求y”=㏑x的通解及过程
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y'' = lnx
y' = ∫ lnx dx = xlnx - ∫ dx = xlnx - x + C₁ = x(lnx - 1) + C₁
y = ∫ [x(lnx - 1) + C₁] dx
= ∫ (lnx - 1) d(x²/2) + C₁x
= (1/2)x²(lnx - 1) - (1/2)∫ x² * (1/x) dx + C₁x
= (1/2)x²(lnx - 1) - (1/2)(x²/2) + C₁x + C₂
= (1/2)x²(lnx - 1) - (1/4)x² + C₁x + C₂
通解为
y = (1/4)x²(2lnx - 3) + C₁x + C₂
y' = ∫ lnx dx = xlnx - ∫ dx = xlnx - x + C₁ = x(lnx - 1) + C₁
y = ∫ [x(lnx - 1) + C₁] dx
= ∫ (lnx - 1) d(x²/2) + C₁x
= (1/2)x²(lnx - 1) - (1/2)∫ x² * (1/x) dx + C₁x
= (1/2)x²(lnx - 1) - (1/2)(x²/2) + C₁x + C₂
= (1/2)x²(lnx - 1) - (1/4)x² + C₁x + C₂
通解为
y = (1/4)x²(2lnx - 3) + C₁x + C₂
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