We are given the second-order differential equation from problem 6 in the image:

$\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 5y = 2e^{-2x}$ with the initial conditions: $x = 0, y = 1, \frac{dy}{dx} = -2.$

Step 1: Solve the homogeneous equation

The homogeneous equation corresponding to the given differential equation is: $\frac{d^2y}{dx^2} + 4\frac{dy}{dx} + 5y = 0.$

The characteristic equation for this differential equation is: $r^2 + 4r + 5 = 0.$

We can solve this quadratic equation using the quadratic formula: $r = \frac{-4 \pm \sqrt{4^2 - 4(1)(5)}}{2(1)} = \frac{-4 \pm \sqrt{16 - 20}}{2} = \frac{-4 \pm \sqrt{-4}}{2} = \frac{-4 \pm 2i}{2}.$

Thus, the roots are: $r = -2 \pm i.$

Therefore, the general solution to the homogeneous equation is: $y_h(x) = e^{-2x}(C_1\cos(x) + C_2\sin(x)).$

Step 2: Solve the particular solution

For the particular solution, we use the method of undetermined coefficients. Since the non-homogeneous term is $2e^{-2x}$, we look for a particular solution of the form: $y_p(x) = Ae^{-2x}.$

Substituting $y_p(x) = Ae^{-2x}$ into the original differential equation:

$\frac{d^2}{dx^2}(Ae^{-2x}) + 4\frac{d}{dx}(Ae^{-2x}) + 5(Ae^{-2x}) = 2e^{-2x}.$

First, compute the derivatives: $\frac{dy_p}{dx} = -2Ae^{-2x}, \quad \frac{d^2y_p}{dx^2} = 4Ae^{-2x}.$

Substitute these into the differential equation: $4Ae^{-2x} + 4(-2Ae^{-2x}) + 5(Ae^{-2x}) = 2e^{-2x}.$

Simplifying: $4Ae^{-2x} - 8Ae^{-2x} + 5Ae^{-2x} = 2e^{-2x},$ $(4 - 8 + 5)Ae^{-2x} = 2e^{-2x},$ $Ae^{-2x} = 2e^{-2x}.$

Thus, $A = 2$, and the particular solution is: $y_p(x) = 2e^{-2x}.$

Step 3: General solution

The general solution to the differential equation is the sum of the homogeneous and particular solutions: $y(x) = y_h(x) + y_p(x) = e^{-2x}(C_1\cos(x) + C_2\sin(x)) + 2e^{-2x}.$ Factor out $e^{-2x}$: $y(x) = e^{-2x}(C_1\cos(x) + C_2\sin(x) + 2).$

Step 4: Apply initial conditions

Use the initial conditions $y(0) = 1$ and $\frac{dy}{dx}(0) = -2$ to solve for $C_1$ and $C_2$.

1. $y(0) = 1$: $y(0) = e^{0}(C_1\cos(0) + C_2\sin(0) + 2) = C_1 + 2 = 1.$ Thus, $C_1 = -1$.

2. $\frac{dy}{dx}(0) = -2$:

First, differentiate $y(x)$: $\frac{dy}{dx} = \frac{d}{dx}\left[e^{-2x}(C_1\cos(x) + C_2\sin(x) + 2)\right].$ Use the product rule: $\frac{dy}{dx} = -2e^{-2x}(C_1\cos(x) + C_2\sin(x) + 2) + e^{-2x}(-C_1\sin(x) + C_2\cos(x)).$ At $x = 0$: $\frac{dy}{dx}(0) = -2e^{0}(C_1\cos(0) + C_2\sin(0) + 2) + e^{0}(-C_1\sin(0) + C_2\cos(0)),$ $-2(C_1 + 2) + C_2 = -2.$ Substitute $C_1 = -1$: $-2(-1 + 2) + C_2 = -2,$ $-2(1) + C_2 = -2,$ $-2 + C_2 = -2.$ Thus, $C_2 = 0$.

Final solution:

The final solution to the differential equation is: $y(x) = e^{-2x}(-\cos(x) + 2).$

Let me know if you'd like further clarification on any steps or details! Here are five related questions to consider:

1. What is the method of undetermined coefficients, and when do we use it?
2. How do we solve second-order linear differential equations with complex roots?
3. How would the solution change if the non-homogeneous term were different, say $2e^{x}$?
4. Can you explain the significance of initial conditions in differential equations?
5. What is the difference between homogeneous and non-homogeneous differential equations?

Tip: When solving differential equations, always check if the non-homogeneous term resembles the solution to the homogeneous equation, as this affects the form of the particular solution.