How to solve the following differential equation $(y^3-2yx^2)dx+(2xy^2-x^3)dy=0.$

Question

I'd love your help with solving this following differential equation: $$(y^3-2yx^2)dx+(2xy^2-x^3)dy=0.$$

I am trying to use the method of finding integrating factor, that is $\frac{1}{Mx+Ny}$. But it is not coming.

Any suggestions?

Answer 1

Just observe that your differential equation is homogeneous, so substitution $y=vx$ will simplify it.

Answer 2

The integrating factor $\mu = xy$ turns the ODE into the exact one $$(xy^4 - 2y^2x^3)\mathrm d x + (2x^2 y^3 - x^4y) \mathrm d y = 0.$$

You can derive this the following way: If the expression

$$\frac{h_x - g_y}{xg - yh} $$

is a function of $xy$ alone (say $f(xy)$), where $g(x, y) = y^3 - 2yx^2$ and $h(x,y) = 2xy^2 - x^3$, then $\mu = \mathrm e^{\int f(u) \mathrm du}$ with $u = xy$. Here,

$$\frac{h_x - g_y}{xg - yh} = \frac{2y^2 - 3x^2 - 3y^2 + 2x^2}{xy^3-2yx^3 -2xy^3+x^3y} = - \frac{x^2 + y^2}{ -(x^3y + xy^3)} = \frac{x^2 + y^2}{ xy(x^2 + y^2)} = \frac{1}{xy}, $$

so when $\frac{1}{xy} = \frac{1}{u}$, one has the integrating factor

$$\mu = \mathrm e^{\int \frac{\mathrm d u}{u} } = \mathrm{e}^{\ln(u)} = xy.$$

Answer 3

Visual inspection of the terms shows that they can be grouped by the degrees structure. This gives an re-arrangement of the equation as $$ 0=(y^3−2yx^2)dx+(2xy^2−x^3)dy=[y^3\,dx+2xy^2\,dy] - [2yx^2\,dx+x^3\,dy] $$ Now as the differential of monomials in general is $$d(x^ay^b)=ax^{a-1}y^bdx+bx^ay^{b-1}dy,$$ these same-degree terms along with their coefficients can be combined into differentials after extracting some factors, $$ 0=y\,d(xy^2)-x\,d(x^2y). $$ This suggests to use $u=x^2y$ and $v=xy^2$ as new variables, the transformation backwards is $x^3=\frac{u^2}{v}$, $y^3=\frac{v^2}u$. The whole expression can be written in the new variables after multiplying with the now easily recognizable integrating factor $xy$ $$ 0=d(v^2-u^2)\implies C=v^2-u^2=x^2y^2(y^2-x^2), $$ which can be solved, for instance, as bi-quadratic equation in $y$.

Answer 4

$$Mx+Ny=3xy(y^2-x^2)$$ $$\frac{y(y^2-2x^2)dx+x(2y^2-x^2)dy}{3xy(y^2-x^2)}=0$$ $$\frac{(y^2-x^2)(ydx+xdy)}{3xy(y^2-x^2)}+\frac{xy(ydy-xdx)}{3xy(y^2-x^2)}=0$$ $$\frac{log{xy}}{3}+\frac{\log(y^2-x^2)}{6}=\frac{logc}{3}$$ $$xy(y^2-x^2)^{1/2}=c$$