The solution of `(dy)/(dx)-y=1, y(0)=1` is given by

Options

A. `xy=-e^(x)`
B. `xy=-e^(-x)`
C. `xy=-1`
D. `y=2e^(x)-1`

Explanation

To solve the differential equation \(\frac{dy}{dx} - y = 1\) with the initial condition \(y(0) = 1\), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \frac{dy}{dx} - y = 1 \]
We can rearrange it to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = y + 1 \]

**Hint:** Rearranging the equation helps in separating the variables for integration.

### Step 2: Separate the variables
Next, we separate the variables \(y\) and \(x\):
\[ \frac{dy}{1 + y} = dx \]

**Hint:** Separating variables allows us to integrate both sides independently.

### Step 3: Integrate both sides
Now, we integrate both sides:
\[ \int \frac{dy}{1 + y} = \int dx \]
The left side integrates to \(\ln|1 + y|\) and the right side integrates to \(x + C\):
\[ \ln|1 + y| = x + C \]

**Hint:** Remember the integral of \(\frac{1}{u}\) is \(\ln|u|\).

### Step 4: Solve for \(y\)
Next, we exponentiate both sides to eliminate the logarithm:
\[ 1 + y = e^{x + C} \]
This can be rewritten as:
\[ 1 + y = e^C e^x \]
Let \(k = e^C\), then:
\[ y + 1 = k e^x \]
Thus, we have:
\[ y = k e^x - 1 \]

**Hint:** Exponentiating helps to express the solution in a more manageable form.

### Step 5: Apply the initial condition
Now we use the initial condition \(y(0) = 1\) to find \(k\):
\[ 1 = k e^0 - 1 \]
This simplifies to:
\[ 1 = k - 1 \implies k = 2 \]

**Hint:** Using initial conditions allows us to find the specific constant in the general solution.

### Step 6: Write the final solution
Substituting \(k\) back into the equation for \(y\):
\[ y = 2 e^x - 1 \]

**Hint:** The final form of the solution is often the most useful for interpretation.

### Conclusion
The solution of the differential equation \(\frac{dy}{dx} - y = 1\) with the initial condition \(y(0) = 1\) is:
\[ \boxed{y = 2 e^x - 1} \]