Given that $$(xy^3+x^2y^7)y'=1$$ satisfies the initial condition $y(\frac{1}{4}) = 1$. Then the value of $y'$ when $y=-1$ is:
- (A) $\frac{4}{3}$.
- (B) $-\frac{4}{3}$.
- (C) $\frac{16}{5}$.
- (D) $-\frac{16}{5}$.
I doubt if it necessary to solve for $y'$ in order to proceed. Please help.