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
Set u = 2x + y
y' = u' - 2
u[u' - 2] = 4x^2 /[u - 2x] + u + 2x or u = 0 即y = -2x
u' - 2 = 4x^2 /[u(u - 2x)] + 1 + 2x/u
= 2x[1/(u - 2x) - 1/u] + 1 + 2x/u
= 2x/(u - 2x) + 1
= u/(u - 2x)
u' = (3u - 4x)/(u - 2x)
齊次ODE
v = u/x代入解
xv' + v = (3v - 4)/(v - 2)
=> xv' = (-v^2 + 5v - 4)/(v - 2) = -(v - 1)(v - 4)/(v - 2)
=> (1/x)dx = [-(1/3)/(v - 1) - (2/3)/(v - 4)]dv
=> c + lnx = (-1/3)ln|v - 1| - (2/3)ln|v - 4|
=> c + lnx = -(1/3)ln|(2x + y)/x - 1| - (2/3)ln|(2x + y)/x - 4|
=> Ax = x(x + y)^(1/3) (y - 2x)^(2/3)
=> C = (x + y) (y - 2x)^2
加上最前面or y + 2x = 0
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謝謝提醒，我漏掉了1
已修正
※ 編輯: Honor1984 (111.243.61.160), 10/24/2018 23:14:27
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