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The image presents a differential equation along with initial conditions, followed by multiple-choice options. The equation is:

\[
y'' - 6y' + 8y = 1
\]
with the initial conditions \( y(0) = 1 \) and \( y'(0) = 7 \).

Let's proceed to solve this differential equation and check which of the options is correct.

### Step 1: Solve the Homogeneous Equation
The corresponding homogeneous equation is:

\[
y_h'' - 6y_h' + 8y_h = 0
\]

The characteristic equation for this is:

\[
r^2 - 6r + 8 = 0
\]

Solving the quadratic equation:

\[
r = \frac{-(-6) \pm \sqrt{(-6)^2 - 4(1)(8)}}{2(1)} = \frac{6 \pm \sqrt{36 - 32}}{2} = \frac{6 \pm 2}{2}
\]

Thus, the roots are \( r = 4 \) and \( r = 2 \).

The general solution to the homogeneous equation is:

\[
y_h(t) = C_1 e^{4t} + C_2 e^{2t}
\]

### Step 2: Find the Particular Solution
Now, for the non-homogeneous part, we assume a particular solution of the form \( y_p = A \), since the right-hand side is a constant.

Substituting \( y_p = A \) into the differential equation:

\[
0 - 0 + 8A = 1 \quad \Rightarrow \quad A = \frac{1}{8}
\]

Thus, the particular solution is \( y_p = \frac{1}{8} \).

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

\[
y(t) = C_1 e^{4t} + C_2 e^{2t} + \frac{1}{8}
\]

### Step 4: Apply Initial Conditions
Using \( y(0) = 1 \):

\[
C_1 e^{4(0)} + C_2 e^{2(0)} + \frac{1}{8} = 1
\]

\[
C_1 + C_2 + \frac{1}{8} = 1 \quad \Rightarrow \quad C_1 + C_2 = 1 - \frac{1}{8} = \frac{7}{8}
\]

Now, using \( y'(0) = 7 \), we first find the derivative of \( y(t) \):

\[
y'(t) = 4C_1 e^{4t} + 2C_2 e^{2t}
\]

Substitute \( t = 0 \):

\[
4C_1 e^{4(0)} + 2C_2 e^{2(0)} = 7 \quad \Rightarrow \quad 4C_1 + 2C_2 = 7
\]

### Step 5: Solve for \( C_1 \) and \( C_2 \)
We now have the system of equations:

1. \( C_1 + C_2 = \frac{7}{8} \)
2. \( 4C_1 + 2C_2 = 7 \)

Multiply the first equation by 2:

\[
2C_1 + 2C_2 = \frac{7}{4}
\]

Now subtract this from the second equation:

\[
(4C_1 + 2C_2) - (2C_1 + 2C_2) = 7 - \frac{7}{4}
\]

\[
2C_1 = \frac{28}{4} - \frac{7}{4} = \frac{21}{4} \quad \Rightarrow \quad C_1 = \frac{21}{8}
\]

Substitute \( C_1 = \frac{21}{8} \) into \( C_1 + C_2 = \frac{7}{8} \):

\[
\frac{21}{8} + C_2 = \frac{7}{8} \quad \Rightarrow \quad C_2 = \frac{7}{8} - \frac{21}{8} = -\frac{14}{8} = -\frac{7}{4}
\]

### Step 6: Final Solution
The solution to the differential equation is:

\[
y(t) = \frac{21}{8} e^{4t} - \frac{7}{4} e^{2t} + \frac{1}{8}
\]

This matches option **(c)**.

---

Do you want more details or have any questions? Here are 5 related questions to consider:

1. How would the solution change if the right-hand side were a function of \(t\)?
2. What is the significance of the roots of the characteristic equation?
3. Can you describe the behavior of the solution as \(t\) approaches infinity?
4. How does the method of undetermined coefficients help in finding the particular solution?
5. What is the role of initial conditions in solving second-order differential equations?

**Tip:** When solving non-homogeneous differential equations, always solve the homogeneous part first and then find a particular solution based on the form of the non-homogeneous term.

y'' - 6y' + 8y = 1, y(0) = 1, y'(0) = 7. Choose the correct solution.

This problem involves solving the differential equation y'' - 6y' + 8y = 1, subject to the initial conditions y(0) = 1 and y'(0) = 7. The solution process includes finding the characteristic equation, solving for the homogeneous and particular solutions, and applying initial conditions to determine the constants.

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