The differential equation $\frac{dy}{dx} + xsin \, 2y = x^3 cos^2 y$ when transformed to linear form becomes
Answer & explanation
Correct answer: option 3
The correct answer is option (3) : $\frac{dz}{dx}+2xz=x^3$
We have,
$\frac{dy}{dx} + x\, sin 2y =x^3\, cos^2\, y $
$⇒sec^2y\frac{dy}{dx}+ (2\, tan \, y) x= x^3$
Putting $tan\, y = 2 $ and $sec^2 \, y \frac{dy}{dx}=\frac{dz}{dx},$ we get
$\frac{dz}{dx}+2xz=x^3$