Find the general solution to the first order linear equation differential equation...

3tx'+ x=12t

I'm studying for an upcoming exam and I am looking for an example of how to do this????

Find the general solution by using an integrating factor:
3tx' + x = 12t
3t(dx / dt) + x = 12t
dx / dt + x / (3t) = 4
dx / dt + P(t)x = f(t)
P(t) = 1 / (3t)
f(t) = 4
I(t) = ℮^[∫ P(t) dx]
I(t) = ℮^[∫ 1 / (3t) dt]
I(t) = ℮^(ln|t| / 3)
I(t) = ℮^(ln|∛t|)
I(t) = ∛t
I(t)y = ∫ I(t)f(t) dt
y∛t = ∫ 4∛t dt
y∛t = 3∛t⁴ + C
y = 3t + C / ∛t

Here is one way to do it. (There are others.)
First solve 3t*dx/dt + x = 0 ----> 3t*dx/dt = -x

This is variables separable leading to
INT 3/x dx = INT -1/t dt ----> 3*ln(x) = -ln(x) + ln(c) = ln(c/t)

ln(x^3) = ln(c/t) ----> x^3 = c/t ----> x = (c/t)^(1/3) = c*t^(-1/3) .... new c.

Now solve for the right hand side. Start with
x = at + b ----> dx/dt = a

Fit these into the differential equation.
3t*dx/dt + x = 3ta + at + b = 4at + b

This must be 12t and so a = 3, b = 0.
Full solution x = c*t^(-1/3) + 3t

Check.
x = c*t^(-1/3) + 3t ----> dx/dt = (-1/3)*c*t^(-4/3) + 3

3t*dx/dt + x = 3t*[(-1/3)*c*t^(-4/3) + 3] + c*t^(-1/3) + 3t = 12t ...... as required