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How to solve (in terms of $y$) $(y+x^4y^2)dx+xdy=0$.

I know I'm supposed to multiply by an integrating factor to turn this equation into an exact equation.

In the previous exercise I proved that in $Mdx+Ndy$ the functions:

  1. $\frac{1}{N}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)$
  2. $\frac{1}{M}\left(\frac{\partial M}{\partial y}-\frac{\partial N}{\partial x}\right)$
  3. If $M=yf(xy)$ and $N=xg(xy)$ then $\frac{1}{xM-yN}$ is an integrating factor

I've tried using 1. and 2., but the calculations are horrible and do not seem to work. I don't think that's the way to go.

About 3. I don't know what $f$ and $g$ I should choose.

how can I go about solving this differential equation?

this isn't homework.

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An integrating factor, as you can check (!), is $$f(x,y) = \frac1{xy(x^4y-3)}\,.$$ Where does this come from, you ask? By noting that the $1$-parameter group on $\mathbb R^2-\{0\}$ $$\{(x,y) \rightsquigarrow (e^t x, e^{-4t}y)\}$$ leaves the differential equation invariant. I was told years ago that it was exactly for the purposes of finding such obscure integrating factors that Sophus Lie "invented" Lie groups.

One then integrates and gets $$\frac{3-x^4y}{xy} = C,$$ which is, indeed, $y=\dfrac3{x^4+Cx}$, as @Amzoti obtained.

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  • $\begingroup$ Welcome to MSE. I'm following your book in my Dif. Geo . course ^_^ $\endgroup$ – Git Gud May 7 '13 at 19:44
  • $\begingroup$ @GitGud, thanks :) I've been around for a while now. Where are you a student? $\endgroup$ – Ted Shifrin May 7 '13 at 20:02
  • $\begingroup$ I'm not from the US, in case you're assuming that. I'm from Europe. If you wish to know more precisely where I'm at, let me know and I'll e-mail you. $\endgroup$ – Git Gud May 7 '13 at 20:09
  • $\begingroup$ @GitGud, I surmised as much. Sure, get in touch. $\endgroup$ – Ted Shifrin May 8 '13 at 14:45
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$(y+x^4y^2)dx+xdy=0$

Subtract $x^4 y^2$ from both sides, yielding:

$y + x \frac{dy}{dx} = -x^4 y^2$

Divide both sides by $-xy^2$, yielding:

$\displaystyle -\frac{1}{xy} - \frac{\frac{dy}{dx}}{y^2} = x^3$

Let $\displaystyle v(x) = \frac{1}{y}$, which yields:

$\displaystyle \frac{dv}{dx} - \frac{v}{x} = x^3$

Choose an integrating factor $\displaystyle \mu = \text{exp}\left(\int -\frac{1}{x} dx\right) = \frac{1}{x}$, so

$\displaystyle \frac{\frac{dv}{dx}}{x} - \frac{v}{x^2} = x^2$

Substitute $\displaystyle -\frac{1}{x^2} = \frac{d}{dx}\frac{1}{x}$, yielding:

$\displaystyle \frac{d}{dx}(\frac{v}{x}) = x^2$

Integrate both sides, yielding:

$\displaystyle v = x(c_1 + \frac{x^3}{3})$

Solve for the original $y$, yielding:

$$\displaystyle y = \frac{3}{x^4 + cx}$$

Note: you should be able to do it the way you suggested, so give that another go.

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  • $\begingroup$ Great work...nice encouragement $\endgroup$ – Namaste May 8 '13 at 0:32
  • $\begingroup$ @amWhy: Thanks for the support! I have to travel tomorrow and Friday for my Thursday talk at the University, so may not be on (hope not), but have a great week! Regards $\endgroup$ – Amzoti May 8 '13 at 0:36

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