Question

The general solution of the differential equation $\cos x \cdot \sin y dx + \sin x \cdot \cos y dy = 0$ is

• A. $\sin x + \sin y = c$
• B. $\sin x \cdot \sin y = c$
• C. $\cos x + \cos y = c$
• D. $\cos x \cdot \cos y = c$

Hint:

When integrating terms that all result in logarithmic functions, it is highly advantageous to also write the arbitrary constant $c$ as $\log c$. This makes combining all terms via log properties seamless.

Answer 1

The Correct Option is B

Step 1: Understanding the Question:
We need to find the general solution to a first-order differential equation using the variable separable method.

Step 2: Detailed Explanation:
The given differential equation is:
$$\cos x \cdot \sin y dx + \sin x \cdot \cos y dy = 0$$ Rearrange the terms to separate the variables $x$ and $y$ on opposite sides:
$$\cos x \cdot \sin y dx = -\sin x \cdot \cos y dy$$ Divide both sides by $(\sin x \cdot \sin y)$ to completely isolate $x$ terms with $dx$ and $y$ terms with $dy$:
$$\frac{\cos x}{\sin x} dx = -\frac{\cos y}{\sin y} dy$$ This simplifies to:
$$\cot x dx = -\cot y dy$$ Integrate both sides:
$$\int \cot x dx = -\int \cot y dy$$ The standard integral of $\cot \theta$ is $\log|\sin \theta|$. Introduce an arbitrary integration constant $\log c$:
$$\log|\sin x| = -\log|\sin y| + \log c$$ Rearrange the logarithmic terms to one side:
$$\log|\sin x| + \log|\sin y| = \log c$$ Apply the logarithmic property $\log a + \log b = \log(ab)$:
$$\log|\sin x \cdot \sin y| = \log c$$ By removing the logarithm from both sides, we get the final general solution:
$$\sin x \cdot \sin y = c$$

Step 3: Final Answer:
The general solution is $\sin x \cdot \sin y = c$, matching option (B).