Solve the ODE: $$(x^{2}-2xy)y'+y^{2}-2xy=0$$ My try:

$\begin{aligned}{c} (x^{2}-2xy)y'+y^{2}-2xy=0 \\ \implies (x^{2}-2xy)\frac{\mathrm{d}y}{\mathrm{d}x}+y^{2}-2xy=0\\ \implies (x^{2}-2xy)\mathrm{d}y+(y^{2}-2xy)\mathrm{d}x=0\\ \implies (x^{2}-2xy)e^{-x}\mathrm{d}y+(y^{2}-2xy)e^{-x}\mathrm{d}x=0\\ \\ \end{aligned}$

but the equation doesn't turn exact. I found no relevant integration factor.

Edit Solve with an exact equation method.


4 Answers 4


$$(x^{2}-2xy)dy+(y^{2}-2xy)dx=0$$ Divide by $x^2y^2$ $$\left(\dfrac 1 {y^{2}}-\dfrac 2 {xy}\right)dy+\left(\dfrac 1 {x^{2}}-\dfrac 2 {xy}\right)dx=0$$ $$\dfrac 1 {y^{2}}dy-\dfrac 2 {xy}d(x+y)+\dfrac 1 {x^{2}}dx=0$$ $$-d(\dfrac 1 {y})-\dfrac 2 {xy}d(x+y)-d(\dfrac 1 x)=0$$ $$d \left(\dfrac {x+y} {xy}\right)+\dfrac 2 {xy}d(x+y)=0$$ Now the integrating factor is obvious. $$\mu (x,y) =\dfrac {xy} {x+y}$$ Multiply by $\mu$ and integrate.


$$y'=\frac{2xy-y^2}{x^2-2xy}=\frac{2\frac{y}{x} -\left(\frac{y}{x}\right)^2}{1-2\frac{y}{x}}$$

$y=ux\rightarrow y'=u+u'x$


$$u'x+u =\frac{2u-u^2}{1-2u}$$

and $$u'x =\frac{u^2 +2u -1}{1-2u}$$

and we obtain

$$\frac{1-2u}{u^2 +2u +1} du =\frac{dx}{x}$$

which is easy to solve.

  • $\begingroup$ the solution which is looked for is an exact equation based. $\endgroup$
    – hash man
    Aug 29, 2020 at 17:32
  • $\begingroup$ I'll add to MotylaNogaTomkaMazura's great solution that this type of equation is called a Homogeneous type differential equation, since it can be written as a function of the variable $u=\frac{y}{x}$. Generally, if a first order differential equation can be written in this form, then use the substitution $y=ux$ in order to get a separable equation. $\endgroup$
    – GSofer
    Aug 29, 2020 at 17:33
  • $\begingroup$ There appears to be an error in the subtraction of $u$, $$\frac{2u-u^2}{1-2u}-u=\frac{2u-u^2}{1-2u}-\frac{u-2u^2}{1-2u}=\frac{u+u^2}{1-2u}.$$ $\endgroup$ Aug 29, 2020 at 19:26

We are given

$$(x^2-2xy)\frac{dy}{dx}+y^2-2xy=0$$ $$\underbrace{(y^2-2xy)}_Mdx+\underbrace{(x^2-2xy)}_Ndy=0$$

by which $M_y = 2y-2x$ is not equal to $N_x=2x-2y$. So, the differential equation is not exact. To make it exact, observe that the differential equation is homogenuous and therefore the integrating factor can be found by


which gives


The new equation is exact since


A summary of general techniques to make non-exact different equations exact is shown here.



This is a homogeneous equation. So substitute $y=tx$ $$x\frac{dt}{dx}+t=\frac{2tx^2-t^2x^2}{x^2-2tx^2}$$ $$x\frac{dt}{dx}+t=\frac{2t-t^2}{1-2t}$$ $$x\frac{dt}{dx}=\frac{t+t^2}{1-2t}$$ $$\frac{dx}{x}=\frac{1-2t}{t+t^2}dt$$ $$\int\frac{dx}{x}=-\int\frac{2t+1-2}{t+t^2}dt$$ Hint [Let $t+t^2=u \Rightarrow (2t+1)dt=du$]

I'll let you integrate


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