※ 引述《rebe212296 (綠豆冰)》之銘言：
: http://ycc.math.fju.edu.tw/t2012s/odes2012s/odes2012sf1.pdf
: 裡面的第三,四,五,八題不會解
: 課本也找不到解答
: 第四題(來源:
: http://www.chegg.com/homework-help/elementary-differential-equations-and-boundary-value-problems-9th-edition-chapter-7.4-problem-3p-solution-9780470383346)
: Let X(1)X(2)...X(n) are the solutions of homogeneous equation
: x' = P(t)x
: Then the Wroskian satisfies the equation
: or
: Integrating both sides: (C is arbitrary constant of integration)
: Taking the exponential of both sides and rearranging:
: W=c e^∫(P11+P22+..+Pnn)dt
: Let X(1)X(2) be the set of fundamental solutions.
: W1=C1e^∫(P11+P22+..+Pnn)dt
: W2=C2e^∫(P11+P22+..+Pnn)dt C1C2不為0 (這一步看不懂)
: Therefore:
: &
: where &
: Dividing W1 by W2 we get:
: Therefore: where is a constant (c2 ≠ 0)
: 第八題看不懂題目
: 跪求解答 謝謝
8.
2 2 2 2
(D - 1)x - 2Dy = 0 (D - 1)(D - 1)x - 2(D - 1)Dy = 0
=>
2 2 2
2Dx + (D - 1)y = 0 4D x + 2D(D - 1)y = 0
兩式相加
4 2
(D + 2D +1)x = 0
4 2 2 2
特徵方程式: r + 2r + 1 = (r + 1) = 0
=> r = ±i (各為二重根)
=> x = (c1+c2t)cos t + (c3+c4t) sin t
x' = c2 cos t - (c1+c2t)sin t + c4sin t + (c3+c4t) cos t
= (c2+c3+c4t)cos t + (c4-c1-c2t)sin t
x'' = c4 cos t -(c2+c3+c4t)sin t -c2sin t + (c4-c1-c2t)cos t
= (2c4-c1-c2t)cos t - (2c2+c3+c4t)sin t
x''-2y' = x
=> y' = (x''- x)/2 = (c4-c1-c2t)cos t - (c2+c3+c4t) sin t
=> y = ∫[(c4-c1-c2t)cos t - (c2+c3+c4t) sin t]dt + c5
= (c4-c1)sin t -c2(tsin t+cos t) + (c2+c3)cos t -c4(-tcos t + sin t) + c5
= (c3+c4t)cos t - (c1+c2t)sin t + c5
y'' = -c2cos t - (c4-c1-c2t)sin t -c4sin t -(c2+c3+c4t)cos t
= -(2c2+c3+c4t)cos t - (2c4-c1-c2t)sin t
y''+2x' = y
=> (c3+c4t)cos t - (c1+c2t)sin t = (c3+c4t)cos t - (c1+c2t)sin t + c5
=> c5 = 0
∴ x = (c1+c2t)cos t + (c3+c4t) sin t
y = (c3+c4t)cos t - (c1+c2t)sin t
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