→ aacvbn:3Q 03/20 14:53
※ 引述《aacvbn (u)》之銘言：
: for the following problem, find ,respectively
: a differential equation having the given function
: as a general solution.
: x x x
: y(x)=e { Acos(2x)+Bsin(2x) } + --- e sin(2x)
: 4
: thanks.
y'' + a1 y' + a0 y = 0
x
yh = e [ A cos2x + B sin2x ]
x(1+2i) x(1-2i)
= A* e + B* e
代入 O.D.E.
2 2
A* ( (1 + 2i) + a1 (1 + 2i) + a0 ) + B* ( (1 - 2i) + a1(1 - 2i) + a0 ) = 0
A , B 不恆為零，故 -3 + 4i + a1 + 2i a1 + a0 = 0
-3 - 4i + a1 - 2i a1 + a0 = 0
a1 + a0 = 3
2 a1 = -4
a1 = -2 , a0 = 5
O.D.E. => y'' - 2y' + 5y = 0
特解
x x
yp = ── e sin(2x) 代入 O.D.E.
4
x (1+2i)x
= Im{── e }
4
1 (1+2i)x 2 (1+2i)x
── [ 0 + 2*(1 + 2i) e + x(1 + 2i) e ] = y''
4
2 + 4i x(-3 + 4i)
1 (1 + 2i)x (1+2i)x
── [ e + x (1 + 2i) e ] = y'
4
1 (1 + 2i)x
── x e = y
4
1 (1+2i)x
y'' - 2y' + 5y = ──(2+4i-3x+4ix-2(1+x+2ix)+5x ) e
4
1 x
= ───(2+4i-3x+4ix-2-2x-4ix+5x) e (cos(2x)+isin(2x))
4
x
e
= ───([2-3x-2-2x+5x]+i[4+4x-4x])(cos2x + isin2x)
4
x
e
= ───(4icos2x)(Imagine part)
4
x
= e cos2x
x
y'' -2y' + 5y = e cos2x
--
--
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◆ From: 114.45.228.96
m = 1 ±2i
2
m-1 = ±2i , m - 2m + 1 = -4
用看的 y''-2y'+5y = 0
特解用看的
x x 1
── e sin(2x) = ─── f
4 L(D)
x x
f = L(D) ── e sin(2x)
4
x x
= e L(D+1) ── sin2x
4
x 2 x
= e [(D+1) -2(D+1) + 5] ── sin2x
4
x 2 x
= e [D + 4] ── sin2x 直接展開也可以
4
or
1 x
─── [cos2x] = ── sin2x
D^2+4 4
x
f = e cos2x
--
※ 編輯: ntust661 來自: 114.45.228.96 (03/20 01:24)