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※ 引述《Highhuman (Ryan)》之銘言: : For a differential equation x^2 y" -3xy' +4y = 0, : (a)use z = lnx to transform such an equation into an equation with constant : coefficients : (b)find the general solution of (a) in terms of x. : 這題爬文後,還是不懂... : 感覺目的好像是利用 z=lnx 把y'和y"換掉而以... : 可以麻煩詳細說明嗎?! @@" 就照做啊... 這種是 Cauchy-Euler type 的 ODE Put Y(z) = y(e^z) z= ln x => y'(x) = Y'(z) * 1/x => y"(x) = Y"(z) * (1/x)^2 - Y'(z) * 1/x^2 Sub. y' y" into the original equation => Y"(z) - Y'(z) - 3 * Y'(z) + 4 Y(z) = 0 解 Y . -- ※ 發信站: 批踢踢實業坊(ptt.cc) ◆ From: 203.101.240.116
Highhuman:略懂!!111.249.226.104 07/03 18:58
Highhuman:general solution 怎寫?!111.249.226.104 07/03 19:23
Eliphalet:Y(z) = a e^(2z) + b z*e^(2z)203.101.240.116 07/03 19:41
Eliphalet:a,b consts203.101.240.116 07/03 19:42
Highhuman:thx111.249.226.104 07/03 19:45