※ 引述《jason021866 (JasonHsu)》之銘言：
: (2coshy+3x)dx+(xsinhy)dy=0
: 請問一下要怎麼解
: 我解出來的積分因子好像不太對
: 謝謝
[e^y+e^(-y)+3x]dx+[(1/2)x(e^y-e^(-y))]dy=0...(1)
令M=e^y+e^(-y)+3x，N=(1/2)x(e^y-e^(-y))]
@表示偏微分
@M @N
─ = e^y-e^(-y) ─ = 1/2(e^y-e^(-y))
@y @x
@M @N
─ ≠ ─
@y @x
@M @N
[─ - ─]/N = 1/x = f(x)
@y @x
∫f(x)dx ∫1/xdx ln│x│
I.F.= e = e = e = x
(1)乘以x得
x[e^y+e^(-y)+3x]dx+[(1/2)(x^2)(e^y-e^(-y))]dy=0必為exact
令Ma=x[e^y+e^(-y)+3x] Na=[(1/2)(x^2)(e^y-e^(-y))]
@ψ
存在ψ(x,y)使得 ─ = Ma
@x
ψ(x,y)=∫Ma dx = ∫x[e^y+e^(-y)+3x]dx
= [e^y+e^(-y)](1/2)x^2 + x^3 + h(y)
@ψ
─ = [(1/2)(x^2)(e^y-e^(-y))] + h'(y) = Na = [(1/2)(x^2)(e^y-e^(-y))]
@y
比較得h'(y)=0
h(y)=k
所以ψ(x,y)= (1/2)x^2[e^y+e^(-y)] + x^3 + k
所求為 (1/2)x^2[e^y+e^(-y)] + x^3 = c
或 (x^2)cosh(y) + x^3 = c
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