Q125 / 200 · Manipal (MET) 2013 · Mathematics

The solution of the equation $\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin y+y \cos y}$ 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
    $y \cos y=x^2 \log x+\frac{x^2}{2}+c$
  • Correct answer
    D
    $y \sin y=x^2 \log x+c$

Solution

$\begin{aligned} & (y \cos y+\sin y) d y=(2 x \log x+x) d x \\ & y \sin y-\int \sin y d y+\int \sin y d x\end{aligned}$ $=x^2 \log x-\int x^2 \cdot \frac{1}{x} d x+\int x d x+c$ $\therefore \quad y \sin y=x^2 \log x+c$

Hence, the correct answer is option: 'D'.

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