Question

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

• A. $(1+\cos x)(1+\sin y)=c$, where $c$ is a constant of integration.
• B. $1+\sin x+\cos y=c$, where $c$ is a constant of integration.
• C. $(1+\sin x)(1+\cos y)=c$, where $c$ is a constant of integration.
• D. $1+\sin x \cos y=c$, where $c$ is a constant of integration.

Hint:

Calculus Tip:When dealing with rational integrals where the numerator is the derivative of the denominator ($\int \frac{f'(x)}{f(x)}dx$), the result is always $\ln|f(x)|$. It is a common and helpful convention to write the arbitrary constant as $\ln|c|$ instead of $c$ when all other terms in the integrated equation are natural logarithms.

Answer 1

The Correct Option is C

Concept:
Differential Equations - Variable Separable Form.

Step 1: Identify the equation type and strategy.
The given equation is $\cos x(1+\cos y)dx - \sin y(1+\sin x)dy = 0$. This first-order differential equation can be solved by separating the variables $x$ and $y$.

Step 2: Separate the variables.
Rearrange the terms to group all $x$ terms with $dx$ and all $y$ terms with $dy$. Moving the $dy$ term to the right side gives $\cos x(1+\cos y)dx = \sin y(1+\sin x)dy$.

Dividing both sides by $(1+\sin x)(1+\cos y)$ yields: $\frac{\cos x}{1+\sin x}dx = \frac{\sin y}{1+\cos y}dy$. Alternatively, expressed as: $\frac{\cos x}{1+\sin x}dx - \frac{\sin y}{1+\cos y}dy = 0$.

Step 3: Set up the integration.
Integrate both sides of the separated equation: $\int \frac{\cos x}{1+\sin x}dx - \int \frac{\sin y}{1+\cos y}dy = 0$.

Step 4: Perform the integration.
For the first integral, the derivative of the denominator $(1+\sin x)$ is exactly the numerator $(\cos x)$.
Thus, the integral is $\ln|1+\sin x|$. For the second integral, the derivative of $(1+\cos y)$ is $-\sin y$. Because of the existing minus sign, $-\int \frac{\sin y}{1+\cos y}dy$ becomes $+\int \frac{-\sin y}{1+\cos y}dy$, which evaluates to $\ln|1+\cos y|$.
Adding a constant of integration $\ln|c|$, the equation becomes: $\ln|1+\sin x| + \ln|1+\cos y| = \ln|c|$.

Step 5: Apply logarithmic properties to find the final general solution.
Use the product property of logarithms, which states that $\ln A + \ln B = \ln(AB)$. Applying this to the left side of our equation gives us $\ln|(1+\sin x)(1+\cos y)| = \ln|c|$.
To eliminate the natural logarithms, we take the antilogarithm (or exponentiate both sides with base $e$).
This cancels out the $\ln$ functions, leaving us with the final algebraic expression: $(1+\sin x)(1+\cos y) = c$. $$ \therefore \text{The general solution is } (1+\sin x)(1+\cos y) = c $$