The solution of the differential equation `sin(x+y)dy =dx` is

Options

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

Answer

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ \sin(x+y) \, dy = dx \]
We can rewrite this as:
\[ \frac{dx}{dy} = \sin(x+y) \]

### Step 2: Introduce a substitution
Let:
\[ t = x + y \]
Then, differentiating both sides with respect to \( y \):
\[ \frac{dt}{dy} = \frac{dx}{dy} + 1 \]
From this, we can express \( \frac{dx}{dy} \):
\[ \frac{dx}{dy} = \frac{dt}{dy} - 1 \]

### Step 3: Substitute into the differential equation
Substituting \( \frac{dx}{dy} \) into the rewritten equation:
\[ \frac{dt}{dy} - 1 = \sin(t) \]
Rearranging gives:
\[ \frac{dt}{dy} = 1 + \sin(t) \]

### Step 4: Separate the variables
Now we can separate the variables:
\[ \frac{dt}{1 + \sin(t)} = dy \]

### Step 5: Integrate both sides
Next, we integrate both sides. To integrate the left side, we will rationalize:
\[ \int \frac{dt}{1 + \sin(t)} = \int dy \]
To rationalize \( \frac{1}{1 + \sin(t)} \), we multiply the numerator and denominator by \( 1 - \sin(t) \):
\[ \frac{1 - \sin(t)}{(1 + \sin(t))(1 - \sin(t))} = \frac{1 - \sin(t)}{\cos^2(t)} \]
Thus, we can rewrite the integral:
\[ \int \frac{1 - \sin(t)}{\cos^2(t)} dt = \int dy \]

### Step 6: Solve the integrals
Now we can split the integral:
\[ \int \sec^2(t) dt - \int \tan(t) dt = y + C \]
The integrals yield:
\[ \tan(t) - \ln|\sec(t) + \tan(t)| = y + C \]

### Step 7: Substitute back for \( t \)
Recall that \( t = x + y \):
\[ \tan(x + y) - \ln|\sec(x + y) + \tan(x + y)| = y + C \]

### Step 8: Rearranging the equation
Rearranging gives us the implicit solution:
\[ \tan(x + y) - y - C = \ln|\sec(x + y) + \tan(x + y)| \]

### Final Result
Thus, the solution to the differential equation is:
\[ \tan(x + y) - y = C + \ln|\sec(x + y) + \tan(x + y)| \]