Q89 / 160 · AP EAMCET 2003 · Mathematics

Solution of $\frac{d y}{d x}=\frac{x \log x^2+x}{\sin y+y \cos y}$ is

  • Correct answer
    A
    $y \sin y=x^2 \log x+C$
  • B
    $y \sin y=x^2+C$
  • C
    $y \sin y=x^2+\log x$
  • D
    $y \sin y=x \log x+C$

Solution

We have, $\frac{d y}{d x}=\frac{x \log x^2+x}{\sin y+y \cos y}$ $\Rightarrow(\sin y+y \cos y) d y=\left(x \log x^2+x\right) d x$ On integrating both sides, we get $-\cos y+y \sin y+\cos y$$=\frac{x^2}{2} \log x^2-\frac{x^2}{2}+\frac{x^2}{2}+C$ $\Rightarrow \quad y \sin y=\frac{x^2}{2} \log x^2+C$ $\therefore \quad y \sin y=x^2 \log x+C$

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

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