Question

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

• A. $y \,\sin\, y = x^2\, \log \,x + C$
• B. $y\, \sin \,y = x^2 + C$
• C. $y \,\sin\, y = x^2 + \log\, x + C$
• D. $y\, \sin \,y = x\, \log\, x + C$

Answer 1

Answer: (A)

Given equation is $\frac{d y}{d x}=\frac{x \log x^{2}+x}{\sin y+y \cos y}$
$\Rightarrow (\sin y+y \cos y) d y=\left(x \log x^{2}+x\right) d x$
On integrating both sides, we get
$\int(\sin y+y \cos y) d y=\int\left(x \log x^{2}+x\right) d x$
$\Rightarrow -\cos y+y \sin y+\cos y=\frac{x^{2}}{2} \log x^{2}-\int \frac{x^{2}}{2} \cdot \frac{1}{x^{2}} 2 x \,d x+\int x d x+C$
$\Rightarrow y \sin y=\frac{x^{2}}{2} \cdot 2 \log x-\int x\, d x+\int x \,d x+C$
$\Rightarrow y \sin y=x^{2} \log x+C$