Show that fraction of differential equation solotions is constant Let  $x_1,x_2,x_3$ be solutions of $\dot{x}(t)+a(t)x(t)=b(t)$


show that$$ \frac{x_2-x_1}{x_3-x_1}= const$$


I tried to isolate x and got to $\frac{x_2-x_1}{x_3-x_1} $ $= \frac{\dot{x_1}-\dot{x_2}}{\dot{x_1}-\dot{x_3}}$
but not sure how to continue.
thanks ahead.
 A: I'll use the Lagrange notation using primes rather than the Newton notation using the overdot. Note that, from the given differential equation,
$$x_i'=b-ax_i$$
for $i=1,2,3$, therefore, using the quotient rule,
$$\begin{align}
\left(\frac{x_2-x_1}{x_3-x_1} \right)' 
 &= \frac{(x_2-x_1)'(x_3-x_1)-(x_3-x_1)'(x_2-x_1)}{(x_3-x_1)^2} \\[2 ex]
 &= \frac{(x_2'-x_1')(x_3-x_1)-(x_3'-x_1')(x_2-x_1)}{(x_3-x_1)^2} \\[2 ex]
 &= \frac{(b-ax_2-b+ax_1)(x_3-x_1)-(b-ax_3-b+ax_1)(x_2-x_1)}{(x_3-x_1)^2} \\[2 ex]
 &= \frac{a(x_1-x_2)(x_3-x_1)-a(x_1-x_3)(x_2-x_1)}{(x_3-x_1)^2} \\[2 ex]
 &= \frac{a(x_1-x_2)(x_3-x_1)-a(x_3-x_1)(x_1-x_2)}{(x_3-x_1)^2} \\[2 ex]
 &= 0
\end{align}$$
Thus $\dfrac{x_2-x_1}{x_3-x_1}$ is a constant.
Of course, this assumes that $x_3-x_1$ is not zero. I'll let you handle that detail.
A: This is one of the cases where we can peek at the solutions. Using
$$
A(t) = \int\limits_{t_0}^t a(\tau) \, d\tau
$$
we can solve the ODE:
\begin{align}
\dot{x} + a(t) \, x(t) & = b(t) \iff \\
\frac{d}{dt} \left(
e^{A(t)} \, x(t)
\right) &= e^{A(t)} \, b(t) \Rightarrow \\
x(t) &= e^{-A(t)} \left(
e^{A(t_0)} \, x(t_0) + 
 \int\limits_{t_0}^t e^{A(\tau)} \, b(\tau) \, d\tau 
\right)
\\
&=  x(t_0) \, e^{-A(t)} + f(t)
\end{align}
where $A(t_0) = 0$ was used. The explicit solution shows that solutions are characterized by their value $x(t_0)$ at some $t_0 \le t$. We get
$$
\frac{x_2(t) - x_1(t)}{x_3(t) - x_1(t)}
=
\frac{x_2(t_0)\,e^{-A(t)}+f(t)- x_1(t_0)\,e^{-A(t)}-f(t)}
{x_3(t_0)\,e^{-A(t)}+ f(t)- x_1(t_0)\,e^{-A(t)}-f(t)} 
= \frac{x_2(t_0) - x_1(t_0)}{x_3(t_0) - x_1(t_0)}
= \text{const}
$$
