Jul 20, 2000 #2 G gepla Guest 2 (dy / dx) = y.tanx ∫ 2dy / y = ∫ tanx.dx 2ln | y | = - ln | cosx | C ln (y ˛) =-ln | cosx | lnc1ln (y ˛) = ln (c1 / | cosx |) y ˛ = c1 / | cosx | y = a √ (c1 / | cosx |) dengan cosx ≠ 0 dan c1> 0
2 (dy / dx) = y.tanx ∫ 2dy / y = ∫ tanx.dx 2ln | y | = - ln | cosx | C ln (y ˛) =-ln | cosx | lnc1ln (y ˛) = ln (c1 / | cosx |) y ˛ = c1 / | cosx | y = a √ (c1 / | cosx |) dengan cosx ≠ 0 dan c1> 0
