Re: [其他] 一階ODE
: 請問第一小題該如何解
: 解答贈微薄p幣 感謝...
(2x + y)y' = (4/y)x^2 + y + 4x
另解
y' = 4x^2 / [y(y + 2x)] + 1 + 2x/(y + 2x)
= 2x[1/y - 1/(y + 2x)] + 1 + 2x/(y + 2x)
= 2(x/y) + 1
令 x = uy代入上式
(u - xu')/ u^2 = 2u + 1
=> x(1/u)' = 2u + 1 - 1/u
令v = 1/u
xv' = 2/v + 1 - v = [2 + v - v^2]/v
=> v/[-(v - 1/2)^2 + (9/4)] dv = (1/x)dx
=> (v - 1/2)/[(9/4) - (v - 1/2)^2]dv + (1/2)/[(9/4) - (v - 1/2)^2] dv = (1/x)dx
=> (-1/2)ln|(9/4) - (v - 1/2)^2| + (1/6)ln|(v + 1)/(2 - v)| = lnx + c
=> (-1/2)ln|(9/4) - (y/x - 1/2)^2| + (1/6)ln|(y + x)/(2x - y)| = lnx + c
=> (-1/2)ln|(y/x + 1)(2 - y/x)| + (1/6)ln|(y + x)/(2x - y)| = lnx + c
=> lnx - (1/2)ln|(y + x)(2x - y)| + (1/6)ln|y + x| - 1/6ln|2x - y| = lnx + c
=> A = (y + x)^(-1/3) (2x - y)^(-2/3)
=> C = (y + x)(2x - y)^2
再加上y + 2x = 0
剩下就看你想整理到什麼程度
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※ 編輯: Honor1984 (111.243.61.160), 10/24/2018 22:34:28
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