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Find the Integrating factor of $y(x^2+y^2)dx - x(x^2+2y^2)dy =0$ I've solved this a bunch of times but I still can't find the I.F. I often get stuck with the $\frac{M_y-N_x}{N}$

$\frac{\partial M}{\partial y}=x^2+3y^2$

$\frac{\partial N}{\partial x}=-3x^2-2y^2$

So after I perform $\frac{M_y-N_x}{N}$ or $\frac{N_x-M_y}{M}$ I still can't get a good Integrating factor, can somebody please help me?

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  • $\begingroup$ "Integrating factors" are from the horror shop of mathematical misconceptions. They have no mathematical content whatsoever. The fact remains that you are given an ODE with no obvious solution, period. $\endgroup$ – Christian Blatter Aug 8 '15 at 18:06
  • $\begingroup$ I don't agree with the statement "Integrating factors are from the horror shop of mathematical misconceptions". $\endgroup$ – JJacquelin Mar 12 '19 at 9:59
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Hint

Just as for the previous post, why don't you start with the almost obvious $y=x \,z(x)$ ? This gives a nice separable equation.

I am sure that you can take from here.

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  • $\begingroup$ I would love to know that, but I wasn't taught on how to perform that, We were just given hand outs and I honestly have no idea what you mean. I just know that process stated above $\endgroup$ – Mickey Aug 8 '15 at 7:31
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$$y(x^2+y^2)dx - x(x^2+2y^2)dy =0 \tag 1$$ This ODE is of homogeneous kind. So, in this case, the simplest way to solve it is the usual change of function $y(x)=xz(x)$ as judiciously suggested by Claude Leibovici.

If definitively you want to use the method of integrating factor, this is possible. But this will be much more difficult in the present case.

An integrating factor $\mu(x,y)$ must transform the above non exact differential $(1)$ into an exact differential : $$\mu y(x^2+y^2) dx - \mu x(x^2+2y^2) dy =0$$ which supposes : $$\frac{\partial}{\partial y}\left(\mu y(x^2+y^2)\right) = \frac{\partial}{\partial x}\left(- \mu x(x^2+2y^2)\right) $$ $$x(x^2+2y^2)\frac{\partial \mu}{\partial x}+ y(x^2+y^2)\frac{\partial \mu}{\partial y}+(4x^2+5y^2)\mu =0 \tag 2$$ Generally, in case of academic exercise, one try some integrating factors on the form $\mu=f(x)$ or $\mu=f(y)$ or $\mu=f(xy)$ or $\mu=f(y/x)$ or other very simple form of function. In the present case, all fail. That is why this case is more difficult than usual.

In fact it requires a partial solving of the PDE $(2)$. Of course the complete solving of the PDE is not necessary : In solving the PDE we would be faced to the solving of Eq.$(1)$, a vicious circle. One only have to find a characteristic curve (with $\mu$ in it's equation).

An arduous calculus leads to $y^5e^{-x^2/(2y^2)}\mu=$constant.

An integrating factor is : $$\mu(x,y)=\frac{e^{x^2/(2y^2)}}{y^5}$$ After this, solving Eq.$(1)$ is easy. $$\frac{e^{x^2/(2y^2)}}{y^5}\left(y(x^2+y^2)dx - x(x^2+2y^2)dy\right) =0$$ This is an exact differential : $\quad d\left(\frac{x}{y^2}e^{x^2/(2y^2)} \right)=0$.

The solution of the ODE $(1)$ on the form of implicit equation is : $$\frac{x}{y^2}e^{x^2/(2y^2)}=C$$ Solving it for $y$ requires a special function, the Lambert W function : $$y(x)=\pm\frac{x}{\sqrt{2W(cx)}}$$

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As this is an exercise, a manually executable transformation into an integrable form supposedly exists. One can thus try to solve it by reverse engineering the construction of this task. Visual inspection of the terms shows that they can be grouped by the degrees structure. This gives an re-arrangement of the equation as $$ [yx^2\,dx-x^3\,dy]+[y^3dx-2xy^2dy]=0. $$ 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, $$ -x^4d(x^{-1}y)-x^2yd(x^{-1}y^2)=0. $$ This suggests to use $u=\frac{y}x$ and $v=\frac{y^2}x$ as new variables, the transformation backwards is $x=\frac{v}{u^2}$, $y=\frac{v}u$. This leads us to the transformed and separated equation $$ u^{-3}\,du+v^{-1}\,dv=0\implies -\frac12u^{-2}+\ln|v|=c. $$ Proceed further like in the answer of JJacquelin.

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