Solution of the differential equation `y_(3)-8y_(2)=0` where `y (0)=(1)/(8), y_(1)(0)=0, y_(2)(0)=1` is equal to

Options

A. `y=(1)/(8) ((e^(8x))/(8)+x-(7)/(8))`
B. `y=(1)/(8) ((e^(8x))/(8) +x+(7)/(8))`
C. `y=(1)/(8) ((e^(8x))/(8)-x+(7)/(8))`
D. none of these

Answer

To solve the differential equation \( y^{(3)} - 8y^{(2)} = 0 \) with the initial conditions \( y(0) = \frac{1}{8} \), \( y'(0) = 0 \), and \( y''(0) = 1 \), we will follow these steps:

### Step 1: Rewrite the Differential Equation
The given differential equation is:
\[ y^{(3)} - 8y^{(2)} = 0 \]
We can factor this as:
\[ y^{(2)}(y' - 8) = 0 \]
This gives us two cases to consider:
1. \( y^{(2)} = 0 \)
2. \( y' - 8 = 0 \)

### Step 2: Solve the First Case \( y^{(2)} = 0 \)
If \( y^{(2)} = 0 \), then integrating gives:
\[ y' = C_1 \quad \text{(where \( C_1 \) is a constant)} \]
Integrating again, we find:
\[ y = C_1 x + C_2 \quad \text{(where \( C_2 \) is another constant)} \]

### Step 3: Solve the Second Case \( y' - 8 = 0 \)
If \( y' - 8 = 0 \), then we have:
\[ y' = 8 \]
Integrating this gives:
\[ y = 8x + C_3 \quad \text{(where \( C_3 \) is another constant)} \]

### Step 4: Combine Solutions
The general solution of the differential equation is a linear combination of the solutions from both cases:
\[ y = C_1 x + C_2 + 8x + C_3 \]
This simplifies to:
\[ y = (C_1 + 8)x + (C_2 + C_3) \]
Let \( C = C_1 + 8 \) and \( D = C_2 + C_3 \), so:
\[ y = Cx + D \]

### Step 5: Apply Initial Conditions
Now we apply the initial conditions to find the constants \( C \) and \( D \).

1. **From \( y(0) = \frac{1}{8} \)**:
\[ D = \frac{1}{8} \]

2. **From \( y'(0) = 0 \)**:
\[ C = 0 \]

3. **From \( y''(0) = 1 \)**:
Since \( y' = C \) and \( y'' = 0 \) for our linear function, we need to check the second derivative condition. Since \( y'' = 0 \) does not satisfy \( y''(0) = 1 \), we must reconsider our solution structure.

### Step 6: Reassess the General Solution
Since we have a repeated root, we should include a term involving \( x^2 \):
\[ y = Ax^2 + Bx + C \]
Now we differentiate:
\[ y' = 2Ax + B \]
\[ y'' = 2A \]
\[ y^{(3)} = 0 \]

### Step 7: Apply the Initial Conditions Again
1. **From \( y(0) = \frac{1}{8} \)**:
\[ C = \frac{1}{8} \]

2. **From \( y'(0) = 0 \)**:
\[ B = 0 \]

3. **From \( y''(0) = 1 \)**:
\[ 2A = 1 \implies A = \frac{1}{2} \]

### Final Solution
Thus, the solution to the differential equation is:
\[ y = \frac{1}{2}x^2 + \frac{1}{8} \]