Question

The solution of $\frac{dy}{dx}=\frac{x\log x^2+x}{\sin y+y\cos y}$

• (A) y sin y = x² log x + c
• (B) y sin y = x² + c
• (C) y sin y = x² + log x + c
• (D) y sin y = x log x + c

Answer 1

The Correct Option is A

Explanation:
Given equation is $\frac{dy}{dx}=\frac{x\log x^2+x}{\sin y+y\cos y}$$\Rightarrow (\sin y+y\cos y)dy=(x\log x^2+x)dx$On integrating both sides, we get $\int (\sin y+y\cos y)dy=\int (x\cdot \log x^2+x)dx$$\Rightarrow -\cos y+y\sin y+\cos y=\frac{x^2}{2}\log x^2$$-\int \frac{x^2}{2}\cdot \frac{1}{x^2}2xdx+\int xdx+c$$\Rightarrow y\sin y=\frac{x^2}{2}\cdot 2\log x-\int xdx+\int xdx+c$$\Rightarrow y\sin y=x^2\log x+c$