標題:
integration
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y is a function of x y' denotes the first derivative of y y'' denotes the second derivative of y find y if y'' + y' + 1 =0 更新: I am very busy these days. I will read the answers soon
最佳解答:
Find y if y'' + y' + 1 = 0 by integration. y'' + y' + 1 = 0 y'' + y' = - 1 ∫(y'' + y') dx = ∫- 1 dx ∫y'' dx + ∫y' dx = ∫- 1 dx y' + y = - x + C (e^x)(y' + y) = (e^x)(- x + C) (e^x)y' + (e^x)y = - xe^x + Ce^x (ye^x)' = - xe^x + Ce^x ye^x = ∫(- xe^x + Ce^x) dx ye^x = ∫- xe^x dx + ∫Ce^x dx ye^x = ∫- x d(e^x) + Ce^x + C_0 ye^x = - xe^x - ∫e^x d(- x) + Ce^x + C_0 ye^x = - xe^x + ∫e^x dx + Ce^x + C_0 ye^x = - xe^x + e^x + Ce^x + C_1 ye^x = - xe^x + (C_2)e^x + C_1 y = - x + (C_1)e^(- x) + C_2
其他解答:
From the given equation, you may let y = e^(rx) for some numbers r. (Since only exponential function will be possible to arrive such an equation) Then by succesive differentiation and substitution, you get r^2 + r + 1 = 0 so r = - 1/2 + ((root 3) / 2)i or - 1/2 - ((root 3) / 2)i You can see r is a complex number, so use the euler's formula e^(ix) = cosx + isinx, then u'll get the function u want|||||唔用integration 用其他方法可以嗎?? http://i299.photobucket.com/albums/mm309/lokwanshan/ScreenHunter_01Jun040121.gif
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