Find the general integral of y'x^3sin y=xy'-2y

Find the general integral of $y'x^3\sin y=xy'-2y$.

Explanation of Method

Inverting to treat $x$ as a function of $y$ gives $\dfrac{dx}{dy}-\dfrac{x}{2y}=-\dfrac{\sin y}{2y}x^3$, a Bernoulli equation ($n=3$); substitute $v=x^{-2}$.

Step 1: Invert and rearrange

From $y'(x^3\sin y-x)=-2y$: $\dfrac{dx}{dy}=\dfrac{x^3\sin y-x}{-2y}$, i.e. $\dfrac{dx}{dy}-\dfrac{x}{2y}=-\dfrac{x^3\sin y}{2y}$.

Step 2: Substitute $v=x^{-2}$

Multiplying by $-2/x^3$ and using $v'=-2x^{-3}dx/dy$: $v'+\dfrac{v}{y}=\dfrac{\sin y}{y}$.

Step 3: Solve the linear equation

With $\mu(y)=y$: $\dfrac{d}{dy}(yv)=\sin y$, so $yv=-\cos y+C$, i.e. $v=\dfrac{C-\cos y}{y}$.

Result

Final answer:

\[\boxed{\,x^2(C-\cos y)=y,\qquad C\in\mathbb{R}}\]