I assume that you mean ln(y) = xy + C then differeniate to get
(1/y)y' = y + xy' multiply through by y to get y' = y^2 + xyy' factor out y' to get
y'(1-xy) = y^2 Then y' = y^2/(1-xy)
Christian D.
asked 04/18/19ln y = xc + C; dy/dx = (y^2)/(1-xy)
I assume that you mean ln(y) = xy + C then differeniate to get
(1/y)y' = y + xy' multiply through by y to get y' = y^2 + xyy' factor out y' to get
y'(1-xy) = y^2 Then y' = y^2/(1-xy)
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