The solution of differential edquation `(dy)/(dx) +1 = cosec (x+y)` is

Options

A. `cos(x+y) + x =c`
B. `cos (x+y)=c`
C. `sin (x+y) +x = c `
D. None of these

Explanation

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

### Step 1: Rewrite the Equation
First, we rewrite the equation to isolate \(\frac{dy}{dx}\):
\[ \frac{dy}{dx} = \csc(x + y) - 1 \]

**Hint:** Rearranging the equation helps to express the derivative in terms of \(x\) and \(y\).

### Step 2: Substitute Variables
Next, we will use the substitution \(v = x + y\). Therefore, we have:
\[ y = v - x \quad \text{and} \quad \frac{dy}{dx} = \frac{dv}{dx} - 1 \]

**Hint:** Substituting \(v\) simplifies the relationship between \(x\) and \(y\).

### Step 3: Substitute into the Equation
Substituting \(\frac{dy}{dx}\) into the equation gives us:
\[ \frac{dv}{dx} - 1 = \csc(v) \]
This simplifies to:
\[ \frac{dv}{dx} = \csc(v) + 1 \]

**Hint:** Make sure to correctly substitute and simplify the terms.

### Step 4: Separate Variables
Now, we separate the variables:
\[ \frac{dv}{\csc(v) + 1} = dx \]

**Hint:** Separating variables allows us to integrate both sides independently.

### Step 5: Integrate Both Sides
Next, we integrate both sides:
\[ \int \frac{dv}{\csc(v) + 1} = \int dx \]

For the left side, we can simplify \(\csc(v) + 1\) to \(\frac{1 + \sin(v)}{\sin(v)}\):
\[ \int \frac{\sin(v)}{1 + \sin(v)} dv = \int dx \]

Using the substitution \(u = 1 + \sin(v)\), we can find:
\[ \int \frac{1}{u} du = \ln|u| + C = \ln|1 + \sin(v)| + C \]

The right side integrates to:
\[ x + C \]

Thus, we have:
\[ \ln|1 + \sin(v)| = x + C \]

**Hint:** When integrating, pay attention to the substitution and the limits of integration.

### Step 6: Solve for \(v\)
Exponentiating both sides gives:
\[ 1 + \sin(v) = e^{x + C} = e^C e^x \]
Let \(K = e^C\):
\[ 1 + \sin(v) = K e^x \]

**Hint:** Remember to exponentiate to eliminate the logarithm.

### Step 7: Substitute Back for \(y\)
Recall that \(v = x + y\):
\[ 1 + \sin(x + y) = K e^x \]
Rearranging gives:
\[ \sin(x + y) = K e^x - 1 \]

**Hint:** Substituting back to the original variables is crucial for the final solution.

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