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

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)=csec(y-x)tan(y-x)`

Explanation

To solve the differential equation

\[ \frac{dy}{dx} = \cos(y - x) + 1, \]

we can follow these steps:

### Step 1: Substitute \( t = y - x \)

Let \( t = y - x \). Then, we can express \( y \) in terms of \( t \) and \( x \):

\[ y = t + x. \]

Now, differentiate both sides with respect to \( x \):

\[ \frac{dy}{dx} = \frac{dt}{dx} + 1. \]

### Step 2: Rewrite the differential equation

Substituting \( \frac{dy}{dx} \) into the original equation gives us:

\[ \frac{dt}{dx} + 1 = \cos(t) + 1. \]

Subtracting 1 from both sides, we have:

\[ \frac{dt}{dx} = \cos(t). \]

### Step 3: Separate variables

Now, we can separate the variables:

\[ \frac{dt}{\cos(t)} = dx. \]

### Step 4: Integrate both sides

Integrating both sides, we get:

\[ \int \sec(t) dt = \int dx. \]

The integral of \( \sec(t) \) is:

\[ \ln | \sec(t) + \tan(t) | + C_1, \]

and the integral of \( dx \) is:

\[ x + C_2. \]

Setting the constants together, we have:

\[ \ln | \sec(t) + \tan(t) | = x + C. \]

### Step 5: Solve for \( t \)

Exponentiating both sides gives:

\[ |\sec(t) + \tan(t)| = e^{x + C} = e^C e^x. \]

Let \( K = e^C \), so we can write:

\[ \sec(t) + \tan(t) = K e^x. \]

### Step 6: Substitute back for \( t \)

Recall that \( t = y - x \). Therefore, we can substitute back:

\[ \sec(y - x) + \tan(y - x) = K e^x. \]

### Final Solution

Thus, the solution of the differential equation is:

\[ \sec(y - x) + \tan(y - x) = C e^x, \]

where \( C \) is a constant.

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