→ littleyinyo:嗯嗯，答案是對的!!謝謝你的解答^^ 06/02 00:34
※ 引述《littleyinyo (I AM YIN)》之銘言：
: 1.Please find the solution of ordinary differential equation
: (x^2+1)y''-2xy'+2y=6(x^2+1)^2
: Boundary condition y(x=0)=-1,y(x=1)=5
: 2.Given an ordinary differential equation as
: d^2y dy
: x------ + 2(x-1)------- +(x-2)y=0
: dx^2 dx
: If boundary condition is given as y(0)=y0,solve for y(x).
: 先謝謝大家囉，變係數實在沒有頭緒..
set y(x)=xv(x) and y'(x)=v(x)+xv'(x) y''(x)=2v'(x)+xv''(x)
2
2 6(x +1)
帶入 DE v'' + --------- v' = ---------
2 x
x(x +1)
set z=v' z'=v''
2
2 6(x +1)
z' + --------- z = ---------
2 x
x(x +1)
2
x
積分因子 I(x)= ---------
2
x +1
x^2+1 6(x^2+1) x^2
z(x)= -------- S -----------*---------- dx
X^2 x X^2+1
x^2+1
z= (3x^2+c) --------- = v'
x^2
c
v= x^3 + (3+c) X - --- + c*
x
y=xv=(x) * ( x^3 + (3+c) X - --- + c* )
x
= x^4 + (3+c) x^2 - c + c*x
y(0)=-1 ---> c=1
y(1)=5 ---> 1+(3+1)-1+c*=5 ---> C*=1
y = x^4 + (3+1) x^2 - 1 + 1x
= x^4 + 4*x^2 - 1 + x
應該沒有錯吧?!
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