$M(x,y)dx + N(x,y)dy=0$ is said to be a perfect differential when

$$\frac{\partial (M(x,y))}{\partial y}=\frac{\partial (N(x,y))}{\partial x}$$

Let $M_y=\frac{\partial (M(x,y))}{\partial y}$ and $N_x=\frac{\partial (N(x,y))}{\partial x}$.

In case if:

$\frac{M_y-N_x}{N(x,y)}$ is only a function of x only (say $f(x)$) then it has a integrating factor $e^{\int f(x)dx}$.

But if $\frac{M_y-N_x}{N(x,y)}$ is a constant then what will be the integrating factor?

Say for this problem: $(axy^2+by)dx+(bx^2y+ax)dy=0$

Is there any other method to solve this differential equation?

  • 2
    $\begingroup$ You can consider any constant function as a function of $x$ as well as a function of $y$. Here you can consider as a function of $x$. $\endgroup$
    – Empty
    Aug 29, 2015 at 3:39

2 Answers 2



If $Mx-Ny\not=0$ and the equation can be written as $f(xy)y\,dx+F(xy)x\,dy=0$ then $\frac{1}{Mx-Ny}$ is an integrating factor of the equation.

Here the equation is $(axy+b)y\,dx+(bxy+a)x\,dy=0$. So , $I.F.=\frac{1}{(a-b)(x^2y^2-xy)}.$

Can you proceed further ?

  • $\begingroup$ Multiplying by the I.F. the second part reduces to $d(x^ay^b)$ which is a perfect differential.But the first part is not reduced. I suggest you to try it on a sheet. $\endgroup$
    – miyagi_do
    Aug 29, 2015 at 7:07
  • $\begingroup$ How do you get that solution? $\endgroup$
    – miyagi_do
    Aug 29, 2015 at 7:08
  • $\begingroup$ This is not the answer.just check your answer by differentiating. You won't get the previous equation. $\endgroup$
    – miyagi_do
    Aug 29, 2015 at 7:22
  • $\begingroup$ yeah, i will see where i went wrong. Please show your method $\endgroup$
    – miyagi_do
    Aug 29, 2015 at 7:32
  • $\begingroup$ I dont understand how are f and M connected? And how do we know that this is the integrating factor. Can you please add it to your answer. $\endgroup$
    – miyagi_do
    Aug 29, 2015 at 11:27

If $a\neq0$ and $b\neq0$ then integrating factor of equation $$(axy^2+by)dx+(bx^2y+ax)dy=0$$ is $$\\\frac{1}{{{x}^{\frac{2 b+a}{b+a}}} {{y}^{\frac{b+2 a}{b+a}}}}$$ and solution $$\\\frac{x^2y^2-xy}{{{x}^{\frac{2 b+a}{b+a}}} {{y}^{\frac{b+2 a}{b+a}}}}=C$$


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