作者Frobenius (▽．(▽×▽φ)=0)
看板Math
标题Re: [微方] ODE求助
时间Sun Apr 3 21:38:31 2011
※ 引述《leo790124 (4浩)》之铭言：
: 题目：y''+(y')^2 = 2 * e^(-y)
: 一开始先用代换令v = y'
: 中间过程要用到Bernoulli eq
: 但是到最後换回去就积不出来了
: 请大大帮帮我解，谢谢
y'' = v' = dv/dx = dv/dy dy/dx = v dv/dy
v dv/dy + v^2 = 2 * e^(-y)
dv/dy + v = 2 * e^(-y) v^(-1)
Let t = v^[1 - (-1)] = v^2 => dt/dy = 2v dv/dy
2 v dv/dy + 2 v^2 = 4 * e^(-y)
dt/dy + 2t = 4 * e^(-y)
t = e^(-∫2dy) { ∫ 4 * e^(-y) e^(∫2dy) dy + C[1] }
= e^(-2y) ( 4 e^y + C[1] ) = 4 e^(-y) + C e^(-2y) = v^2 = (y')^2
y' = [4 e^(-y) + C[1] e^(-2y)]^(1/2) = dy/dx
dy/ [4 e^(-y) + C[1] e^(-2y)]^(1/2) = dx
e^y dy/ [4 e^(-y) e^(2y) + C[1] e^(-2y)e^(2y)]^(1/2) = dx
e^y dy/ (4e^y + C[1])^(1/2) = dx
(1/4) d(4e^y + C[1]) /(4e^y + C[1])^(1/2) = dx
(1/4) (4e^y + C[1])^(-1/2) d(4e^y + C[1]) = dx
2 (1/4) (4e^y + C[1])^(1/2) = x + C[2]
(1/2) (4e^y + C[1])^(1/2) = x + C[2]
(1/4) (4e^y + C[1]) = ( x + C[2] )^2
e^y + C[1]/4 = ( x + C[2] )^2
e^y = ( x + C[2] )^2 - C[1]/4
y(x) = ㏑ | ( x + C[2] )^2 - C[1]/4 |
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1F：推 leo790124 :感激不尽，我还正在努力理解中...谢谢 04/03 22:04
2F：推 leo790124 :跪拜!! 04/03 22:09