Real and complex solutions of cubic implicit equation

Question

I have faced this differential problem: $(y'(x))^3 = 1/x^4$.

From the fundamental theorem of algebra i know there exist 3 solutions $y_1$, $y_2$, $y_3$, but formally how can I procede to deduce that?

Answer 1

If $y'(x)^3=\frac{1}{x^4}$, then $y(x)=-3 x^{-1/3}+c$ ........

Answer 2

Hint: Write $$\frac{dy}{dx}=x^{-4/3}$$ Use the factorization of $$a^3-b^3=(a-b)(a^2+ab+b^2)$$ This is $$(y'(x)-x^{-4/3})(y'(x)^2+y'(x)x^{-4/3}+x^{-8/3})$$

Answer 3

Every number has three cube roots. You only used one of them.

In particular, if a is real number then the three cube roots are $a^{1/3}$, a real number, $a^{1/3}(cos(2\pi/3)+ isin(2\pi/3))$, and $a^{1/3}(cos(4\pi/3)+ isin(4\pi/3))$.

Given $(y'(x))^3= \frac{1}{x^4}= x^{-4}$, we have $y'(x)= x^{-4/3}$, $y'(x)= x^{-4/3}(cos(2\pi/3)+ isin(2\pi/3))$, and $y'(x)= x^{-4/3}(cos(4\pi/3)+ isin(4\pi/3))$.