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※ 引述《k0184990 (追隨夢想..)》之銘言: : Solve the following initial value problem: : dy/dx-2y^2+3y=1, y( 0 )=1 : ans: y=1 : 請問這題該如何解呢? y'=2y^2-3y+1 , dy/(2y^2-3y+1) = dx , dy/[(2y-1)(y-1)] = dx [1/(y-1) -2/(2y-1)]dy = dx ln[y-1] - ln[2y-1] = x + c ln[(y-1)/(2y-1)] = x + c (y-1)/(2y-1) = C*exp^(x) y(0)=1 , 0 = C => (y-1)/(2y-1) = 0 , y=1 -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 114.37.175.176
ntust661:嗚...感覺最近又要開始 ODE 了 0.0 07/07 10:02
mike7689:ODE是重頭戲阿~(準備重考,衝刺中)...XD 07/07 11:34