作者LuisSantos (^______^)
看板trans_math
標題Re: [考古] 元智的兩題怪題目...
時間Sun Jul 3 01:19:24 2005
※ 引述《jonerty (寶爺)》之銘言：
: 1.
: find the general solution of y"+2y'+y=e^-x.
: 2.
: find a function y=f(x) satisfying (x^2)y'-2xy-y=0
: (x不等於0) and f(1)=1/e
: 請各位大大幫忙解惑一下...請問各位大大有誰能助我一臂之力...
1.
y"+2y'+y=e^-x
y"+2y'+y = 0 的特徵方程式為 m^2 + 2*m + 1 = 0 => m = -1 , -1
y_c = ( c_1*x + c_2 )*e^-x
The UC set of e^-x is S = {e^-x}
因為 {e^-x} 被包含在y_c裡
所以用 S' = {(x^2)*e^-x} 來代替 S
令 y_p = A*(x^2)*e^-x 代入 y"+2y'+y=e^-x
得 A*(x^2 - 4x + 2)*e^-x + A*(-2*x^2 + 4x)*e^-x + A*(x^2)*e^-x = e^-x
=> 2A*e^-x = e^-x => A = 1/2
所以 y_p = (1/2)*(x^2)*e^-x
因此 y"+2y'+y=e^-x 的一般解為
y = y_c + y_p
= ( c_1*x + c_2 )*e^-x + (1/2)*(x^2)*e^-x
= [ (1/2)*(x^2) + c_1*x + c_2 ]*e^-x
2.
(x^2)y'-2xy-y=0 and f(1)=1/e
dy
(x^2)---- - (2x+1)y = 0
dx
1 2x + 1
=> ---dy - --------dx = 0
y x^2
=> ∫(1/y)dy - ∫(2x + 1)/(x^2) dx = 0
=> ln|y| - [2ln|x| - (1/x) ] = c
因為 f(1)=1/e 所以 ln|1/e|- [2ln|1| - (1/1) ] = c
因此 c = 0
所以 ln|y| - [2ln|x| - (1/x) ] = 0
=> ln|y| = 2ln|x| - (1/x)
=> ln|y| = ln|(x^2)| + lne^(-1/x)
=> ln|y| = ln|(x^2)| + lne^(-1/x)
=> ln|y| = ln|(x^2)*e^(-1/x)| => y = (x^2)*e^(-1/x)
--
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◆ From: 61.66.173.21
※ 編輯: LuisSantos 來自: 61.66.173.21 (07/03 01:59)
※ 編輯: LuisSantos 來自: 61.66.173.21 (07/03 03:44)
1F：推 jonerty:第一題不是人寫的...因為都看不懂... 59.121.155.199 07/03
2F：→ jonerty:第2題原來就醬而已..是我想太多了..感謝大大指教 59.121.155.199 07/03