$(4x + 3y – 4) dx + (3x – 7y – 3) dy = 0$

Solution:

Given, $(4 x+3 y-4) d x+(3 x-7 y-3) d y=0$

$\therefore \mathrm{M}=(4 x+3 y-4)$ and $N=(3 x-7 y-3)$

Differentiating M by y and N by x, we get,

$\frac{d M}{d y}=3 \quad$ And $\quad \frac{d N}{d x}=3$

$\therefore \frac{d M}{d y}=\frac{d N}{d x}$

$\therefore$ The given equations are exact.

For solution,

$\int M d x=\int(4 x+3 y-4) d x$

$\int M d x=2 x^{2}+3 x y-4 x$

$\int(T e r m \ i s \ N $Free from $x)=\int-7 y-3 d y$

$=\frac{-7 y^{2}}{2}-3 y$

$\therefore$ The final solution is,

$2 x^{2}+3 x y-4 x-\frac{7 y^{2}}{2}-3 y=c$

$4 x^{2}+6 x y-8 x-7 y^{2}-6 y=c$