標題:
ordinary differential equation
發問:
1) y' = x secy Does y = arcsin (0.5x^2 + C) equal y = arcsin (x^2 + C) ? The former is myu answer while the latter is the answer from the book. 2) y' = y(x+2y)/x(2x+y)
最佳解答:
其他解答:
ordinary differential equation
發問:
1) y' = x secy Does y = arcsin (0.5x^2 + C) equal y = arcsin (x^2 + C) ? The former is myu answer while the latter is the answer from the book. 2) y' = y(x+2y)/x(2x+y)
最佳解答:
此文章來自奇摩知識+如有不便請留言告知
1) y' = x secy 1/sec y dy/dx = x cos y dy = x dx ∫cos y dy = ∫x dx sin y = 0.5x^2 +C y = arcsin(0.5x^2 + C) Checking, y' = [1-(0.5x^2+C)]^-0.5?x = x secy [1] If y = arcsin(x^2 + C), y' would be [1-(x^2+C)]^-0.5?2x = 2x secy ≠ x secy The answer from the book should be incorrect [1] Consider a right triangle with 1 as hypotenuse and (0.5x^2+C) as opposite side, then [1-(0.5x^2+C)]^-0.5 would be the adjacent side. So sec y = [1-(0.5x^2+C)]^-0.5 2) f(x,y) = y(x+2y) is of degree 2 and g(x,y) = x(2x+y) is of degree 2 The differential equation is homogeneous Let wx = y, where w is a function of x, then y' = w+w'x y' = y(x+2y)/x(2x+y) w+w'x = wx(x+2wx)/x(2x+wx) w'x = wx^2(1+2w)/x^2(2+w) - w x dw/dx = [w(1+2w)-w(2+w)]/(2+w) dw/dx = (w+2w^2-2w-w^2)/[x(2+w)] (2+w)/(w^2-w) dw = dx/x ∫[-2/w + 3/(w-1)] dw = ∫dx/x [2] -2ln w + 3ln(w-1) = ln x +C -2ln(y/x) + 3ln(y/x-1) = ln x +C [2] Let (2+w)/(w^2-w) = A/w + B/(w-1) Then A(w-1)+Bw = (2+w) Let w=1, Then B=3 Let w=0, Then -A=2 -> A=-2 Therefore (2+w)/(w^2-w) = -2/w + 3/(w-1) If you are not satisfied with the solution or plan to remove the question, please notify me. 2013-07-02 19:17:12 補充： For question 2, direct substitution shows that y=0 is also a solution.其他解答:
文章標籤
全站熱搜
留言列表
