Initial value problem y'-y/x=x/(1+4*x^2)

I need help with the following example; It is initial value problem.

y'-y/x=x/(1+4*x^2)

your help is much appreciated.

y' + (-1 / x) y = x / (1 + 4x^2)

Well, this isn't an IVP since there aren't any initial conditions. I'll give you the general solution though; start with the integrating factor:

u(x) = e^( ∫ (-1 / x) dx ) = e^(- ln(x)) = e^(ln(1 / x)) = 1/x

Multiplying through:

(1 / x)y' + (-1 / x^2) y = 1 / (1 + 4x^2)
[(1 / x) y] ' = 1 / (1 + 4x^2)

∫ [(1 / x) y] ' dx = ∫ 1 / (1 + 4x^2) dx

In the right integral:

u^2 = 4x^2
u = 2x
du = 2 dx
dx = (1/2) du

∫ [(1 / x) y] ' dx = (1/2) ∫ 1 / (1 + u^2) du
(1 / x) y = (1/2) arctan(u)
(1 / x) y = (1/2) arctan(2x) + C
y = (x / 2) arctan(2x) + Cx

Done! (well, sort of)