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The image contains an initial value problem of a third-order linear differential equation with constant coefficients, and the task is to find the solution given the initial conditions.

### The differential equation is:
\[
y^{(3)} + 9y'' + 49y' + 21y = 0
\]

### The general solution is given as a combination of exponential and trigonometric functions:
\[
y(x) = c_1 e^{-4x} \cos(-5x) + c_2 e^{-4x} \sin(-5x) + c_3 e^{1x}
\]

### Initial conditions:
- \( y(0) = 4 \)
- \( y'(0) = 24 \)
- \( y''(0) = -106 \)

### Step-by-Step Solution:

1. **General solution**:
   \[
   y(x) = c_1 e^{-4x} \cos(5x) + c_2 e^{-4x} \sin(5x) + c_3 e^{x}
   \]
   The solution involves three constants \( c_1 \), \( c_2 \), and \( c_3 \) that we need to determine using the initial conditions.

2. **Find \( y(0) \)**:
   \[
   y(0) = c_1 e^{0} \cos(0) + c_2 e^{0} \sin(0) + c_3 e^{0}
   = c_1 + c_3
   \]
   Using \( y(0) = 4 \), we get:
   \[
   c_1 + c_3 = 4 \quad \text{(Equation 1)}
   \]

3. **Find \( y'(x) \)**:
   Differentiate the general solution:
   \[
   y'(x) = -4c_1 e^{-4x} \cos(5x) - 5c_1 e^{-4x} \sin(5x) 
   + (-4c_2 e^{-4x} \sin(5x) + 5c_2 e^{-4x} \cos(5x)) 
   + c_3 e^{x}
   \]
   At \( x = 0 \):
   \[
   y'(0) = -4c_1 + 5c_2 + c_3
   \]
   Using \( y'(0) = 24 \), we get:
   \[
   -4c_1 + 5c_2 + c_3 = 24 \quad \text{(Equation 2)}
   \]

4. **Find \( y''(x) \)**:
   Differentiate \( y'(x) \) again:
   \[
   y''(x) = 16c_1 e^{-4x} \cos(5x) + 40c_1 e^{-4x} \sin(5x)
   + (-16c_2 e^{-4x} \sin(5x) - 40c_2 e^{-4x} \cos(5x)) + c_3 e^{x}
   \]
   At \( x = 0 \):
   \[
   y''(0) = 16c_1 - 40c_2 + c_3
   \]
   Using \( y''(0) = -106 \), we get:
   \[
   16c_1 - 40c_2 + c_3 = -106 \quad \text{(Equation 3)}
   \]

5. **Solving the system of equations**:
   The three equations are:
   1. \( c_1 + c_3 = 4 \)
   2. \( -4c_1 + 5c_2 + c_3 = 24 \)
   3. \( 16c_1 - 40c_2 + c_3 = -106 \)

   Solving this system yields the constants:
   - \( c_1 = 1 \)
   - \( c_2 = -3 \)
   - \( c_3 = 5 \)

6. **Final solution**:
   Substituting these values into the general solution:
   \[
   y(x) = e^{-4x} \cos(5x) - 3 e^{-4x} \sin(5x) + 5 e^{x}
   \]

---

Would you like further details on the solution steps, or do you have any specific questions?

Here are five related questions for deeper understanding:
1. What method is used to solve linear differential equations with constant coefficients?
2. How do you apply initial conditions to find unknown constants in a solution?
3. Can the solution be verified by substituting it back into the original differential equation?
4. What is the significance of the exponential and trigonometric terms in the solution?
5. How would the solution change if the initial conditions were different?

**Tip**: When solving differential equations, it's helpful to double-check each derivative to ensure that no steps are skipped in applying initial conditions.

Solve the initial value problem y(3) + 9y'' + 49y' + 21y = 0 with the solution set {e^(-4x)cos(-5x), e^(-4x)sin(-5x), e^(1x)} given y(0) = 4, y'(0) = 24, and y''(0) = -106.

Learn how to solve a third-order linear differential equation with initial conditions. This solution covers key concepts such as exponential and trigonometric functions, and applying initial conditions to solve for constants. Suitable for college-level mathematics.

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