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设于 y1 线性无关的解 y2 = u(x)e^x, 代入原齐次微分方程
(2x-1)(u''+2u'+u) - (2x+1)(u'+u) + 2u = 0,
(2x-1)u'' + (2x-3)u' = 0,
令 u' = p, 则 (2x-1)p' = -(2x-3)p,
dp/p = -(2x-3)dx/(2x-1) = [-1+2/(2x-1)]dx
lnp = -x+ln(2x-1)+lnC1
u' = p = C1(2x-1)e^(-x)
u = C1∫(2x-1)e^(-x)dx = -C1∫(2x-1)de^(-x)
= -C1(2x-1)e^(-x) + 2C1∫e^(-x)dx
= -C1(2x-1)e^(-x) - 2C1e^(-x) + C2
= -C1(2x+1)e^(-x) + C2
取 C1 = -1， C2 = 0，得 u = (2x+1)e^(-x),
y2 = 2x + 1,
通解 y = C1e^x + C2(2x+1)
(2x-1)(u''+2u'+u) - (2x+1)(u'+u) + 2u = 0,
(2x-1)u'' + (2x-3)u' = 0,
令 u' = p, 则 (2x-1)p' = -(2x-3)p,
dp/p = -(2x-3)dx/(2x-1) = [-1+2/(2x-1)]dx
lnp = -x+ln(2x-1)+lnC1
u' = p = C1(2x-1)e^(-x)
u = C1∫(2x-1)e^(-x)dx = -C1∫(2x-1)de^(-x)
= -C1(2x-1)e^(-x) + 2C1∫e^(-x)dx
= -C1(2x-1)e^(-x) - 2C1e^(-x) + C2
= -C1(2x+1)e^(-x) + C2
取 C1 = -1， C2 = 0，得 u = (2x+1)e^(-x),
y2 = 2x + 1,
通解 y = C1e^x + C2(2x+1)
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