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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.

Solve number 6 from the image: d^2y/dx^2 + 4dy/dx + 5y = 2e^-2x, given that x=0, y=1, and dy/dx=-2

This problem involves solving a second-order differential equation with both homogeneous and particular solutions using the method of undetermined coefficients. It also includes applying initial conditions to find specific constants for the final solution.

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