The integrating factor of `(d ^(2) y )/( dx ^(2)) - y = e ^(x) ` is ` e ^(-x)` then its solution is

Options

A. `yr ^(-x) = x +c`
B. `ye ^(x) = x + c`
C. `ye ^(x) = 2x + c`
D. `ye ^(-x) = 2x +c`

Answer

To solve the differential equation \(\frac{d^2y}{dx^2} - y = e^x\) using the integrating factor \(e^{-x}\), we will follow these steps:

### Step 1: Rewrite the equation in standard form
The given equation is already in a standard linear form:
\[ \frac{d^2y}{dx^2} - y = e^x \]

### Step 2: Identify \(p\) and \(q\)
From the standard form \(\frac{d^2y}{dx^2} + py = q\), we identify:
- \(p = -1\)
- \(q = e^x\)

### Step 3: Calculate the integrating factor
The integrating factor \(I\) is given by:
\[ I = e^{\int p \, dx} = e^{\int -1 \, dx} = e^{-x} \]

### Step 4: Multiply the entire equation by the integrating factor
Multiply both sides of the equation by \(e^{-x}\):
\[ e^{-x} \frac{d^2y}{dx^2} - e^{-x}y = e^{-x} e^x \]
This simplifies to:
\[ e^{-x} \frac{d^2y}{dx^2} - e^{-x}y = 1 \]

### Step 5: Rewrite the left-hand side
The left-hand side can be rewritten as the derivative of a product:
\[ \frac{d}{dx}\left(e^{-x} \frac{dy}{dx}\right) = 1 \]

### Step 6: Integrate both sides
Integrate both sides with respect to \(x\):
\[ \int \frac{d}{dx}\left(e^{-x} \frac{dy}{dx}\right) \, dx = \int 1 \, dx \]
This gives:
\[ e^{-x} \frac{dy}{dx} = x + C \]
where \(C\) is the constant of integration.

### Step 7: Solve for \(\frac{dy}{dx}\)
Multiply both sides by \(e^x\):
\[ \frac{dy}{dx} = (x + C)e^x \]

### Step 8: Integrate again
Now, integrate both sides:
\[ y = \int (x + C)e^x \, dx \]
Using integration by parts, we can evaluate this integral. Let \(u = x + C\) and \(dv = e^x dx\):
\[ y = (x + C)e^x - \int e^x \, dx = (x + C)e^x - e^x + D \]
where \(D\) is another constant of integration.

### Step 9: Final solution
Thus, we can express the solution as:
\[ y = (x + C - 1)e^x + D \]

### Summary of the solution
The solution to the differential equation \(\frac{d^2y}{dx^2} - y = e^x\) is:
\[ y = (x + C - 1)e^x + D \]