if y=y(x) is a solution of
y' = y/x + x/y and y(1)=2
then y(e) is ?
so I used the homogeneous equation method and subbed v=y/x then found y' = xv' +v and subbed those into the original equation to get
xv' + v = v +1/v
what do i do from here?
v = y/x ? so y = v x
then
y' = v + x dv/dx
v + x dv/dx = v + 1/v agree with you so far
x dv/dx = 1/v
v dv = dx/x
(1/2) v^2 = ln x + c
(1/2) y^2/x^2 = ln x + c
(1/2) y^2 = x^2 ln x + c x^2
when x = 1 , y = 2
(1/2)(4) = ln 1 + c(1) but ln 1 = 0
c = 2
(1/2) y ^2 = x^2 ln x + 2 x^2
now if x = e, ln x = 1
(1/2) y^2 = e^2 + 2e^2 = 3 e^2
y^2 = 6 e^2
y = e sqrt 6