(1+lnx) dx + (1+lny)dy = 0
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(xx·yy) = C
xx·yy = eC = D (D – arbitrary constant)
Answer
x*lnx + y*lny = C or xx·yy = D
Comments
Leave a comment