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20. I = ∫<0, 1>x^4dx + (1/ln3)∫<0, 1>xd(3^x) + (1/3)∫<0, 1>xde^(3x)
= [x^5/5]<0, 1> + (1/ln3)[x 3^x]<0, 1> - (1/ln3)∫<0, 1>3^xdx
+ (1/3)[x e^(3x)]<0, 1> - (1/3)∫<0, 1>e^(3x)dx
= 1/5 + 3/ln3 - [1/(ln3)^2][3^x]<0, 1> + e^3/3 - (1/9)[e^(3x)]<0, 1>
= 1/5 + 3/ln3 - 2/(ln3)^2 + e^3/3 - (1/9)(e^3-1)
= 14/45 + 3/ln3 - 2/(ln3)^2 + 2e^3/9
21. 微分方程两边同除以 cosxcosy, 得
tanydy = tanxdx, - ln(cosy) = - ln(cosx) - lnC
cosy = Ccosx, y(0) = π/4 代入得
C = 1/√2, 则特解为 cosy = (1/√2)cosx
= [x^5/5]<0, 1> + (1/ln3)[x 3^x]<0, 1> - (1/ln3)∫<0, 1>3^xdx
+ (1/3)[x e^(3x)]<0, 1> - (1/3)∫<0, 1>e^(3x)dx
= 1/5 + 3/ln3 - [1/(ln3)^2][3^x]<0, 1> + e^3/3 - (1/9)[e^(3x)]<0, 1>
= 1/5 + 3/ln3 - 2/(ln3)^2 + e^3/3 - (1/9)(e^3-1)
= 14/45 + 3/ln3 - 2/(ln3)^2 + 2e^3/9
21. 微分方程两边同除以 cosxcosy, 得
tanydy = tanxdx, - ln(cosy) = - ln(cosx) - lnC
cosy = Ccosx, y(0) = π/4 代入得
C = 1/√2, 则特解为 cosy = (1/√2)cosx
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