Question

(6x-5y+4)dy+(y-2x-1)dx=0​

Answer 1

(6x - 5y + 4) dy + (y - 2x - 1) dx = 0

(6x - 5y + 4) dy = (2x - y + 1) dx

dy/dx = (2x - y + 1) / (6x - 5y + 4)

Let X = x - a and Y = y - b. We want to find constants a and b such that

dY/dX = (a rational function)

where the numerator and denominator on the right side are free of constant terms. Substituting x and y in the equation, we have

dY/dX = (2 (X + a) - (Y + b) + 1) / (6 (X + a) - 5 (Y + b) + 4)

dY/dX = (2X - Y + 2a - b + 1) / (6X - 5Y + 6a - 5b + 4)

Then we solve for a and b in the system,

2a - b + 1 = 0

6a - 5b + 4 = 0

==>   a = -1/4 and b = 1/2

With these constants, the equation reduces to

dY/dX = (2X - Y) / (6X - 5Y)

Now substitute Y = VX and dY/dX = X dV/dX + V :

X dV/dX + V = (2X - VX) / (6X - 5VX)

The equation becomes separable after some simplification:

X dV/dX + V = (2 - V) / (6 - 5V)

X dV/dX = (2 - V) / (6 - 5V) - V

X dV/dX = (2 - V - (6 - 5V)) / (6 - 5V)

X dV/dX = (4V - 4) / (6 - 5V)

- (5V - 6) / (4V - 4) dV = 1/X dX

Integrate both sides:

-5/4 V + 1/4 ln|4V - 4| = ln|X| + C

Extract a constant from the logarithm on the left:

-5/4 V + 1/4 (ln(4) + ln|V - 1|) = ln|X| + C

-5/4 V + 1/4 ln|V - 1| = ln|X| + C

-5V + ln|V - 1| = 4 ln|X| + C

Get this back in terms of Y :

-5Y/X + ln|Y/X - 1| = 4 ln|X| + C

Now get the solution in terms of y and x :

-5 (y - 1/2)/(x + 1/4) + ln|(y - 1/2)/(x + 1/4) - 1| = 4 ln|x + 1/4| + C

With some manipulation of constants and logarithms, and a bit of algebra, we can rewrite this solution as

-5 (4y - 2)/(4x + 1) + ln|(4y - 4x - 3)/(4x + 1)| = 4 ln|x + 1/4| + 4 ln(4) + C

-5 (4y - 2)/(4x + 1) + ln|(4y - 4x - 3)/(4x + 1)| = 4 ln|4x + 1| + C

-5 (4y - 2)/(4x + 1) + ln|4y - 4x - 3| - ln|4x + 1| = 4 ln|4x + 1| + C

-5 (4y - 2)/(4x + 1) + ln|4y - 4x - 3| = 5 ln|4x + 1| + C