Question

The transformed equation of $\frac{dy}{dx}+x\sin 2y=x^3{\cos }^{2}y$ is

• A. $\frac{du}{dx}+\frac{u}{x^2}=x$
• B. $\frac{du}{dx}+xu=\frac{x^3}{2}$
• C. $\frac{du}{dx}+2xu=x^3$
• D. $\frac{du}{dx}+\frac{u}{x}=x^2$

Answer 1

The correct option is C $\frac{du}{dx}+2xu=x^3$

Given :
$\frac{dy}{dx}+x\sin 2y=x^3{\cos }^{2}y\Rightarrow {\sec }^{2}y\frac{dy}{dx}+2x\tan y=x^3$
put $\tan y=u\Rightarrow {\sec }^{2}y\frac{dy}{dx}=\frac{du}{dx}$
The equation becomes,
$\frac{du}{dx}+2xu=x^3$