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.