consider the function:

f(x) = (1/3)(x^2+4)^{3/2}

df/dx = (1/3)(3/2) 2x (x^2+4)^{1/2} = x[(x^2 + 4)^(1/2)]

So you can use this to integrate your DE you have:

y = (1/3)(x^2+4)^{3/2} + C,

when x=0 we require:

1 = (1/3)(4)^{3/2} + C

so:

C = 1-8/3 = -5/3

and so:

y = (1/3)(x^2+4)^{3/2} - 5/3

RonL