推 ezWang:感謝e大>_<~ 11/05 23:19
※ 引述《ezWang (ez王)》之銘言：
: 1.
: solve xz''+2z'=(9x+6)e^(3x+2y)
: z'是z對x做偏微分
: z''是z對x做兩次偏微分
: 答案:z=c1(y)+c2(y)/x + e^(3x+2y)
z = z_h + z_p
z_h滿足 xz''+2z'=0 => x^2z''+2xz'=0為Cauchy-Euler Equation
z_h = Cx^m代入, m*(m-1)*x^m + 2*m*x^m = 0,
m(m+1)=0, m=0, -1
=> z_h = c1(y)*x^0 + c2(y)*x^(-1) = c1(y) + c2(y)/x
這裡把y視為常數 所以積分常數可以是y的函數
z_p滿足xz''+2z'=(9x+6)e^(3x+2y)
令z_p = A*e^(3x+2y)
z_p' = 3A*e^(3x+2y)
z_p'' = 9A*e^(3x+2y) 代入
(9Ax+6A)*e^(3x+2y) = (9x+6)e^(3x+2y)
比較得A=1, z_p = e^(3x+2y)
z = z_h+z_p = c1(y) + c2(y)/x + e^(3x+2y)
: 2.
: x
: solve y''-2y'+y= e /(1+x^2)
: yp算到一半卡住
: x x -1
: 答案的yp=0.5e ln(x^2+1)+xe tan x
: 請教各位~
: 謝謝~
答案有錯
y = y_h + y_p
y_h = c1*e^x + c2*x*e^x
y_p有記公式就代公式，沒記就推一下，這邊直接代公式
| y1 y2 | | e^x x*e^x |
W = | | = | | = e^(2x)
| y1' y2'| | e^x x*e^x+e^x |
u1' = -x*e^x*e^x/(1+x^2)/e^(2x) = -x/(1+x^2)
u1 = ∫-x/(1+x^2)dx = ∫-0.5/(1+x^2) d(1+x^2) = -0.5 ln(x^2+1)
u2' = e^x*e^x/(1+x^2)/e^(2x) = 1/(1+x^2)
u2 = ∫1/(1+x^2)dx = atan(x) (令x=tanθ可得)
y_p = y1*u1 + y2*u2
= -0.5*e^x*ln(x^2+1) + x*e^x*atan(x)
前面那項差個負號
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