I am struggling with an exercise. Can you please give me a hint?


Show that the solution curves of the differential equation:

$\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$, are of the form $x^3+y^3=3Cxy$.

I tried the substitution $u=y/x \rightarrow y=xu, \frac{dy}{dx}=u+x\frac{du}{dx}$.

Hence I get:


This gives:




So I can atleast reduce the problem to a seperable differential equation, but I am not able to integrate the left side up. Do you have any tips?


Use partial fractions

$$\begin{align} \frac{2u^{3} - 1}{u^{4} + u} &= \frac{2u^{3} - 1}{u(u^{3} + 1)} \\ &= \frac{A}{u} + \frac{Bu^{2} + Cu + D}{u^{3} + 1} \\ \implies 2u^{3} - 1 &= (A + B)u^{3} + Cu^{2} + Du + A \\ \end{align}$$

Equating coefficients

$$\begin{align} A &= -1 \\ B &= 3 \\ \end{align}$$


$$\begin{align} \frac{2u^{3} - 1}{u^{4} + u} &= \frac{3u^{2}}{u^{3} + 1} - \frac{1}{u} \end{align}$$


$$\begin{align} \int \bigg(\frac{3u^{2}}{u^{3} + 1} - \frac{1}{u} \bigg)du &= - \int \frac{1}{x} dx \\ \implies \ln(u^{3} + 1) - \ln(u) &= -\ln(x) + K_1 \\ \implies \ln \bigg( \frac{u^{3} + 1}{u} \bigg) &= - \ln(x) + K_1 \\ \implies \frac{u^{3} + 1}{u} &= \frac{K_2}{x} \\ \end{align}$$


$$\begin{align} u &= \frac{y}{x} \\ \implies \frac{y^{2}}{x^{2}} + \frac{x}{y} &= \frac{K_2}{x} \\ \implies y^{3} + x^{3} &= K_2xy \end{align}$$

  • $\begingroup$ No worries mate. $\endgroup$ – Mattos Feb 5 '15 at 16:06

Of course, it is possible to solve the ODE, but it is not what is wanted.

The question is : Show that the solution is on the given form.

Proving that the given solution is correct is not the same as finding the solution : It is a waste of time to try to solve the ODE since it is already solved and the solution known.

It is sufficient to bring back the known solution : $x^3+y^3=3Cxy$ into the ODE $\frac{dy}{dx}=-\frac{y(2x^3-y^3)}{x(2y^3-x^3)}$ and show that it agrees. The differentiation of $x^3+y^3=3Cxy$ leads to : $$3x^2 dx+3y^2dy=3Cy dx+3Cx dy$$ $$\frac{dy}{dx}=-\frac{3x^2+3Cy}{3y^2-3Cx}$$ From $x^3+y^3=3Cxy$ we have $3C=\frac{x^3+y^3}{xy}$. Use it to eliminate $C$.

  • $\begingroup$ Thanks, but I have a question. You have shown that if we have an y-function, satisfying the last equation, then it satisfies the original differential-equation. But is the converse nececarrily true?, that is, if the solution satisifies the differential equation, then it satisifies the given equation? I mean, showing that the solution of a given differential equation satisfies a condition, is not the same as showing that all solutions that satisfies that condition also satisfies the differential equation? $\endgroup$ – user119615 Feb 5 '15 at 16:02
  • 1
    $\begingroup$ I mean, haven't you shown the opposite implication of what they asked? $\endgroup$ – user119615 Feb 5 '15 at 16:05
  • $\begingroup$ You do not show that the solution satisfies a condition (by the way, what condition are you talking about ?). You show that the set of functions, given on the implicite form $x^3+y^3=3Cxy$ whit any $C$, satisfies the ODE. So they are the solutions of the ODE. $\endgroup$ – JJacquelin Feb 5 '15 at 16:10
  • $\begingroup$ The condition was $x^3+y^3=Cxy$. What I meant was that you are right that the equations that satisfy $x^3+y^3=Xcy$, satisfies the differential equation. But theoretically there might be more solutions of the differential equation that does not satisfy $x^3+y^3=Cxy$, so by solving the differential equation you show that indeed all solutions satisfy $x^3+y^3=Cxy$. $\endgroup$ – user119615 Feb 5 '15 at 16:16
  • $\begingroup$ If the original wording of the problem has been well rewriten, I understand that what is wanted is to verify that the given implicite function is solution of the given ODE. That's all. But, of course, I could be wrong in translating the wording of the question. $\endgroup$ – JJacquelin Feb 5 '15 at 16:21

$\bf hint:$ so you need to integrate $\int \frac{2u^3 - 1}{u^4 + u}$ we can do this by using partial fraction. here is how it goes $$ \frac{2u^3 - 1}{u^4 + u} = \frac{1}{2}\left(\frac{4u^3 +1}{u^4 + u} - \frac{3}{u(u+1)(u^2-u+1)}\right)$$

now you need to find the constants $A, B, C$ and $D$ so that $$ \frac{3}{u(u+1)(u^2-u+1)} = \frac{A}{u} + \frac{B}{u+1} + \frac{C(2u-1)}{u^2-u+1} + \frac{D}{u^2 - u + 1} $$ is an identity.

finding $A, B$ are easier. just by looking at the behaviour near $u = 0, u = -1.$ they turn out to be $A = 3, B = -1.$


here is a cheap way of doing this knowing that $x^3 + y^3 = Cxy$ is a solution. we will make the change of variables $$u = x^3+y^3, t = xy$$

the deifferential equation satisfied by $u$ and $t$ are:

$$\frac{du}{dx} = 3x^2 + 3y^2 \frac{dy}{dx} = \frac{3(y^6-x^6)}{x(2y^3 - x^3}, \frac{dt}{dx} = x\frac{dy}{dx} + y = \frac{3y(y^3-x^3)}{2y^3 - x^3} $$

we can divide it out and get a separable equation $$\frac{du}{dt} = \frac{u}{t} $$ which has solution of the form $$u = Ct, x^3 + y^3 = Cxy$$


Left hand side simplifies by integration:

$ \int \dfrac {2 u^3 - 1} {u(u^3 +1)} du = log \dfrac{(1+u^3)}{u} + const. $


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