# Exact Differential Equations

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

#### Solutions Collecting From Web of "Exact Differential Equations"

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 ?