Problem with solving ODE: $t \frac{dx}{dt} + (1+ \alpha t)x = t$ For $\alpha \in \mathbb{R}$ and $t> 0$, I have the inhomogeneous equation:
$t \frac{dx}{dt} + (1+ \alpha t)x = t$. 
I am asked to make it homogeneous and then determine the general solution. However, I am confused since the equation is not of the general form: 
$\frac{dx}{dt} + a(t)x = b(t)$.
I know I should set $b(t)$ equal to $0$ to make an equation homogeneous. My confusion is about $\frac{dx}{dt}$ being multiplied by $t$. 
Am I supposed to change the equation into $\frac{dx}{dt} + \frac{(1+ \alpha t)x}{t} = 1$, and then into $\frac{dx}{dt} + \frac{(1+ \alpha t)x}{t} = 0$? This does not seem right since there is no longer a function of $t$ on the right hand side (when the right hand side equals 1).
 A: First thing to notice
$$
\frac{d}{dt}tx = x + t\frac{dx}{dt}
$$
which we can look at the original equation we can see that the above is embedded into the equation.
$$
t\frac{dx}{dt}  + x + atx = t
$$
or
$$
\frac{d}{dt}tx  + atx = t
$$
we can now set $u = tx$
$$
\frac{du}{dt} + au = t
$$
which you can now proceed with.
A: *

*Let solve first for $\alpha=0$.


We get $\displaystyle t\dot x+x=t$ and by dividing by $t$ (since $t>0$) $\displaystyle \quad\dot x+\frac xt=1$
The homogeneous equation can be rearranged $\displaystyle \frac{\dot x}x=-\frac 1t$ whose solutions are $\displaystyle x=\frac ct$
Since $x=t$ is an easy particular solution, we get $$x=t+\frac ct$$


*

*Let solve now for $\alpha\neq 0$.


By the same operation we get $\displaystyle \dot x+\left(\frac{1+\alpha t}t\right)x=1$
The homogeneous equation $\frac{\dot x}x=-\frac 1t-\alpha\quad$ integrates to $\quad\displaystyle x=c\times\frac 1t\times e^{-\alpha t}=\frac{c\,e^{-\alpha t}}t$
But there is no obvious particular solution for the equation with RHS that I can see, so we'll use the variation of the constant method.
$\displaystyle t\dot x+(1+\alpha t)x=\dot c\, e^{-\alpha t}=t\iff \dot c=t\,e^{\alpha t}\iff c=\frac{\alpha t-1}{\alpha^2}e^{\alpha t}$
Finally a particular solution is given by $\displaystyle x=\frac 1\alpha-\frac 1{\alpha^2t}$
And the general solution is $$x=\frac 1{\alpha^2}\left(\alpha+\frac{c e^{-\alpha t}-1}t\right)$$
