Question

The solution of the equation $\frac{d y}{d x}=\frac{x(2 \log x+1)}{\sin -y+y \cos y}$

• 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. $y \sin y=x^{2} \log x+c$

Answer 1

Answer: (D)

$(y \cos y+\sin y) d y=(2 x \log x+x) d x$
$y \sin y-\int \sin y\, d y+\int \sin\, y \,d y$
$=x^{2} \log x-\int x^{2} \cdot \frac{1}{x} d x+\int x d x+c$
$\therefore y \sin y=x^{2} \log x+c$