What is the solution of the differential equation `(dy)/(dx) = cos (y - x) + 1 ?`

Options

A. `e^(x) [sec (y - x) - tan (y - x)] = c`
B. `e^(x) [sec (y - x) + tan (y -x)]= c `
C. `e^(x) sec (y - x) tan(y - x) = c`
D. `e^(x) = cosec (y - x) tan (y - x)`

Answer

To solve the differential equation \(\frac{dy}{dx} = \cos(y - x) + 1\), we can follow these steps:

### Step 1: Rewrite the equation
We start with the given differential equation:
\[ \frac{dy}{dx} = \cos(y - x) + 1 \]

### Step 2: Introduce a substitution
Let \(v = y - x\). Then, we can express \(y\) in terms of \(v\):
\[ y = v + x \]
Now, differentiate \(y\) with respect to \(x\):
\[ \frac{dy}{dx} = \frac{dv}{dx} + 1 \]

### Step 3: Substitute into the differential equation
Substituting \(\frac{dy}{dx}\) into the original equation gives:
\[ \frac{dv}{dx} + 1 = \cos(v) + 1 \]
Subtracting 1 from both sides, we have:
\[ \frac{dv}{dx} = \cos(v) \]

### Step 4: Separate variables
We can separate the variables by rewriting the equation as:
\[ \frac{dv}{\cos(v)} = dx \]

### Step 5: Integrate both sides
Now, we integrate both sides:
\[ \int \sec(v) \, dv = \int dx \]
The integral of \(\sec(v)\) is \(\ln |\sec(v) + \tan(v)| + C\), and the integral of \(dx\) is \(x + C\). Therefore, we have:
\[ \ln |\sec(v) + \tan(v)| = x + C \]

### Step 6: Exponentiate both sides
To eliminate the logarithm, we exponentiate both sides:
\[ |\sec(v) + \tan(v)| = e^{x + C} = e^C e^x \]
Let \(k = e^C\), then:
\[ \sec(v) + \tan(v) = k e^x \]

### Step 7: Substitute back for \(v\)
Recall that \(v = y - x\). Thus, we substitute back:
\[ \sec(y - x) + \tan(y - x) = k e^x \]

### Final Solution
This gives us the implicit solution to the differential equation:
\[ \sec(y - x) + \tan(y - x) = k e^x \]