工程數學Initial value problems題目

Question

Solve the initial value problem. Check that your answer satisfies the ODE as well the initial Conditions. (Show the details of your work.)


y''-25y=0 y(0)=0, y'(0)=40

越快越好

感恩

Answer 1

法一:輔助方程式解法
(1)輔助(特徵)方程式 m² - 25 = 0 ==> m = 5, -5
故通解為 y = A e^(5x) + B e^(-5x)
(2)求 A, B
y(0) = 0 => A + B = 0
y'(0)=40 => 5A - 5B = 40
解聯立得 A = 4, B = -4
故 y = 4 [ e^(5x) - e^(-5x) ]
(3)Check
y = 4 [ e^(5x) - e^(-5x) ]
y"=100[ e^(5x) - e^(-x) ]
代入 y" - 25 y 得 0,
且y(0)=4(1-1)=0 , y'(0)= 20(1+1)=40

故 y = 4 [ e^(5x) - e^(-5x) ]

法二: 直接猜測解法
(猜)設 y=e^(mx) 滿足 y" - 25y =0 (註: m未知)
代入 y" - 25 y =0, 則
(m² - 25)e^(mx) = 0, 故只要 m²- 25= 0 (即 m = 5, -5), 則
y=e^(5x), y=e^(-5x)均為 y" - 25y = 0之解
又e^(5x), e^(-5x)線性獨立, 且原ODE為線性齊次ODE,
故通解為 y = A e^(5x) + B e^(-5x)

註: A, B之計算, 與法一同