`"Solve the following differential equations"` :
`cos^(-1)(dy/dx)=x+y`.

Answer

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

### Step 1: Rewrite the equation
We start with the given equation:
\[ \cos^{-1}\left(\frac{dy}{dx}\right) = x + y \]
Taking the cosine of both sides, we have:
\[ \frac{dy}{dx} = \cos(x + y) \]

### Step 2: Separate variables
We can separate the variables \(y\) and \(x\):
\[ dy = \cos(x + y) \, dx \]
This can be rewritten as:
\[ \frac{dy}{\cos(x + y)} = dx \]

### Step 3: Introduce a substitution
Let \(b = x + y\). Then, differentiating both sides with respect to \(x\):
\[ \frac{db}{dx} = 1 + \frac{dy}{dx} \]
Substituting \(\frac{dy}{dx} = \cos(b)\) into the equation gives:
\[ \frac{db}{dx} = 1 + \cos(b) \]
This can be rearranged to:
\[ \frac{db}{1 + \cos(b)} = dx \]

### Step 4: Simplify the left-hand side
Using the identity \(1 + \cos(b) = 2\cos^2\left(\frac{b}{2}\right)\), we can rewrite the left-hand side:
\[ \frac{db}{2\cos^2\left(\frac{b}{2}\right)} = dx \]
This simplifies to:
\[ \frac{1}{2} \sec^2\left(\frac{b}{2}\right) db = dx \]

### Step 5: Integrate both sides
Integrating both sides:
\[ \frac{1}{2} \int \sec^2\left(\frac{b}{2}\right) db = \int dx \]
The integral of \(\sec^2\) is:
\[ \frac{1}{2} \cdot 2 \tan\left(\frac{b}{2}\right) = x + C \]
Thus, we have:
\[ \tan\left(\frac{b}{2}\right) = x + C \]

### Step 6: Substitute back for \(b\)
Recalling that \(b = x + y\), we substitute back:
\[ \tan\left(\frac{x + y}{2}\right) = x + C \]

### Final Solution
The final solution to the differential equation is:
\[ \tan\left(\frac{x + y}{2}\right) = x + C \]