By a suitable substitution or differentiation, reduce…

By a suitable substitution or differentiation, reduce $(x^2-1)y'\sin y+2x\cos y=2x-2x^3$ to linear form and find its general integral.

Step 1: Substitute $u=\cos y$

Let $u=\cos y$, so $u'=-\sin y\cdot y'$, i.e. $y'\sin y=-u'$. The equation becomes:

\[(x^2-1)(-u')+2xu=2x-2x^3=-2x(x^2-1).\]

Step 2: Rearrange to standard linear form

Multiply by $-1$: $(x^2-1)u'-2xu=2x(x^2-1)$. Dividing by $x^2-1$:

\[u'-\frac{2x}{x^2-1}u=2x.\]

Step 3: Solve the linear equation in $u$

The integrating factor is $\mu(x)=e^{-\int \frac{2x}{x^2-1}dx}=e^{-\ln|x^2-1|}=\dfrac{1}{x^2-1}$, so $\dfrac{d}{dx}\!\left(\dfrac{u}{x^2-1}\right)=\dfrac{2x}{x^2-1}$, giving

\[\frac{u}{x^2-1}=\ln|x^2-1|+C\quad\Longrightarrow\quad u=(x^2-1)\big(\ln|x^2-1|+C\big).\]