x'-t*x=x^(1/2), x=x(t). We use this notation at college so in x and y it is y'-x*y=y^(1/2)
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?
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?