We are tasked with solving the non-homogeneous second-order differential equation:

\[
y'' + 10y' + 21y = 21e^{0t}, \quad y(0) = 8, \quad y'(0) = 5
\]

### Step 1: Solve the homogeneous equation
The associated homogeneous equation is:

\[
y_h'' + 10y_h' + 21y_h = 0
\]

The characteristic equation is:

\[
r^2 + 10r + 21 = 0
\]

Solve this quadratic equation using the quadratic formula:

\[
r = \frac{-10 \pm \sqrt{10^2 - 4(1)(21)}}{2(1)} = \frac{-10 \pm \sqrt{100 - 84}}{2} = \frac{-10 \pm \sqrt{16}}{2} = \frac{-10 \pm 4}{2}
\]

So, the roots are:

\[
r_1 = \frac{-10 + 4}{2} = -3, \quad r_2 = \frac{-10 - 4}{2} = -7
\]

Thus, the general solution to the homogeneous equation is:

\[
y_h(t) = C_1 e^{-3t} + C_2 e^{-7t}
\]

### Step 2: Solve the particular solution
For the non-homogeneous term \( 21e^{0t} = 21 \), we propose a particular solution of the form \( y_p(t) = A \), since the right-hand side is a constant.

Substitute \( y_p(t) = A \) into the original equation:

\[
0 + 10(0) + 21A = 21
\]

This gives:

\[
21A = 21 \quad \Rightarrow \quad A = 1
\]

Thus, the particular solution is:

\[
y_p(t) = 1
\]

### Step 3: General solution
The general solution to the differential equation is the sum of the homogeneous and particular solutions:

\[
y(t) = y_h(t) + y_p(t) = C_1 e^{-3t} + C_2 e^{-7t} + 1
\]

### Step 4: Apply initial conditions
We use the initial conditions \( y(0) = 8 \) and \( y'(0) = 5 \) to find \( C_1 \) and \( C_2 \).

1. **Condition 1: \( y(0) = 8 \)**

\[
y(0) = C_1 e^{-3(0)} + C_2 e^{-7(0)} + 1 = C_1 + C_2 + 1 = 8
\]

This simplifies to:

\[
C_1 + C_2 = 7 \quad \text{(Equation 1)}
\]

2. **Condition 2: \( y'(0) = 5 \)**

First, find \( y'(t) \):

\[
y'(t) = -3C_1 e^{-3t} - 7C_2 e^{-7t}
\]

Now, apply \( t = 0 \):

\[
y'(0) = -3C_1 - 7C_2 = 5
\]

This gives:

\[
-3C_1 - 7C_2 = 5 \quad \text{(Equation 2)}
\]

### Step 5: Solve the system of equations
We now solve the system of equations:

1. \( C_1 + C_2 = 7 \)
2. \( -3C_1 - 7C_2 = 5 \)

From the first equation, solve for \( C_1 \):

\[
C_1 = 7 - C_2
\]

Substitute this into the second equation:

\[
-3(7 - C_2) - 7C_2 = 5
\]

Simplify:

\[
-21 + 3C_2 - 7C_2 = 5
\]
\[
-21 - 4C_2 = 5
\]
\[
-4C_2 = 26 \quad \Rightarrow \quad C_2 = -\frac{26}{4} = -\frac{13}{2}
\]

Now substitute \( C_2 = -\frac{13}{2} \) into \( C_1 + C_2 = 7 \):

\[
C_1 - \frac{13}{2} = 7 \quad \Rightarrow \quad C_1 = 7 + \frac{13}{2} = \frac{14}{2} + \frac{13}{2} = \frac{27}{2}
\]

### Final solution
Thus, the solution to the differential equation is:

\[
y(t) = \frac{27}{2} e^{-3t} - \frac{13}{2} e^{-7t} + 1
\]

Would you like any additional details or have any questions?

### Related questions:
1. How do we handle initial conditions if the differential equation is more complex?
2. Can you explain how the form of the particular solution is chosen for different right-hand side functions?
3. What are some methods to solve differential equations when the characteristic equation has complex roots?
4. How would the solution change if the right-hand side involved a trigonometric function?
5. How does the Laplace transform method compare to the method of undetermined coefficients?

### Tip:
Always verify the form of your particular solution by substituting it back into the differential equation to avoid errors.

Explore the step-by-step solution to the non-homogeneous second-order differential equation y'' + 10y' + 21y = 21e^(0t) with initial conditions y(0) = 8 and y'(0) = 5. This detailed guide covers key concepts like homogeneous solutions, the quadratic formula, and the method of undetermined coefficients.

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