The task is to find $a$ and $b$ such that $u=x^ay^b$ is an integrating factor (IF) for the ODE $ydx+x(xy-1)dy=0$.

Attempt: For $u$ to be an IF for the given ODE, the multiplication of ODE by the IF must be an exact ODE. That is, $$ x^ay^{b+1}dx + (x^{a+1}y^b)(xy-1)dy=0 $$ must be exact. This necessitates the condition $M_y=N_x$ which translates into $$(b+1)x^ay^b = (a+1)x^ay^b(xy-1) + x^{a+1}y^{b+1}.$$ Expanding the expression on the right and equalizing the coefficients with the left side leads to the conditions $a+b=-2$ and $a = -2$ which in turn gives $b=0$.

The solution manual however disagrees and says $a=-3$ and $b=1$. I'd be grateful if you could help me figure out why my answer is wrong.

  • 1
    $\begingroup$ You are right $a=-2$ and $b=0$. Probably there is a typo in the non-exact ODE. $\endgroup$
    – JJacquelin
    Feb 10, 2021 at 11:24
  • $\begingroup$ Thanks! I repeated the same approach for a few other problems and got the correct answer. There must've been a typo with this one indeed. $\endgroup$
    – User32563
    Feb 11, 2021 at 8:52

1 Answer 1


You can also approach from the other direction. With this type of integrating factor, the total degree of coefficient and differential changes in the same way for all terms. To get complete differentials, one can group terms of the same degree structure. Here this results in $$ 0=[y\,dx-x\,dy]+x^2y\,dy=-x^2\,d(y/x)+\frac12x^2\,d(y^2). $$ From this one can see directly that dividing by $x^2$ removes all obstacles to integration.


You must log in to answer this question.