Let $f$ be an arbitrary continuously differentiable function. If the ordinary differential equation $$(3y^2-x)f(x+y^2)+2y(y^2-3x)f(x+y^2)y'=0$$
is exact. Then which of the following are true
1) $xf'(x)-3f(x)=0$
2) $xf'(x)+3f(x)=0$
3) $f'(x)+3f(x)=0$
4) $f'(x)-3f(x)=0$
I tried using $Mdx+Ndy=0$, but I did not get any option. Please help
The second option seems correct to me ..
I used f for $f(y^2+x)$ for simplicity except in the last line
$$(3y^2-x)f(x+y^2)dx+2y(y^2-3x)f(x+y^2)dy=0$$
$$Pdx+Qdy=0$$
$$\partial_y P= f\times 6y+2yf' \times (3y^2-x)$$ $$\partial_x Q= f\times (-6y)+f' \times 2y(y^2-3x)$$ Equate both expressions for exactness $$\partial_y P=\partial_x Q$$
$$ f\times 6y+2yf' \times (3y^2-x)- f\times (-6y)-f' \times 2y(y^2-3x)=0$$ $$12yf+4y(y^2+x)f'=0$$ $$(y^2+x)f'=-3f$$ $$(y^2+x)f'(y^2+x)=-3f(y^2+x) \implies xf'(x)=-3f(x)$$