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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.

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4 Answers 4

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$$(x^{2}-2xy)dy+(y^{2}-2xy)dx=0$$ 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.

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

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

hence

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

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.

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  • $\begingroup$ 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
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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

$$\mu(x)=\frac{1}{xM+yN}=-\frac{1}{xy(x+y)}$$

which gives

$$\underbrace{\left(-\frac{y}{x(x+y)}+\frac{2}{x+y}\right)}_{{M}^{*}}dx+\underbrace{\left(-\frac{x}{y(x+y)}+\frac{2}{x+y}\right)}_{N^{*}}dy=0$$

The new equation is exact since

$${M^{*}}_y=-\frac{3}{(x+y)^2}={N^{*}}_x$$

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

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

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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