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The differential equation $\frac{dy}{dx} + xsin \, 2y = x^3 cos^2 y$ when transformed to linear form becomes |
$\frac{dz}{dx}+\frac{z}{x^2}=x$ $\frac{dz}{dx}+zx=\frac{x^3}{2}$ $\frac{dz}{dx}+2xz=x^3$ $\frac{dz}{dx}-\frac{z}{x}=x^2$ |
$\frac{dz}{dx}+2xz=x^3$ |
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 $\frac{dz}{dx}+2xz=x^3$ |
