1. ## non exact equation

(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

2. 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).

3. i cant get a coefficient i cant transform it to be exact
how to solve it
?

i dont know what transformations to do

4. for both (P_y-Q_x)/P and (P_y-Q_x)/Q i dont get a function of one variablr i cant make it exact

5. 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: $(a_1x+b_1y+c_1)dx+(a_2x+b_2y+c_2)dy=0
$

Let $L_1 : a_1x+b_1y+c_1=0$ and $L_2 : a_2x+b_2y+c_2=0$

Now, There are two cases:

1) If $\dfrac{a_1}{a_2}=\dfrac{b_1}{b_2}$ 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) $\dfrac{a_1}{a_2} \neq \dfrac{b_1}{b_2}$.

First step:
Find the intersection point between L1 & L2 by solving the following two equations:

$a_1x+b_1y+c_1=0$ ... (1)

$a_2x+b_2y+c_2=0$ ... (2)

Assume the intersection point is (h,k).

Second step:
Substitute $x=u+h$ and $y=v+k$.
Clearly $dx=du$ and $dy=dv$.
These two substituion will transform the equation to a homogeneous equation.

$(2x-4y+6)dx+(x+y-3)dy=0$
Clearly, $\dfrac{2}{1} \neq\dfrac{-4}{1}$.