nonlinear, nonhomogeneous ODE 1. order Solve $x'(t)-\dfrac{a}{t}x(t)=b(t),~a=const,~x(0)=0$. 
Homogeneous Solution:
$\dfrac{x'(t)}{x(t)}=\dfrac{a}{t}\quad|\int\\
\ln(x(t))=a\ln(t)+c,~c=const\quad| e\\
x(t)=t^ae^c$
Is that correct?
No clue how to figure the particular solution or the general solution respectively.
edit: Thanks  @agha for the helpfull comment. I try to write down the entire solution. With a slighty different Noation.
Let $t\in[0,t_f]$
$$(1)~ x'(t) - \dfrac{a}{t}x(t) = b(t)$$
$$(2)~ x_h(t) = Kt^a$$
Particular Solution:
$$x_s(t) = c(t)t^a~\text{with}~ x'(t) = c'(t)t^a + c(t)at^{a-1}$$
Substitute in (1)
$$c'(t)t^a + c(t)at^{a-1} - \dfrac{a}{t}c(t)t^a = b(t)$$
$$c'(t) = \dfrac{b(t)}{t^a}\quad |\int{}$$
$$c(t) = \int{\dfrac{b(t)}{t^a}dt} + K'$$
It follows for $x_s(t)$
$$(3)~ x_s(t) = \left(\int{\dfrac{b(t)}{t^a}dt} + K'\right)t^a$$
The general solution reads (2)+(3), i.e.
$$(4)~ x(t) = Kt^a + \left(\int{\dfrac{b(t)}{t^a}dt} + K'\right)t^a = \left(\int{\dfrac{b(t)}{t^a}dt} + K''\right)t^a\\
\text{where}~ K''=K+K'$$
With the initial conditon $x(0)=0$ we might identify $K''$ 
$$(5)~ x(0) = 0 = \int{\dfrac{b(t)}{t^a}dt}\bigg|_{t_f} + K''\\
K'' = -\int{\dfrac{b(t)}{t^a}dt}\bigg|_{t_f}$$
Thus the solution reads:
$$(6)~x(t) = \left(\int{\dfrac{b(t)}{t^a}dt} - \int{\dfrac{b(t)}{t^a}dt}\bigg|_{t_f}\right)t^a$$
Am I right?

You can use general method - Variation of parameters.
Let $e^C=D$, so you have solution of homogeneous ODE. $x(t)=Dt^{a}$ ($D$ is a constant). Now by variation of parameter solution of non-homogeneous ODE is $x(t)=D(t)t^{a}$($D$ is a function).Next:
$$x'(t)=(D(t)t^{a})'=D'(t)t^{a}+aD(t)t^{a-1}$$
Now substitute to equation:
$$x'(t)-\frac{a}{t}x(t)=(D'(t)t^{a}+aD(t)t^{a-1})-\frac{a}{t}D(t)t^{a}=D'(t)t^{a}=b(t)$$
So:
$$D'(t)=\frac{b(t)}{t^{a}}$$ 
Finally:
$$D(t)=\int \frac{b(t)}{t^{a}} dt $$
