Jan 25, 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 ˛ = γ1 / | cosx | y = √ α (C1 / | cosx |) με ≠ 0 cosx και Γ1> 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 ˛ = γ1 / | cosx | y = √ α (C1 / | cosx |) με ≠ 0 cosx και Γ1> 0