Solve the given initial value problem y''+2y'+5y=0 y(0)=3 y'(0)=-2

To solve the initial value problem (IVP) given by the differential equation \( y'' + 2y' + 5y = 0 \) with initial conditions \( y(0) = 3 \) and \( y'(0) = -2 \), follow these steps:

1. **Find the characteristic equation**:
   The characteristic equation for the differential equation \( y'' + 2y' + 5y = 0 \) is obtained by assuming a solution of the form \( y = e^{rt} \). This gives:
   \[
   r^2 + 2r + 5 = 0
   \]

2. **Solve the characteristic equation**:
   The characteristic equation \( r^2 + 2r + 5 = 0 \) is a quadratic equation. Solve for \( r \) using the quadratic formula \( r = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a} \):
   \[
   r = \frac{-2 \pm \sqrt{4 - 20}}{2} = \frac{-2 \pm \sqrt{-16}}{2} = \frac{-2 \pm 4i}{2} = -1 \pm 2i
   \]
   So, the roots are \( r_1 = -1 + 2i \) and \( r_2 = -1 - 2i \).

3. **Form the general solution**:
   Since the roots are complex, the general solution to the differential equation is:
   \[
   y(t) = e^{-t}(C_1 \cos(2t) + C_2 \sin(2t))
   \]

4. **Apply initial conditions to determine constants \( C_1 \) and \( C_2 \)**:
   - Use \( y(0) = 3 \):
     \[
     y(0) = e^{0}(C_1 \cos(0) + C_2 \sin(0)) = C_1 = 3
     \]

   - Use \( y'(0) = -2 \):
     First, find \( y'(t) \):
     \[
     y'(t) = \frac{d}{dt} \left( e^{-t} (3 \cos(2t) + C_2 \sin(2t)) \right)
     \]
     Using the product rule, \( y'(t) \) becomes:
     \[
     y'(t) = e^{-t} \left( -3 \cos(2t) - 2 \cdot 3 \sin(2t) + C_2 (2 \cos(2t) - \sin(2t)) \right)
     \]
     \[
     y'(t) = e^{-t} \left( -3 \cos(2t) - 6 \sin(2t) + 2C_2 \cos(2t) - C_2 \sin(2t) \right)
     \]
     \[
     y'(t) = e^{-t} \left( (2C_2 - 3) \cos(2t) + (-6 - C_2) \sin(2t) \right)
     \]
     Use \( y'(0) = -2 \):
     \[
     y'(0) = (2C_2 - 3) \cos(0) + (-6 - C_2) \sin(0) = 2C_2 - 3 = -2
     \]
     \[
     2C_2 - 3 = -2 \implies 2C_2 = 1 \implies C_2 = \frac{1}{2}
     \]

5. **Write the specific solution**:
   Substitute \( C_1 \) and \( C_2 \) back into the general solution:
   \[
   y(t) = e^{-t} \left( 3 \cos(2t) + \frac{1}{2} \sin(2t) \right)
   \]

Thus, the solution to the initial value problem is:
\[
y(t) = e^{-t} \left( 3 \cos(2t) + \frac{1}{2} \sin(2t) \right)
\]

Would you like details on any step or have any questions?

Here are five related questions:
1. How would the solution change if the initial condition \( y(0) \) was different?
2. What is the effect of changing the coefficients in the differential equation?
3. Can you solve a similar IVP with non-homogeneous terms?
4. How does the presence of complex roots affect the form of the general solution?
5. What if the characteristic equation had repeated roots?

**Tip:** When solving second-order differential equations, always start by finding the characteristic equation, as it provides the foundation for the general solution.

Step-by-step solution for solving the initial value problem y'' + 2y' + 5y = 0 with initial conditions y(0) = 3 and y'(0) = -2. Learn how to find the characteristic equation, solve for complex roots, and apply initial conditions to find the specific solution. Ideal for first-year undergraduate mathematics students.

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