The solution of ` y'-y=1, y(0) =1 ` is given by y(x) =

Options

A. `-exp .(x) `
B. `- exp .(-x)`
C. `-1`
D. `2 exp . (x) -1`

Answer

To solve the differential equation \( y' - y = 1 \) with the initial condition \( y(0) = 1 \), we can follow these steps:

### Step 1: Rewrite the Equation
We start with the given equation:
\[ y' - y = 1 \]
We can rewrite \( y' \) as \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} - y = 1 \]

### Step 2: Rearrange the Equation
Rearranging gives us:
\[ \frac{dy}{dx} = y + 1 \]

### Step 3: Separate Variables
We can separate the variables by moving all terms involving \( y \) to one side and \( x \) to the other:
\[ \frac{dy}{y + 1} = dx \]

### Step 4: Integrate Both Sides
Now we integrate both sides:
\[ \int \frac{dy}{y + 1} = \int dx \]
The left-hand side integrates to \( \ln |y + 1| \) and the right-hand side integrates to \( x + C \):
\[ \ln |y + 1| = x + C \]

### Step 5: Solve for \( y \)
To solve for \( y \), we exponentiate both sides:
\[ |y + 1| = e^{x + C} = e^C e^x \]
Let \( K = e^C \), then:
\[ y + 1 = K e^x \]
Thus,
\[ y = K e^x - 1 \]

### Step 6: Apply the Initial Condition
We apply the initial condition \( y(0) = 1 \):
\[ 1 = K e^0 - 1 \]
This simplifies to:
\[ 1 = K - 1 \implies K = 2 \]

### Step 7: Write the Final Solution
Substituting \( K \) back into the equation for \( y \):
\[ y = 2 e^x - 1 \]

### Final Answer
Thus, the solution to the differential equation is:
\[ y(x) = 2 e^x - 1 \]