Question

solve differential equation sqrt 1+x2+y2+x2y2 + xy dy/dx=0

Answer 1

We are not given any initial conditions.

ODE given :
       √[1 + x² + y² + x²y²]  + x y * dy/dx = 0     ---- (1)
=>   √(1+x²) * √(1+y²) + x y * dy/dx = 0
=>   y * dy / √(1+y²)  =  - dx *√(1+x²)  / x     ----- (2)

==   RHS:
Let    x = tan t.    dx = sec² t dt
LHS of Eq (2):   - sec² t dt * sec t /tan t
           =  - dt / [sin t * cos² t] = - Sin t * dt /[ sin² t * cos² t ]
      
Let    cos t = z.   Then RHS =  dz /[z² (1 - z²)       Use partial fractions.

RHS = dz * [ 1/z²  + 1/(1-z²) ]
        =  dz / z²  + 1/2* dz /(1-z)  + 1/2* dz/(1+z)

Integrating RHS:
      -1/z + 1/2 Ln | (1+z)/(1-z) |  + K
  =  - sec t  + 1/2  Ln | (1+cos t)/(1-cos t) |  + K
  =  - √(1+x²)  + 1/2 Ln  {  [√(1+x²) + 1] / [ √(1+x²) - 1 ] }  + K
  =  - √(1+x²)  + 1/2  Ln  [ 2+x² + 2√(1+x²) ]  -   Ln x   +  K 
                          By rationalizing the denominator

=== LHS:
Integration of LHS of equation (2) gives:  √(1+y²)

So the solution is:
   √(1+y²) = - √(1+x²)  + 1/2 * Ln [2+x²+ 2√(1+x²) ]  - Ln x + K

We can square both sides and then find RHS for y on the LHS.