推 k0184990:感謝喔:) 08/23 21:38
※ 引述《k0184990 (追隨夢想..)》之銘言：
: 這題想了很久無方向下筆...
: 題目:
: xy''(x)-y'(x)- 4x^3 y(x) = 0 by setting x=t^(1/2)
: 請給個方向吧:)
x = t^(1/2) => t = x^2
dy dy d^2 y dy d^2 y
---- = 2x ---- , ------- = 2 ---- + 4(x^2) -------
dx dt d(x^2) dt d(t^2)
代入原ODE中
dy d^2 y dy
=> 2x ---- + 4(x^3) ------- - 2x ---- - 4(x^3)y = 0
dt d(t^2) dt
d^2 y
=> ------- - y = 0
d(t^2)
=> y = c1(e^t) +c2[e^(-t)] = c1[e^(x^2)] + c2[e^(-x^2)]
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