Re: [工數] ODE
※ 引述《TsungMingC (TMC)》之銘言：
: solve the following D.E.
: y
: y'-e = cosx
: could someone please give me a hint to solve the nonlinear problem
: thanks
dy y dy y dx 1 e^(-y)
-- - e = cosx => -- = e + cosx => -- = ------------ = ----------------- = x'
dx dx dy e^y + cosx 1 + e^(-y) cosx
x' [ 1 + e^(-y) cosx ] = e^(-y)
x' + e^(-y) x' cosx - e^(-y) = 0
兩邊同乘 e^(sinx) => x' e^(sinx) + e^(-y + sinx) x' cosx - e^(-y + sinx) = 0
x' e^(sinx) - 1' e^(sin1) + d [e^(-y + sinx)]/dx = 0 ( 1' = 0 )
兩邊同時積分
x(y) sin(τ)
∫ e dτ + e^( - y + sin [ x(y) ] ) = C
1
x sin(τ)
∫ e dτ + e^( - y + sinx ) = C
1
x sin(τ)
e^( - y + sinx ) = C - ∫ e dτ
1
x sin(τ)
e^( y - sinx ) = 1 / [ C - ∫ e dτ ]
1
x sin(τ)
e^y = e^sinx / [ C - ∫ e dτ ]
1
x sin(τ)
y(x) = sin(x) - ㏑ | C - ∫ e dτ |
1
1 sin(τ)
y(1) = sin(1) - ㏑ | C - ∫ e dτ | = sin(1) - ㏑ | C |
1
if C = e^sin(1) => y(1) = 0 ; 另外 y'(1) = cos1 + 1
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◆ From: 118.161.243.220
※ 編輯: Frobenius 來自: 118.161.243.220 (04/03 18:02)
※ 編輯: Frobenius 來自: 118.161.243.220 (04/03 19:05)
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