推 PuddingRina:推薦這篇文章 ○(*￣︶￣*)○ 05/01 15:58
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推 redpig: 推薦這篇文章 ○(*￣︶￣*)○ 05/01 17:26
※ 引述《redpig (耶耶耶~)》之銘言：
: 不好意思 正在學 工程數學 中
: 所以似乎有點簡單 但是我不太會解耶
: Z' = -Z + 1 + t
: Z' = -3Z = -3 -4t
: Z' = 2Z - 2 + e3t
: Z' = 6Z + 2 + 3e3t
: 這類的題目要怎麼解阿 = =
: 謝謝各位大大解惑~
這種類型的都是同一種...不過這是比較簡單的
Consider the following o.d.e
y' + p(x) y = f(x) (*)
with initial condition y(0) = c.
Observing that for any two differentiable function u and v, we always have
(uv)'=u'v + uv'.
Hence we may try to solve (*) by multiplying some differentaible function u
so that
u y ' + p(x) y u = v '
for some function v.
Remember that
d x x
- exp(∫p(t)dt) = p(x) exp(∫p(t)dt).
dx 0 0
x
Multiply (*) by exp(∫p(t)dt), we have
0
d x x
- [exp(∫p(t)dt)y(x)] = exp(∫p(t)dt) f(x) (**)
dx 0 0
Integrating (**), we get
x x t
exp(∫p(t)dt)y(x) - c = ∫ exp(∫p(s)ds)f(t)dt
0 0 0
Hence
x x x t
y(x) = exp(∫-p(t)dt) c +exp(∫-p(t)dt)∫exp(∫p(s)ds)f(t)dt.
0 0 0 0
Matrix valued Ode case, the solution is also simailar.
If U : [0,a]-> M_n(R) is a differentiable curve on M_n(R), and U satiesfies
U'(t) - A U(t) = B(t), (*)
with initial condition U(0) = C , and A belong to M_n(R) and B:[0,a]-> M_n(R)
is continuous.
Hence we can also multiply by exp(-tA), and solve it by using a similar way
as before.
It's important to know the Jordan form theory to solve exp(-tA).
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