(2x-4y+6)dx+(x+y-3)dy=0
i cant make integration cofficient
because
(P_y-Q_x)/P is not a function of x and so is the other one
??
how to solve it
If my memory serves me right, first you would use an affine transformation to reduce the equation down to a homogeneous equation. Then you would do a second transformation which would transform the coefficients to a single variable (Elementary Differential Equations by Rainville and Bedient covers this type of equation).
It can not be converted to an exact equation.
This equation is called non homogeneous first order equation with linear coefficient.
which has the general form: $\displaystyle (a_1x+b_1y+c_1)dx+(a_2x+b_2y+c_2)dy=0
$
Let $\displaystyle L_1 : a_1x+b_1y+c_1=0$ and $\displaystyle L_2 : a_2x+b_2y+c_2=0$
Now, There are two cases:
1) If $\displaystyle \dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$ then L1 & L2 are parallel.
Your equation does not satisfy this case. Tell me If you want to know how to deal with that case.
2) $\displaystyle \dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$.
First step:
Find the intersection point between L1 & L2 by solving the following two equations:
$\displaystyle a_1x+b_1y+c_1=0$ ... (1)
$\displaystyle a_2x+b_2y+c_2=0$ ... (2)
Assume the intersection point is (h,k).
Second step:
Substitute $\displaystyle x=u+h$ and $\displaystyle y=v+k$.
Clearly $\displaystyle dx=du$ and $\displaystyle dy=dv$.
These two substituion will transform the equation to a homogeneous equation.
So, for your equation:
$\displaystyle (2x-4y+6)dx+(x+y-3)dy=0$
Clearly, $\displaystyle \dfrac{2}{1} \neq\dfrac{-4}{1}$.
i.e. you will apply the procedure of the second case.
Can you do that?