Question

The solution of the differential equation \( \frac{dy}{dx}+y\log y\cot x=0 \) is

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

Hint:

For separable differential equations, keep all \(y\)-terms with \(dy\) and all \(x\)-terms with \(dx\), then integrate both sides.

Answer 1

The Correct Option is C

Step 1: Write the given differential equation.
\[ \frac{dy}{dx}+y\log y\cot x=0. \]

Step 2: Rearrange the equation.
\[ \frac{dy}{dx}=-y\log y\cot x. \]

Step 3: Separate variables.
\[ \frac{dy}{y\log y}=-\cot x\,dx. \]

Step 4: Integrate both sides.
\[ \int \frac{dy}{y\log y}=-\int \cot x\,dx. \]

Step 5: Evaluate the integrals.
We know:
\[ \int \frac{dy}{y\log y}=\log(\log y), \] and
\[ \int \cot x\,dx=\log(\sin x). \]
Therefore:
\[ \log(\log y)=-\log(\sin x)+C. \]

Step 6: Simplify logarithmic expression.
\[ \log(\log y)+\log(\sin x)=C. \]
\[ \log(\sin x\log y)=C. \]
Taking exponential on both sides:
\[ \sin x\log y=c. \]

Step 7: Final conclusion.
Thus, the solution is:
\[ \boxed{\sin x\log y=c}. \]