Re: [工數] 常微分方程
※ 引述《ru18284 (夯哥)》之銘言:
: 請教各位大大兩個工數問題
: 謝謝囉
: 1.
: 求解 Y〞+Y = (x-1) cosx
: 2.
: 求解 Y〞+6Y'+9Y = e^-3x / (x^2+1)
: 如果不要用逆運算子法 求解得出來嘛? 那該怎麼解呢?
: 因為小弟我對這兩題解答的逆運算方法看攏無 = =
: 感謝各位高手!!!!!
(D^2+1)y=(x-1)cosx
y_h=Acosx+Bsinx
令y_p=(ax+b)cos(x)+(cx+d)sin(x)
dy_p
── = acos(x)-(ax+b)sin(x)+csin(x)+(cx+d)cos(x)
dx
(y_p)"=[(y_p)']'=-asin(x)-asin(x)-(ax+b)cos(x)+ccos(x)+ccos(x)-(cx+d)sin(x)
=2ccos(x)-2asin(x)-(ax+b)cos(x)-(cx+d)sin(x) 代回ODE
2ccos(x)-2asin(x)-(ax+b)cos(x)-(cx+d)sin(x)+acos(x)-(ax+b)sin(x)
+csin(x)+(cx+d)cos(x)=(x-1)cos(x)
(-a+c)xcos(x)+(-c-a)xsin(x)+(2c-b+a+d)cos(x)+(-2a-d-b+c)sin(x)=x(cosx)-cos(x)
比較係數得
-a+c=1 ........(1) (1)-(2)=> 2c=1, c=0.5, a=-0.5
-c-a=0 ........(2) 代入(3)&(4)
2c-b+a+d=-1....(3) => 1-b+0.5+d=-1 => 2b-2d=5 ....(5) (5)+(6)=> b=1
-2a-d-b+c=0....(4) => -1-d-b+0.5=0 => 2b+2d=-1....(6) d=-3/2
得y_p=(ax+b)cos(x)+(cx+d)sin(x) =(-0.5x+1)cos(x)+(0.5x-3/2)sin(x)
2. Y〞+6Y'+9Y = e^-3x / (x^2+1)
exp(-3x)
(D+3)(D+3)y= ─────
x^2+1
y_h=(c1+c2x)exp(-3x)
1
y_p= ──── exp(-3x)/(x^2+1)
(D+3)^2
1 1
= exp(-3x) ─── ───
D^2 x^2+1
1
=exp(-3x)── ln(x^2+1)+xarctan(x)
2
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◆ From: 111.243.129.38
※ 編輯: zi6ru04zpgji 來自: 111.243.129.38 (11/21 21:21)
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