<s>[INST] How to solve this first order ODE right Here is ODE of first order:
$$ y'=- \frac{x^2}{y^3} $$
with $y(0)=1 , y(0)=-1 $
Can my method be the separation method? If I use it, it doesn't work out:
$ \frac{dy}{dx}= -x^2 \frac{1}{y^3} $
$ \leftrightarrow y^3 dy = -x^2 dx $
By integration I get:
$$ \int y^3 dx = \int -x^2 dx $$
$$ \leftrightarrow \frac{1}{4} y^4+c = - \frac{1}{3} x^3+c $$
$$ \leftrightarrow y= \mp \sqrt[4]{ - \frac{4}{3} x^3 +c } $$
Where is my mistake?
 [/INST] You are correct. If $y\not=0$ then
$$y'=- \frac{x^2}{y^3}\leftrightarrow \int y^3 dy = \int -x^2 dx \leftrightarrow \frac{1}{4} y^4= - \frac{1}{3} x^3+c  \leftrightarrow y= \pm \sqrt[4]{ - \frac{4}{3} x^3 +c }. $$
Now for $y(0)=1$, we get
$$ y(x)= \color{blue}{+}\sqrt[4]{1 - \frac{4}{3} x^3},$$
whereas for $y(0)=-1$ we find
$$ y(x)= \color{blue}{-}\sqrt[4]{1 - \frac{4}{3} x^3}.$$
Both solutions, which are symmetric with respect to the line $y=0$, are defined in the half-line $(-\infty,(3/4)^{1/3})$.
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