標題:
Solve: y'' - 4y' + 4y = 2e^2x .
發問:
想知道怎樣計這條數學題Solve: y'' - 4y' + 4y = 2e^2x .
最佳解答:
y'' - 4y' + 4y = 2e^2x Note that it is a non-homogeneous second order linear equation. The associated homogeneous equation: - has constant coefficients - the non-homogeneous term g(x)=2e^(2x) is in a special form g(x) = P(x)e^(αx)+cos(βx) or g(x) = P(x)e^(αx)+sin(βx). We have P(x)=2, α=2, β=0 Hence we use the method of undetermined coefficient. The characateristic equation: r^2 - 4r + 4 = 0 (r - 2)^2 = 0 r = 2 The general solution is given as: y[g] = Ae^(2x) + Bxe^(2x) where A, B are some constants The particular solution is given as: y[p] = C(x^s)e^(αx) where C is a constant and s counts how many times α+iβ is a root of the character equation. α+iβ=2, hence s=2. y[p] = C(x^2)e^(2x) y[p]' = 2C(x^2+x)e^(2x) y[p]'' = 2C(2x^2+4x+1)e^(2x) Put this into our equation: 2C(2x^2+4x+1)e^(2x) - 8Cx(x^2+x)e^(2x) + 4C(x^2)e^(2x) = 2e^2x C(2x^2+4x+1) - 4C(x^2+x) + 2Cx^2 = 1 C = 1 Hence y[p] = (x^2)e^(2x) The general solution is given by: y = y[g] + y[p] = Ae^(2x) + Bxe^(2x) + (x^2)e^(2x)
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其他解答:
let p(x)=p^2-4p+4 p(x)=0 p=2(repeated) Yc=Ae^2x+Bxe^2x Yp=x^2*2e^2x/p''(2)=x^2e^2x Y=Yc+Yp=Ae^2x+Bxe^2x+x^2e^2x
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