The equation $(6$xy$^(4)+4$x$^3)$y$^2))$ dx $+(6$x$^2)$y$^3)-1)$ dy $=0$ in differential form M $$ dx $+$ N $$ dy $=0$is not exact. Indeed, we have M $$ y $$ N $$ x $=$ For this exercise we can find an integrating factor which is a function of x $$alone since M $$ y $$ N $$ xN $=1$ can be considered as a function of x $$alone$.$ Namely we have $\backslash$ mu $($ x $)=$ e $^$ x Multiplying the original equation by the integrating factor we obtain a new equation Mdx $+$ Ndy $=0$where M $=5$ e $^$ xe $^(-3$ y $)-(20$ x $^4$ y $^5+4$ sin $($ x $))$ N $=-20$ x $^5$ y $^4-15$ e $^$ xe $^(-3$ y $)$ Which is exact since My $=-15$ e $^$ xe $^(-3$ y $)-100$ x $^4$ y $^4$ Nx $=-15$ e $^$ xe $^(-3$ y $)-100$ x $^4$ y $^4$ are equal. This problem is exact. Therefore an implicit general solution can be written in the form F $($ x $,$ y $)=$ C $$where F $($ x $,$ y $)=$