Question

The solution of the equation \(\frac{dy}{dx} = \frac{x(2\log x + 1)}{\sin y + y\cos y}\) is

• A. \(y\sin y = x^2\log x + \frac{x^2}{y} + c\)
• B. \(y\cos y = x^2(\log x + 1) + c\)
• C. \(y\cos y = x^2\log x + \frac{x^2}{2} + c\)
• D. \(y\sin y = x^2\log x + c\)

Hint:

Separate variables and integrate by parts where needed.

Answer 1

The Correct Option is D

Step 1: Formula / Definition}
\[ (\sin y + y\cos y)dy = x(2\log x + 1)dx \]
Step 2: Calculation / Simplification}
LHS: \(\int (\sin y + y\cos y)dy = -\cos y + y\sin y + \cos y = y\sin y\)
RHS: \(\int x(2\log x + 1)dx = \int (2x\log x + x)dx\)
\(= x^2\log x - \int x dx + \frac{x^2}{2} = x^2\log x - \frac{x^2}{2} + \frac{x^2}{2} = x^2\log x\)
\(\therefore y\sin y = x^2\log x + c\)
Step 3: Final Answer
\[ y\sin y = x^2\log x + c \]