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.


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.

  • $\begingroup$ 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


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.