Solving a differential equation, substitution has been recommended I have the differential equation, $x$ is a function of $t$:

$xx'=-a(t+b)+cx$ where $a,b$ and $c$ are constants

We have been told that using the homogeneous substitution method might be useful, but when I sub in $v=\frac xt$ I get:

$vt(v+v't)=-a(t+b)+cvt $

this seems more complicated.

This is part of a larger question which was solve the following system of equations:

$x'=1$  which implies $x=t+x_0$
$y'=0$ which implies $y=y_0$
$z'=3y -\frac{2xy^2}{z}$ so subbing in the values above gives $z'=3y_0-\frac{2y_0^2(t+x_0)}{z}$

As you can see the equation that I generalised at the start refers to the third equation. I'm not sure if this context helps in answering the question.
$xx'=-a(t+b)+cx$

$x'=-\dfrac{a(t+b)}{x}+c$

This is an ODE of the form $x'=f\left(\dfrac{x}{t+b}\right)$ .

In fact the substitution $v=\dfrac{x}{t+b}$ can leads the ODE to becomes a separable ODE.

Let $v=\dfrac{x}{t+b}$ ,

Then $x=(t+b)v$

$x'=(t+b)v'+v$

$\therefore(t+b)v'+v=-\dfrac{a}{v}+c$

$(t+b)v'=-\dfrac{a}{v}+c-v$

$(t+b)\dfrac{dv}{dt}=-\dfrac{v^2-cv+a}{v}$

$\dfrac{dt}{t+b}=-\dfrac{v}{v^2-cv+a}dv$

$\int\dfrac{dt}{t+b}=-\int\dfrac{v}{v^2-cv+a}dv$

If you think $\int\dfrac{v}{v^2-cv+a}dv$ is too tedious to be simplified, you may leave it as for the solution, however in fact $\int\dfrac{v}{v^2-cv+a}dv$ is just a standard integral of the form http://en.wikipedia.org/wiki/List_of_integrals_of_rational_functions#Integrands_of_the_form_xm_.2F_.28a_x2_.2B_b_x_.2B_c.29n, means that leaving it as for the solution tends to be unreasonable.
This by no means gets all solutions but here are a whole heap of them:

Guessing $x(t) = et + f$ allowing for coefficients in $\mathbb{C}$, we see that this equation reads:

$$ e^2 t + ef + a(t + b) - cet - cf = 0 $$

$$ (e^2 + a - ce)t + (ef - ab - cf) = 0$$

The first is a quadratic in $e$ so you can solve that, the second is then a linear equation in $f$ which you can solve provided $e - c \neq 0$.  

If $e = c$, then $c + \sqrt{c^2 - 4a} = 2c$ meaning $a=0$. But then we just have $x' = c$ in which case $x = ct + x_0$ is a solution