Find the general solution (x+2y-1)dx-(2x+y-5)dy=0

Answer 1

∂(x+2y-1)/∂y gives 2;

∂[-(2x+y-5)]/∂x or ∂[-2x-y+5]/∂x gives -2.

2 ≠ -2, so (x+2y-1)dx − (2x+y-5)dy = 0 is not an Exact Differential Equation.

Explore a double-transformation method.

Simultaneous solution of x+2y=1 & 2x+y=5 gives (x,y)=(3,-1).

The transformation x = x'+3 & y = y'-1 (with dx=dx' & dy=dy') rewrites the equation as

(x' + 3 + 2y' - 2 -1)dx' - (2x' + 6 + y' - 1 - 5)dy' = 0 which reduces to

(x' + 2y')dx' - (2x' + y')dy' = 0.

A second transformation of y'=vx' (with dy'=vdx'+x'dv) will give (x'+2vx')dx' - (2x'+vx')(vdx'+x'dv) = 0.

Rewrite as x'dx'+2vx'dx'-2vx'dx'-2x'²dv-v²x'dx'-vx'²dv = 0 and simplify to

x'dx'-2x'²dv-v²x'dx'-vx'²dv = 0 or x'(1-v²)dx' - x'²(2+v)dv = 0.

Division of both sides of the equation by x' will yield

(1-v²)dx' - x'(2+v)dv = 0 or (1-v²)dx' - x'(2+v)dv = 0. Then dx'/x' = (2+v)dv/(1-v²).

Further rearrange as 2dv/(1-v²) + vdv/(1-v²) - dx'/x' = 0 and then integrate to

2(0.5) ln |(v+1)/(v-1)| - (0.5) ln |1-v²| - ln |x'| = C₁. Multiplication of both sides of the equation by 2

will yield 2 ln |(v+1)/(v-1)| - ln |1-v²| - 2 ln |x'| = C. With v equal to y'/x', take the last equation

to 2 ln |(y'+x')/x'÷(y'-x')/x'| - ln |(x'²-y'²)/x'²| - 2 ln |x'| = C. This simplifies to

ln |(y'+x')²/(y'-x')² • x'²/(x'²-y'²) • 1/x'²| = C or ln |(x'+y')/(x'-y')³| = C.

Write y' as (y+1) and x' as (x-3) to obtain ln |(x+y-2)/(x-y-4)³| = C.

Answer 2

first, re-arrange the problem to be x+2y-1+(2x+y-5)dy/dx=0

second, let M=x+2y-1 and N=2x+y-5 and define Fx=M and Fy=N

third, show My=2 and Nx=2, thus the problem is an EXACT ODE

fourth, find integral of Mdx holding y constant = x²/2+2xy-x+h(y) = F

fifth, find Fy= 2x+h' = N = 2x+y-5 ... solve for h'=y-5 and h=y²/2-5y

sixth, solution is F=x²/2+2xy-x+y²/2-5y=constant