4
$\begingroup$

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

$\endgroup$
  • $\begingroup$ Your equation is equivalent with $$(1-xy)dx+(x^2-y)dy=0,$$ but $(1-xy)'_{y}=-x \neq 2x = (x^2-y)'_y$ Maybe we could find integration factor $\endgroup$ – Cortizol Mar 4 '13 at 21:59
  • $\begingroup$ Why do you think there is an explicit solution in known functions? $\endgroup$ – GEdgar Mar 4 '13 at 22:00
  • $\begingroup$ Maybe .I tried the method too but I could not find integration factor. Thanks for advice $\endgroup$ – Mathlover Mar 4 '13 at 22:04
  • $\begingroup$ Check this solution. $\endgroup$ – Mhenni Benghorbal Mar 4 '13 at 22:28
  • $\begingroup$ @MhenniBeghorbal I would like to learn methods to get that result $\endgroup$ – Mathlover Mar 4 '13 at 22:35
1
$\begingroup$

$y'=\dfrac{1-xy}{y-x^2}$

$(y-x^2)y'=1-xy$

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 $u=\dfrac{x^3-1}{v-x}$

$u'=\dfrac{3x^2(v-x)-(x^3-1)(v'-1)}{(v-x)^2}$

$\therefore\dfrac{x^3-1}{v-x}\dfrac{3x^2(v-x)-(x^3-1)(v'-1)}{(v-x)^2}=-3x\dfrac{x^3-1}{v-x}-x^3+1$

$\dfrac{3x^2(v-x)-(x^3-1)(v'-1)}{(v-x)^2}=-3x-(v-x)$

$3x^2(v-x)-(x^3-1)(v'-1)=-3x(v-x)^2-(v-x)^3$

$(x^3-1)(v'-1)=3x^2(v-x)+3x(v-x)^2+(v-x)^3$

$(x^3-1)v'-x^3+1=3x^2(v-x)+3x(v-x)^2+(v-x)^3$

$(x^3-1)v'=x^3+3x^2(v-x)+3x(v-x)^2+(v-x)^3-1$

$(x^3-1)v'=(x+v-x)^3-1$

$(x^3-1)v'=v^3-1$

$\endgroup$
  • $\begingroup$ Thanks a lot for wonderful answer. I wonder if we did not get help of wolframalpha how we could find the transform. $\endgroup$ – Mathlover Mar 5 '13 at 1:50
1
$\begingroup$

Equation $$zz'=-3xz-x^3+1$$ is Abel equation of the second kind. You can find here some solvable Abel equation.

Good luck with finding your equation!

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.