How do I solve dx/x2+y2=dy/2xy=dz/(x+y) 2?

Answer 1

Given equation : dx/(x^2+y^2) = dy/(2xy):  Solution y = 1/(2e^C₁) [1± (4e^(2C₁)x^2+1) ^½]

Steps:

1.    Divide by dx:                                     1/(x^2+y^2) = (dy/dx)/(2xy) 

2.    Rewrite dy/dx = y’:                           1/(x^2+y^2) = y’/(2xy)

3.    Isolate y’:                                          y’ = 2xy/(x^2+y^2)

4.    Substitute y =vx:                               (vx)’ = 2xvx/(x^2+(vx)^2)

5.    Simplify:                                           (vx)’ = 2v/(1+v^2)

6.    Differentiate left side::                      xv’+ v = 2v/(1+v^2)

7.    Rewrite first order ODE:                  [(1+v^2)/(v-v^3)]v’ = 1/x

8.    Solve by integrating both sides:       lnv – ln(v+1) – ln(v-1) = lnx +C₁                    

9.    Isolate v:                                           v = 1/(2xe^C₁) [1± (4e^(2C₁)x^2+1) ^½]

10. Change v to y, v=y/x:                         y = 1/(2e^C₁) [1± (4e^(2C₁)x^2+1) ^½]

There is no solution for dy/(2xy) = dz/(x+y)^2

Cheers! Happy July 4th

Answer 2

Given:

dx dy dz

------------ = ------------------------- = ------------------- = k (constant)

x² + y² 2xy (x+y)²

dx + dy = k(x² + y²) + k(2xy) = k(x² + y²+2xy) =k(x+y)² = dz

or d(x+y) = dz or z= x+y+ c _, is a constant. ------(1)

Again, 2xy dx = (x² + y²) dy or 2xydx -x² dy = y²dy

or yd(x²) - x²dy = y²dy

yd(x²) - x²dy

or ------------------- = dy or d(x²/y) = dy

y²

Or (x²/y) = y + a , where a is a constant

Or x² = y² + ay .........................(ii)