Answer to Question #206290 in Differential Equations for mike

Question

(3x+y-4)dx+(x+y-2)dy=0

Answer 1

Solution

Stationary point from the system  3x+y-4 = 0, x+y-2 = 0

Subtracting this equations 2x-2 = 0 => x=1, y=1

New variables X = x-1, Y = y-1

$\frac{dY}{dX}=-\frac{3x\ +y\ -4}{x\ +\ y\ -\ 2}=-\frac{3X+Y}{X+Y}=-\frac{Y/X+3}{Y/X+1}$

For z = Y/X or Y = z*X expression of the derivative is $\frac{dY}{dX}=\frac{dz}{dX}X+z$

So $\frac{dz}{dX}X+z=-\frac{z+3}{z+1}$   =>  $\frac{dz}{dX}X=-\frac{z^2+2z+3}{z+1}$

According to separable variable method  $\frac{\left(z+1\right)dz}{z^2+2z+3}=-\frac{dX}{X}$  and $\int\frac{\left(z+1\right)dz}{z^2+2z+3}=-\int\frac{dX}{X}$

After multiplication by 2: ln(z²+2z+3)+2ln|X|=lnC

(z²+2z+3)*X²=C => Y²+2X*Y+3X² = C => (y-1)²+2(x-1)(y-1)+3(x-1)² = C

Answer

(y-1)²+2(x-1)(y-1)+3(x-1)² = C