Re: [微方] ODE求助

[Frobenius (▽．(▽×▽φ)=0)]

※ 引述《leo790124 (4浩)》之銘言：
: 題目：y''+(y')^2 = 2 * e^(-y)
: 一開始先用代換令v = y'
: 中間過程要用到Bernoulli eq
: 但是到最後換回去就積不出來了
: 請大大幫幫我解，謝謝

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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推 leo790124   :感激不盡，我還正在努力理解中...謝謝
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