Solve the following differential equations :
`cos(x+y)dy=dx`

Answer

To solve the differential equation \( \cos(x+y) \, dy = dx \), we will follow these steps:

### Step 1: Rewrite the equation
We start with the given equation:
\[ \cos(x+y) \, dy = dx \]
We can rearrange this to express \( dy \) in terms of \( dx \):
\[ dy = \frac{dx}{\cos(x+y)} \]

### Step 2: Substitute \( t = x + y \)
To simplify the equation, we can use the substitution:
\[ t = x + y \implies y = t - x \]
Now, we differentiate \( t \) with respect to \( x \):
\[ \frac{dt}{dx} = 1 + \frac{dy}{dx} \]
From the rearranged equation, we have:
\[ \frac{dy}{dx} = \frac{1}{\cos(t)} \]
Substituting this into the derivative of \( t \):
\[ \frac{dt}{dx} = 1 + \frac{1}{\cos(t)} \]

### Step 3: Rearranging the equation
Now we can rearrange this to separate the variables:
\[ \frac{dt}{dx} = 1 + \sec(t) \]
This can be rewritten as:
\[ \frac{dt}{1 + \sec(t)} = dx \]

### Step 4: Integrate both sides
Next, we integrate both sides:
\[ \int \frac{dt}{1 + \sec(t)} = \int dx \]
The left side can be simplified using the identity \( \sec(t) = \frac{1}{\cos(t)} \):
\[ \int \frac{\cos(t)}{1 + \cos(t)} \, dt = x + C \]

### Step 5: Solve the integral
To solve the integral on the left, we can use a substitution or recognize it as a standard integral. The integral simplifies to:
\[ \int \frac{\cos(t)}{1 + \cos(t)} \, dt = \ln |1 + \cos(t)| + C \]
Thus, we have:
\[ \ln |1 + \cos(t)| = x + C \]

### Step 6: Substitute back for \( t \)
Substituting back \( t = x + y \):
\[ \ln |1 + \cos(x+y)| = x + C \]

### Step 7: Exponentiate to solve for the function
Exponentiating both sides gives:
\[ |1 + \cos(x+y)| = e^{x+C} = e^C e^x \]
Let \( K = e^C \), where \( K \) is a constant:
\[ 1 + \cos(x+y) = K e^x \]

### Final Solution
Thus, the solution to the differential equation is:
\[ \cos(x+y) = K e^x - 1 \]