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

Options

A. `y + tan (x + y) = c`
B. `y - tan { (x+y)/(2)} = c`
C. `y + tan {((x + y))/(2)} = c`
D. `y + tan{((x + y))/(2)} = c`

Answer

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

### Step 1: Rewrite the equation
The given differential equation is:
\[ \frac{dy}{dx} = \sec(x + y) \]
We can rewrite \(\sec\) in terms of cosine:
\[ \frac{dy}{dx} = \frac{1}{\cos(x + y)} \]

### Step 2: Cross-multiply
Cross-multiplying gives us:
\[ \cos(x + y) \, dy = dx \]

### Step 3: Substitute \(v = x + y\)
Let \(v = x + y\). Then, differentiating both sides with respect to \(x\) gives:
\[ \frac{dv}{dx} = 1 + \frac{dy}{dx} \]
From this, we can express \(\frac{dy}{dx}\) as:
\[ \frac{dy}{dx} = \frac{dv}{dx} - 1 \]

### Step 4: Substitute into the equation
Substituting \(\frac{dy}{dx}\) into the equation gives:
\[ \cos(v) \left(\frac{dv}{dx} - 1\right) = 1 \]
Expanding this, we have:
\[ \cos(v) \frac{dv}{dx} - \cos(v) = 1 \]
Rearranging gives:
\[ \cos(v) \frac{dv}{dx} = 1 + \cos(v) \]

### Step 5: Separate variables
Now, we can separate the variables:
\[ \frac{dv}{1 + \cos(v)} = \frac{dx}{\cos(v)} \]

### Step 6: Integrate both sides
Integrating both sides, we have:
\[ \int \frac{dv}{1 + \cos(v)} = \int \sec(v) \, dx \]
Using the identity \(1 + \cos(v) = 2 \cos^2\left(\frac{v}{2}\right)\), we can rewrite the left side:
\[ \int \frac{dv}{2 \cos^2\left(\frac{v}{2}\right)} = \frac{1}{2} \int \sec^2\left(\frac{v}{2}\right) \, dv \]
This integrates to:
\[ \tan\left(\frac{v}{2}\right) + C \]

### Step 7: Solve for \(y\)
Now we have:
\[ \tan\left(\frac{v}{2}\right) = x + C \]
Substituting back \(v = x + y\):
\[ \tan\left(\frac{x + y}{2}\right) = x + C \]

### Step 8: Rearranging the equation
Rearranging gives us:
\[ y = 2 \tan^{-1}(x + C) - x \]

### Final Solution
Thus, the solution to the differential equation \(\frac{dy}{dx} = \sec(x + y)\) is:
\[ y = 2 \tan^{-1}(x + C) - x \]

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