Answer 1

Given, `(4x+3y-4)dx+(3x-7y-3)dy=0` 

∴ `M=(4x+3y-4) and (3x-7y-3)` 

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

`(dM)/dy=3`    And `(dN)/dx=3`

∴` (dM)/dy=(dn)/dx`

∴ The given equations are exact.
For solution, 

`int M  dx=int (4x+3y-4)dx`

`int M  dx = 2x^2+3xy-4x` 

`int" (Term is N free from x)"=int-7y-3 dy` 

= `(-7y^2)/2-3y` 

∴ The final solution is,  

`2x^2+3xy-4x-(7y^2)/2-3y=c` 

`4x^2+6xy-8x-7y^2-6y=c`