Solution of the differential equation `cos^(2)(x-y)(dy)/(dx)=1`

Options

A. y+tan(x-y)=c
B. `-y+tan(x-y)=c`
C. y+cot(x-y)=c
D. `-y+cot(x-y)=c`

Answer

To solve the differential equation \( \cos^2(x - y) \frac{dy}{dx} = 1 \), we can follow these steps:

### Step 1: Rewrite the equation
We start by rewriting the given equation in a more manageable form. We can express it as:
\[ \frac{dy}{dx} = \frac{1}{\cos^2(x - y)} \]

### Step 2: Substitute \( u = x - y \)
Let’s make the substitution \( u = x - y \). Then, we have:
\[ y = x - u \quad \Rightarrow \quad \frac{dy}{dx} = 1 - \frac{du}{dx} \]

### Step 3: Substitute \( \frac{dy}{dx} \) into the equation
Now, substituting \( \frac{dy}{dx} \) into the rewritten equation gives:
\[ 1 - \frac{du}{dx} = \frac{1}{\cos^2(u)} \]

### Step 4: Rearranging the equation
Rearranging the equation yields:
\[ -\frac{du}{dx} = \frac{1}{\cos^2(u)} - 1 \]
\[ -\frac{du}{dx} = \frac{1 - \cos^2(u)}{\cos^2(u)} = \frac{\sin^2(u)}{\cos^2(u)} \]

### Step 5: Separate variables
We can separate the variables:
\[ -\frac{du}{\sin^2(u)} = \frac{dx}{\cos^2(u)} \]

### Step 6: Integrate both sides
Now we integrate both sides:
\[ \int -\frac{du}{\sin^2(u)} = \int \frac{dx}{\cos^2(u)} \]
The left side integrates to:
\[ \int -\frac{du}{\sin^2(u)} = \cot(u) + C_1 \]
The right side integrates to:
\[ \int \sec^2(u) dx = \tan(u) + C_2 \]

### Step 7: Combine results
Combining the results gives us:
\[ \cot(u) = \tan(x) + C \]
where \( C = C_2 - C_1 \).

### Step 8: Substitute back for \( u \)
Now, substituting back \( u = x - y \):
\[ \cot(x - y) = \tan(x) + C \]

### Step 9: Final result
This is the implicit solution of the differential equation:
\[ \cot(x - y) - \tan(x) = C \]