Question

General Solution of the differential equation:
\(cos\,x(1+cos\,y)dx-sin\,y(1+sin\,x)dy=0\) is:

• A. \((1+cos\,x)(1+sin\,y)\)= c
• B. \(1+sin\,x+cos\,y=c\)
• C. \((1+sin\,x)(1+cos\,y)=c\)
• D. \(1+sin\,x.cos\,y=c\)

Answer 1

The Correct Option is C

Let's solve the given differential equation step by step.
1. Separate the Variables:
The given equation is: \[ \cos(x)(1 + \cos(y)) \, dx - \sin(y)(1 + \sin(x)) \, dy = 0 \] Rearranging the terms, we get: \[ \cos(x)(1 + \cos(y)) \, dx = \sin(y)(1 + \sin(x)) \, dy \] Dividing both sides by their respective terms: \[ \frac{\cos(x)}{1 + \sin(x)} \, dx = \frac{\sin(y)}{1 + \cos(y)} \, dy \]

 

2. Integrate Both Sides:
Now we can integrate both sides of the equation: \[ \int \frac{\cos(x)}{1 + \sin(x)} \, dx = \int \frac{\sin(y)}{1 + \cos(y)} \, dy \]

3. Solve the Integrals:
* Left side integral:
Let \( u = 1 + \sin(x) \). Then, \( du = \cos(x) \, dx \). The integral becomes: \[ \int \frac{1}{u} \, du = \ln|u| = \ln|1 + \sin(x)| \] 

* Right side integral:
Let \( v = 1 + \cos(y) \). Then, \( dv = -\sin(y) \, dy \). The integral becomes: \[ \int \frac{-1}{v} \, dv = -\ln|v| = -\ln|1 + \cos(y)| \]

4. Combine the Results:
Combining the results of the integrals, we get: \[ \ln|1 + \sin(x)| = -\ln|1 + \cos(y)| + C \] Adding \( \ln|1 + \cos(y)| \) to both sides: \[ \ln|1 + \sin(x)| + \ln|1 + \cos(y)| = C \] Using the property of logarithms: \[ \ln| (1 + \sin(x)) (1 + \cos(y)) | = C \]

5. Remove the Logarithm:
Exponentiating both sides to remove the logarithm: \[ (1 + \sin(x)) (1 + \cos(y)) = e^C \] Let \( e^C = K \), where \( K \) is a positive constant: \[ (1 + \sin(x)) (1 + \cos(y)) = K \]