Question

The solution of the differential equation \( x \frac{dy}{dx} = \cot y \) is

• A. \( y \cos x = c \)
• B. \( x \cos y = c \)
• C. \( \log(x \cos y) = c \)
• D. \( \log(y \cos x) = c \)

Hint:

Be careful with separating variables—always check whether it's a direct separable form and apply identities (like \( \int \tan y \, dy = -\ln|\cos y| \)) correctly.

Answer 1

The Correct Option is B

We are given the differential equation: \[ x \frac{dy}{dx} = \cot y \] Step 1: Rearranging the equation
Divide both sides by \(x\) to isolate \( \frac{dy}{dx} \): \[ \frac{dy}{dx} = \frac{\cot y}{x} \] Step 2: Separate the variables
\[ \sin y \cos y \frac{dy}{\sin^2 y} = \frac{dy}{\tan y} = \frac{dx}{x} \Rightarrow \tan y \, dy = \frac{dx}{x} \] Actually, let's directly rearrange: \[ \cot y \, dy = \frac{dx}{x} \] Step 3: Integrate both sides
\[ \int \cot y \, dy = \int \frac{dx}{x} \] Recall that: \[ \int \cot y \, dy = \ln|\sin y| + C_1 \quad \text{and} \quad \int \frac{1}{x} dx = \ln|x| + C_2 \] So: \[ \ln|\sin y| = \ln|x| + c \quad \text{(where } c = C_2 - C_1 \text{)} \] Step 4: Exponentiate both sides
\[ |\sin y| = A|x| \quad \text{where } A = e^c \] \[ x \frac{dy}{dx} = \cot y \Rightarrow \tan y \, dy = \frac{dx}{x} \text{ is incorrect. Let's fix it.} \] Now integrate: \[ \int \tan y \, dy = \int \frac{dx}{x} \] \[ \int \tan y \, dy = -\ln|\cos y| \quad \text{and} \quad \int \frac{1}{x} dx = \ln|x| \] \[ -\ln|\cos y| = \ln|x| + c \Rightarrow \ln\left(\frac{1}{\cos y}\right) = \ln|x| + c \Rightarrow \ln\left(\frac{1}{\cos y}\right) = \ln|Ax| \quad \text{where } A = e^c \] \[ \frac{1}{\cos y} = Ax \Rightarrow x \cos y = \text{constant} \] Thus, the required solution is: \[ x \cos y = c \]