How do you solve dy/dx= (x+y+1) / (x-y-1)?

Answer 1

(x-y-1)dy = (x+y+1)dx or (x+y+1)dx - (x-y-1)dy = 0. Solve the system x+y+1 = 0 & x-y-1 =0 to obtain (x,y) = (0,-1). Then write x=h=0 & y=k=-1. The transformation [x =x'+h=x'; y=y'+k=y'-1; dx=dx'; dy=dy'] reduces the second given equation to (x'+y'-1+1)dx' - (x'-(y'-1)-1)dy' = 0 or (x'+y')dx' - (x'-y')dy' = 0.

Next use the transformation [y' = vx'; dy' = vdx' + x'dv] to obtain ((x'+vx')dx' - (x'-vx')(vdx'+x'dv) = 0) and multiply through the equation by (1/x') which gives (1+v)dx' - (1-v)(vdx'+x'dv) = 0.

Rewrite last equation as dx'+vdx'-vdx'-x'dv+v²dx'+vx'dv = 0 which reduces to dx'/x' - dv + v²dx'/x' +vdv = 0. Rearrange as dx'(1+v²)/x' + dv(v-1) = 0 or dx'/x' = -(v-1)dv/(v²+1). Set last equation as (dx'/x') = (1/(v²+1) - v/(v²+1))dv and integrate to ln |x'| = arctan v - 0.5ln |v²+1| + C.

Transform x' and v to acquire ln |x| = arctan ((y+1)/x) - 0.5ln |((y+1)²/x²)+1|+C.

Multiply through last equation by 2 and arrange terms to reach 2ln |x|+ln |((y+1)²+x²)/x²|+C = 2arctan ((y+1)/x).

Combining natural logarithm terms gives final answer of ln ((y+1)²+x²) - 2arctan ((y+1)/x) = C.

Answer 2

dy/dx=(x+y+1)/(x-y-1) let y= v-1 then dy/dx= dv/dx so dv/dx = (x+v)/(x-v) let v=xu(x) now dv/dx = x du/dx +u

xdu/dx + u = ( x+ xu)/( x-xu) = (1+u)/(1-u) so du/dx= 1/x( -u +(u+1)/(1-u)) = 1/x(u² +1)/(1-u) so du/dx((u-1)/(-u²-1))= 1/x. Integrating both sides Int (du(u-1)/(-u²-1))= Int dx/x =log|x| + C. so Int -du/ (-u²-1) + Int udu/(-u² -1)

equals tan^-1u - (1/2) log|u² +1| and this equals log|x| + C

so u=v/x and v=y+1 are substituted into the expression to get tan^-1(y+1)/x - (1/2)log| (y+1)²/x² +1|= log|x| + C as a solution

Answer 3

Make the substitution u = y+1 so that du = dy and

du/dx=(x+u)(x-u)

Then

x du - u du = x dx + u dx

x du - u dx =x dx + u du

If you divide by xy, you will get a solution that makes no sense.

Divide by x² + y² to get

(x du - u dx)/(x² + y²) = (x dx + u du)/(x² + y²)

The left side is the derivative of arctan (u/x) and the right side is the derivative of (1/2)ln(x² + y²) which makes the solution

arctan[(y+1)/x] =(1/2)ln[x² + (y+1)²] + C

This is a technically correct mechanical solution; I am not sure what it means!