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標題:

app math DE 4條easy 10mark

發問:

dy/dx+y=e^-x y(-1)=3 dy/dx-xy=x y(0)=1/2 dy/dx-(x/x^2+1)y=0 y(0)=3 dy/dx+3y/x=[(1+3x^3)^1/2]/x y(1)=1

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(1) Integrating factor = exp(∫dx) = ex So: ex(dy/dx) + yex = 1 d(yex)/dx = 1 yex = x + C x = -1, y = 3, giving C = 3e-1 + 1 yex = x + 3e-1 + 1 y = xe-x + 3e-1-x + e-x (2) Integrating factor = exp(-∫xdx) = e-(x^2)/2 So: e-(x^2)/2(dy/dx) - xye-(x^2)/2 = xe-(x^2)/2 d[ye-(x^2)/2]/dx = xe-(x^2)/2 ye-(x^2)/2 = ∫xe-(x^2)/2dx =∫e-(x^2)/2d(x2/2) = - e-(x^2)/2 + C y = - 1 + Ce(x^2)/2 x = 0, y = 1/2, giving C = 3/2 y = - 1 + [3e(x^2)/2]/2 (3) Integrating factor = exp[-∫xdx/(x2 + 1)] = exp[-(1/2)∫d(x2 + 1)/(x2 + 1)] = exp[-(1/2) ln (x2 + 1)] = 1/√(x2 + 1) So: (dy/dx)/√(x2 + 1) - xy/√(x2 + 1) = 0 d[y/√(x2 + 1)]/dx = 0 y/√(x2 + 1) = C x = 0, y = 3, giving C = 3 y = 3√(x2 + 1) (4) Integrating factor = exp(∫3/xdx) = x3 So: x3 dy/dx + 3yx2 = x2√(1 + 3x3) d(yx3)/dx = x2√(1 + 3x3) yx3 = ∫x2√(1 + 3x3)dx = (1/9)∫√(1 + 3x3)d(1 + 3x3) = (2/27)√(1 + 3x3)3 + C x = 1, y = 1, giving C = 11/27 yx3 = [2√(1 + 3x3)3 + 11]/27 y = [2√(1 + 3x3)3 + 11]/(27x3)

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