The solution of the differential equation $(x^2 y^2 -1)dy + 2xy^3 dx= 0$ is:
a) $1+x^2y^2 = cx$
b) $1+x^2 y^2 = cy$
c) $y=0$
d) $y = -\dfrac{1}{x^2}$
By converting the equation into exact differential form and integrating I got option b as the answer and it's correct.
But I would like to know why option $c$ is not a valid solution.
If we rearrange the differential equation we get:
$\dfrac {dy}{dx}= \dfrac{2xy^3}{1- x^2 y^2}$
$\implies y' = 0$ for $y = 0$
Also, originally too for $y=0 $ the slope $\dfrac{dy}{dx}= 0$.
Then why is $y = 0$ not a solution?
Answer given is only option $b$.
Best Answer
You can view the solution $y = 0$ as a limit of the more general solution in (b). Specifically, one of the possible solutions is $$ 1 - c y + x^2 y^2 = 0 \quad \Rightarrow \quad y = \frac{c - \sqrt{c^2 - 4 x^2}}{2 x^2} $$ But in the limit $c \to \infty$, and for any value of $x$, we have $$ \lim_{c \to \infty} \left[c - \sqrt{c^2 - 4 x^2} \right] = \lim_{c \to \infty} \left[\frac{4 x^2}{c + \sqrt{c^2 - 4 x^2}} \right] = 0 $$ and so $y \to 0$ for all values of $x$. Thus, the solution $y = 0$ can be viewed as a special case of the more general solution $cy = 1 + x^2 y^2$ in the limit $c \to \infty$.