4
$\begingroup$

$M(x,y)dx + N(x,y)dy=0$ 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?

$\endgroup$
  • 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 '15 at 3:39
5
$\begingroup$

Remark:

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 ?

$\endgroup$
  • $\begingroup$ 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. $\endgroup$ – yasir Aug 29 '15 at 7:07
  • $\begingroup$ How do you get that solution? $\endgroup$ – yasir Aug 29 '15 at 7:08
  • $\begingroup$ This is not the answer.just check your answer by differentiating. You won't get the previous equation. $\endgroup$ – yasir Aug 29 '15 at 7:22
  • $\begingroup$ yeah, i will see where i went wrong. Please show your method $\endgroup$ – yasir Aug 29 '15 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. $\endgroup$ – yasir Aug 29 '15 at 11:27

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.