
\]
\[
r = \frac{7 \pm \sqrt{-13951}}{500}.
\]

Since the discriminant is negative, the roots will be complex. Let’s express the roots as:

\[
r = \frac{7}{500} \pm i \frac{\sqrt{13951}}{500}.
\]

### Step 2: General solution form
The general solution to the differential equation will be of the form:

\[
y(t) = e^{\frac{7}{500} t} \left( C_1 \cos\left(\frac{\sqrt{13951}}{500} t \right) + C_2 \sin\left(\frac{\sqrt{13951}}{500} t \right) \right).
\]

### Step 3: Apply initial conditions
Using \( y(0) = 7 \) and \( y'(0) = 8 \), we can determine the constants \( C_1 \) and \( C_2 \).

1. **At \( t = 0 \), the solution becomes:**

   \[
   y(0) = e^0 \left( C_1 \cos(0) + C_2 \sin(0) \right) = C_1 = 7.
   \]

   So, \( C_1 = 7 \).

2. **Now, differentiate \( y(t) \) to find \( y'(t) \):**

   \[
   y'(t) = \frac{7}{500} e^{\frac{7}{500}t} \left( 7 \cos\left(\frac{\sqrt{13951}}{500} t\right) + C_2 \sin\left(\frac{\sqrt{13951}}{500} t\right)\right)
   + e^{\frac{7}{500}t} \left(-7 \sin\left(\frac{\sqrt{13951}}{500} t\right) + C_2 \cos\left(\frac{\sqrt{13951}}{500} t\right)\right).
   \]

   At \( t = 0 \):

   \[
   y'(0) = \frac{7}{500} \cdot 7 + C_2 \cdot \frac{\sqrt{13951}}{500}.
   \]

   Given that \( y'(0) = 8 \), we can solve:

   \[
   \frac{49}{500} + C_2 \cdot \frac{\sqrt{13951}}{500} = 8.
   \]
   
   Multiply the entire equation by 500 to simplify:

   \[
   49 + C_2 \cdot \sqrt{13951} = 4000,
   \]
   \[
   C_2 \cdot \sqrt{13951} = 3951,
   \]
   \[
   C_2 = \frac{3951}{\sqrt{13951}}.
   \]

### Step 4: Final solution
Substituting the values of \( C_1 \) and \( C_2 \) into the general solution, we get:

\[
y(t) = 7 e^{\frac{7}{500}t} \cos\left(\frac{\sqrt{13951}}{500} t \right) + \frac{3951}{500 \sqrt{13951}} e^{\frac{7}{500}t} \sin\left(\frac{\sqrt{13951}}{500} t \right).
\]

This matches the structure of your answer. However, it seems there might be a minor formatting or simplification issue in how the final form was submitted, which could be why the platform marked it incorrect.

Would you like to go over any of the steps in more detail, or need help simplifying the expression?

Here are 5 related questions to deepen your understanding:
1. How do you interpret the role of complex roots in solving differential equations?
2. What happens to the solution if the discriminant of the characteristic equation is positive or zero?
3. How do initial conditions influence the constants in the solution?
4. Can you explain the behavior of the solution as \( t \to \infty \)?
5. How would the solution change if the coefficient of \( y' \) was positive instead of negative?

**Tip:** Always check whether the final form of the solution is simplified and matches the expected format on your platform.

Find y as a function of t if 250y'' - 7y' + 14y = 0, with initial conditions y(0) = 7 and y'(0) = 8.

Learn how to solve the second-order linear differential equation 250y'' - 7y' + 14y = 0 with initial conditions y(0) = 7 and y'(0) = 8. This problem covers solving characteristic equations, identifying complex roots, and applying initial conditions to find the constants. Suitable for undergraduate students.

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