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3. y'+y/x = cosx
y = e^(-∫dx/x)[∫cosx e^(∫dx/x) dx + C]
= e^(-∫dx/x)[∫cosx e^(∫dx/x) dx + C]
= (1/x) (∫xcosxdx + C) = (1/x) (∫xdsinx + C)
= (1/x) (xsinx-∫sinxdx+C) = (1/x) (xsinx+cosx+C)
y(π/2) = 0 代入得 0 = (2/π)(π/2+C), 得 C = -π/2
故得 y = (1/x) (xsinx+cosx-π/2)
4. 联立解 y = x^2, y = 1, 得交点 (-1, 1), (1, 1), 由对称性
S = 2∫<0, 1>(1-x^2)dx = 2[x-x^3/3]<0, 1> = 4/3
Vy = ∫<0, 1>πx^2dy = π∫<0, 1>ydy = π[y^2/2]<0, 1> = π/2
y = e^(-∫dx/x)[∫cosx e^(∫dx/x) dx + C]
= e^(-∫dx/x)[∫cosx e^(∫dx/x) dx + C]
= (1/x) (∫xcosxdx + C) = (1/x) (∫xdsinx + C)
= (1/x) (xsinx-∫sinxdx+C) = (1/x) (xsinx+cosx+C)
y(π/2) = 0 代入得 0 = (2/π)(π/2+C), 得 C = -π/2
故得 y = (1/x) (xsinx+cosx-π/2)
4. 联立解 y = x^2, y = 1, 得交点 (-1, 1), (1, 1), 由对称性
S = 2∫<0, 1>(1-x^2)dx = 2[x-x^3/3]<0, 1> = 4/3
Vy = ∫<0, 1>πx^2dy = π∫<0, 1>ydy = π[y^2/2]<0, 1> = π/2
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