Question

(x − 4y -9)dx + (4x + y − 2)dy = 0

Answer 1

Solution

Let’s find stationary point from the system

x − 4y - 9 = 0

4x + y – 2 = 0

From this system 17x – 17 = 0 => x = 1, y = -2

Change of variables: X = x-1, Y = y+2

In new variables equation is: $\frac{dY}{dX}=-\frac{x\ -\ 4y\ -9}{4x\ +\ y\ -\ 2}=-\frac{X-4Y}{4X+Y}=\frac{4Y/X-1}{Y/X+4}$

Let z = Y/X => Y = z*X =>  $\frac{dY}{dX}=\frac{dz}{dX}X+z$  => $\frac{dz}{dX}X+z=\frac{4z-1}{z+4}$  => $\frac{dz}{dX}X=-\frac{z^2+1}{z+4}$

$\frac{\left(z+4\right)dz}{z^2+1}=-\frac{dX}{X}$  => $\int\frac{\left(z+4\right)dz}{z^2+1}=-\int\frac{dX}{X}$  => (1/2)*ln(z²+1)+4arctan(z)=-ln|X|+C

C = (1/2)*ln(z²+1)+4arctan(z)+ln|X| = (1/2)*ln(Y²/ X²+1)+4arctan(Y/X)+ln|X| = (1/2)*ln[(y+2)²/(x-1)²+1]+4arctan[(y+2)/(x-1)]+ln|x-1|

Answer

C = (1/2)*ln[(y+2)²/(x-1)²+1]+4arctan[(y+2)/(x-1)]+ln|x-1|