Question

 (1+lnx) dx + (1+lny)dy = 0

Answer 1

Solution

(1+lnx) dx = - (1+lny)dy

∫(1+lnx) dx = - ∫ (1+lny)dy +C

x*lnx = -y*lny + C (C – arbitrary constant)

x*lnx  + y*lny = C

Or in another form ln(x^x·y^y) = C

x^x·y^y = e^C = D   (D – arbitrary constant)

Answer

x*lnx + y*lny = C or x^x·y^y = D