x'-t*x=x^(1/2), x=x(t). Here is what i tried.

Let there be z = x^(1-1/2) = x^(1/2) => x=z^2

x'=2*z*z'

2*z*z'-t*z^2 = z | :z => 2*z' -t*z = 1 |:2 => z' -(t*z)/2 = 1/2

z' - (t*z)/2 = 0 => z' = (t*z)/2 => dz/dt = (t*z)/2 => dz/z = (t/2)*dt | integrate => ln|z| = (t^2)/4 + C => z = e^((t^2)/4 +C)

z0 = e^((t^2)/4 +C)

zp = e^((t^2)/4 +C(t)) --- I need to find C(t)

z'p= e^((t^2)/4 +C(t)) *( t/2 + C'(t))

z' -(t*z)/2 = 1/2

e^((t^2)/4 +C(t)) *( t/2 + C'(t)) - ( (t/2)*e^((t^2)/4 +C(t)) = 1/2 => e^((t^2)/4 +C(t)) * C'(t) = 1/2 --- Here i only needed C'(t) then integrate to find C(t)--- but C(t) appears too, can you tell me where is my mistake?