Find the general solution of each of the following differential equations:
`(dy)/(dx)+sin(x+y)=sin(x-y)`

Explanation

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

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

### Step 2: Use the sine addition and subtraction formulas
We can use the sine addition and subtraction formulas to rewrite \(\sin(x+y)\) and \(\sin(x-y)\):
\[ \sin(x+y) = \sin x \cos y + \cos x \sin y \]
\[ \sin(x-y) = \sin x \cos y - \cos x \sin y \]

### Step 3: Substitute these formulas into the equation
Substituting these into the differential equation gives:
\[ \frac{dy}{dx} + (\sin x \cos y + \cos x \sin y) = (\sin x \cos y - \cos x \sin y) \]

### Step 4: Simplify the equation
Now, we can simplify the equation:
\[ \frac{dy}{dx} + \sin x \cos y + \cos x \sin y = \sin x \cos y - \cos x \sin y \]
This simplifies to:
\[ \frac{dy}{dx} + 2\cos x \sin y = 0 \]

### Step 5: Rearrange the equation
Rearranging gives us:
\[ \frac{dy}{dx} = -2\cos x \sin y \]

### Step 6: Separate the variables
Now we separate the variables:
\[ \frac{dy}{\sin y} = -2\cos x \, dx \]

### Step 7: Integrate both sides
Integrating both sides, we have:
\[ \int \frac{dy}{\sin y} = \int -2\cos x \, dx \]
The left side integrates to \(\ln|\tan(\frac{y}{2})|\) and the right side integrates to \(-2\sin x + C\):
\[ \ln|\tan(\frac{y}{2})| = -2\sin x + C \]

### Step 8: Solve for y
Exponentiating both sides gives:
\[ \tan\left(\frac{y}{2}\right) = e^{-2\sin x + C} = Ce^{-2\sin x} \]
where \(C\) is a constant.

### Step 9: Final form of the solution
Thus, the general solution of the differential equation is:
\[ y = 2\tan^{-1}(Ce^{-2\sin x}) \]