Find $\frac{dy}{dx}$ when $y=-1$ of the differential equation $(xy^3 +x^2y^7)\frac{dy}{dx} =1$

Question

Suppose a solution of the differential equation

$(xy^3 +x^2y^7)\frac{dy}{dx} =1$

satisfies the initial condition $y(1/4)=1$. Then the value of $\frac{dy}{dx}$ when $y=-1$

is

(A) $4/3$

(B) $-4/3$

(C) $16/5$

(D)$-16/5$

Cannot separate the two variables and hence cannot come to a solution. It is a 1 mark MCQ question in an exam. There must be some quick method to reach to the answer.

Answer 1

Consider the inverse function $x(y)$. $$\frac{dx}{dy}=xy^3 +x^2y^7 \quad\text{with condition}\quad x(1)=\frac14$$ By inspection only, changing the sign of the variable $y$ changes nothing. Thus the function $x(y)$ is even. As a consequence : $$x(-y)=x(y)\quad\text{especially}\quad x(-1)=x(1)=\frac14$$ So, at point $(y=-1\:,\:x=\frac14)$ : $$\frac{dx}{dy}=\frac14(-1)^3 +\left(\frac14\right)^2(-1)^7=-\frac14-\frac{1}{16}=-\frac{5}{16}$$ $$\frac{dy}{dx}=-\frac{16}{5}$$ $$\text{The correct answer is : (D).}$$

Answer 2

Simple hint $$(xy^3 +x^2y^7)\frac{dy}{dx} =1$$ It's Bernouilli's equation $$(xy^3 +x^2y^7) =x'$$ $$x'-xy^3 =x^2y^7$$ $$\frac {x'}{x^2}-\frac {y^3}x =y^7$$ Substitute $z=\frac 1x$ $$z'+zy^3 =-y^7$$ It's easy to solve now