Question 404148
{{{du/dx = x/y}}}


{{{du = (x/y)dx}}} Integrate both sides:


{{{int(1, du) = int(x/y, dx)}}}


{{{u = (x^2/2y) + C}}}


The other problem is slightly different. The second problem {{{du/dx = y/x}}} is the same, except there is an ln function involved:


{{{du = (y/x)dx}}}


{{{u = int(y/x, dx) = y ln (abs(x)) + C}}}