工程數學問題求解

y' = 1 + y/x + (y/x)^2

dy/dx = 1 + y/x + (y/x)^2

Let u = y/x, y = ux

dy/dx = u + xdu/dx

So, the differential equation becomes:

u + xdu/dx = 1 + u + u^2

xdu/dx = 1 + u^2

Separating variables,

S du/(1 + u^2) = S dx / x

tan^-1u = lnx + C, where C is a constant

tan^-1(y/x) = lnx + C

y/x = tan(lnx + C)

y = x tan(lnx + C)

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題目是-(y/x)^2??

2010-07-08 23:55:15 補充：
ln((x+y)/(x-y))=2lnx+c

y' = 1 + y/x + (y/x)^2

dy/dx = 1 + y/x + (y/x)^2

Let u = y/x, y = ux

dy/dx = u + xdu/dx

So, the differential equation becomes:

u + xdu/dx = 1 + u + u^2

xdu/dx = 1 + u^2

Separating variables,

S du/(1 + u^2) = S dx / x

tan^-1u = lnx + C, where C is a constant

tan^-1(y/x) = lnx + C

y/x = tan(lnx + C)

y = x tan(lnx + C)