<s>[INST] Differential equation $y'=\frac{1-xy}{y-x^2}$ I tried many thing but I could not find a method to solve the  differential equation $$y'=\frac{1-xy}{y-x^2}$$
$$y'=-x+\frac{1-x^3}{y-x^2}$$
$Z=y-x^2$
$$Z'+2x=-x+\frac{1-x^3}{Z}$$
$$Z(Z'+3x)=1-x^3$$
Could you please give me hint which method can be used for such first order differential equations?
Thanks a lot for advice and answers
 [/INST] $y'=\dfrac{1-xy}{y-x^2}$
$(y-x^2)y'=1-xy$
Let $u=y-x^2$ ,
Then $y=u+x^2$
$y'=u'+2x$
$\therefore u(u'+2x)=1-x(u+x^2)$
$uu'+2xu=1-xu-x^3$
$uu'=-3xu-x^3+1$
Try to solve this ODE by Wolfram Alpha, you will discover that the general solution is implicitly expressed by $x$ and $x+\dfrac{x^3-1}{u}$ .
Since Wolfram Alpha discover that the substitution $v=x+\dfrac{x^3-1}{u}$ leads the ODE becomes a separable ODE:
Let $v=x+\dfrac{x^3-1}{u}$ ,
Then $u=\dfrac{x^3-1}{v-x}$
$u'=\dfrac{3x^2(v-x)-(x^3-1)(v'-1)}{(v-x)^2}$
$\therefore\dfrac{x^3-1}{v-x}\dfrac{3x^2(v-x)-(x^3-1)(v'-1)}{(v-x)^2}=-3x\dfrac{x^3-1}{v-x}-x^3+1$
$\dfrac{3x^2(v-x)-(x^3-1)(v'-1)}{(v-x)^2}=-3x-(v-x)$
$3x^2(v-x)-(x^3-1)(v'-1)=-3x(v-x)^2-(v-x)^3$
$(x^3-1)(v'-1)=3x^2(v-x)+3x(v-x)^2+(v-x)^3$
$(x^3-1)v'-x^3+1=3x^2(v-x)+3x(v-x)^2+(v-x)^3$
$(x^3-1)v'=x^3+3x^2(v-x)+3x(v-x)^2+(v-x)^3-1$
$(x^3-1)v'=(x+v-x)^3-1$
$(x^3-1)v'=v^3-1$
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