Differential Equation

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

y = integrate [(x^2 + y^2) / 2xy]dx

= 1/2y integrate [x + (y^2)/x] dx

= 1/2y [(x^2)/2 + (y^2)ln(x)] + C

2y^2 = (x^2)/2 + (y^2)ln(x) + C

(y^2)[2 - ln(x)] = (x^2)/2 + C

y^2 = [(x^2) + 2C]/ 2[2-ln(x)]

y = squareRoot {[(x^2) + 2C] / 2[2-ln(x)]}

where C is a contant; integrate 1/x dx = ln(x); integrate x dx = (x^2)/2

Thanks

2006-12-12 11:25:05 補充：
If your want to find C, the question should state some conditions. e.g. when x = 0, y = sth... Thanks

Rewrite dy/dx = (x^2 + y^2) / 2xy as
0= (x^2 + y^2)dx - 2xydy
Multiplying an integrating factor x^(-2) and solve the p.d.e. below
0= (1 + (y/x)^2)dx - 2(y/x)dy
We have
x^2 - y^2 = kx .....(*) where k is an arbitrary constant

This result has been checked by differentiating (*)

2006-12-19 20:33:39 補充：
Check the answer!!

dy/dx = (x^2+y^2)/2xy
dy/dx = [(x+y)^2-2xy]/2xy
dy/dx=(x+y)^2 -1 -----(1)
let z=x+y
dz/dx = 1+dy/dx
sub it into (1), we have
dz/dx-1 = (z)^2 -1
dz/dx = z^2
dz/z^2 = dx
integrate both sides, we have
- z^-1=x+C
-(x+y)^-1 = x+C
-1=(x+y)x + C(x+y)
-1=x^2 +Cx +(x+C)y
y=-(x^2+Cx+1)/(x+C)


2006-12-12 12:44:23 補充：
對唔住!一時漏了東西，現在重新來過dy/dx = (x^2+y^2)/2xy --------(1)let z=y/xdz/dx = -y/x^2 + (1/x)dy/dxthereforedy/dx=x(dz/dx +y/x^2) = xdz/dx + z ------(2)

2006-12-12 12:45:23 補充：
sub it into (1), then we havexdz/dx + z = (x/2y + y/2x) = (1/2)(1/z + z)xdz/dx = (1/2)(1/z -z) = (1/2)((1-z^2)/z)2zdz/(1-z^2) = dx/x(1/(1-z) – 1/(1+z))dz = dx/x

2006-12-12 12:45:41 補充：
Integrate both sides, we have-ln(1-z) – ln(1+z) = ln(x) + C-ln(1-z^2)= ln(Kx)1/(1-z^2)=Kx1/(1-(y/x)^2)=Kx1=Kx(1-(y/x)^2)1=Kx – Kx(y^2/x^2)(K/x)y^2=Kx-1y^2=(Kx-1)/Kxy= √[(Kx-1)/Kx]

2006-12-12 12:48:54 補充：
where K is a constant