Solve $y'+xy=y^3$

Question

solving it using reduction to linear from Bernoulli equation. $y'+xy=y^3$.

should I use first $u=y^{1-3}$

Then, what will be the remaining procedure.

Answer 1

$u=y^{1-3}=y^{-2}=>y=u^{-0.5}=>y'=-u^{-3/2}u'/2$ $$ \\-u^{-3/2}u'/2+xu^{-1/2}=u^{-3/2} \\u\equiv0 \;\text or\; u'-2xu=-1 \\u_0'=2xu_0 \\\ln|u_0|=x^2+const \\u_0=ce^{x^2}(c=const) \\u=f(x)e^{x^2} \\u'=f'e^{x^2}+2xfe^{x^2} \\u'-2xu=f'e^{x^2}=-1 \\f'=e^{-x^2} \\f=\int e^{-x^2}dx \\y=\frac{1}{\sqrt{u}}=\dfrac{1}{\sqrt{e^{x^2}\int e^{-x^2}dx}} $$ Answer $y=\dfrac{1}{\sqrt{e^{x^2}\int e^{-x^2}dx}}$ or $y\equiv 0$