Question

Solution of $\frac{dy}{dx}=\frac{x(2\log x+1)}{\sin y+y\cos y}$

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

Answer 1

The correct option is B $y\sin y=x^2\log x+c$

$(\sin y+y\cos y)\frac{dy}{dx}=2x\log x+x$

$\Rightarrow \frac{d}{dx}\left(y\sin y\right)=\frac{d}{dx}\left(x^2\log x\right)$

[By observation we see than $\sin y+y\cos y$ in $\frac{d}{dy}$ $y\sin y$]

and $2x\log x+x$ in $\frac{d}{dx}{x}^2\log x$]

$\Rightarrow \int d\left(y\sin y\right)=\int d\left(x^2\log x\right)+c$

$\Rightarrow y\sin y=x^2\log x+c$