考虑二阶微分方程 $\frac{d^{2}x}{dt^{2}} + a\frac{dx}{dt} = b,$ 其中 $a$ 和 $b$ 为非零常数，且 $\frac{dx}{dt} \ne \dfrac{b}{a}$。

1. Find an expression for the second derivative in terms of $y$.

Given $y = \frac{dx}{dt}$, we differentiate both sides with respect to $t$:
$\frac{dy}{dt} = \frac{d^{2}x}{dt^{2}}$

2. Substitute into the original equation and rearrange to show the required form.

Substitute $y = \frac{dx}{dt}$ and $\frac{dy}{dt} = \frac{d^{2}x}{dt^{2}}$ into the given second order differential equation:
$\frac{dy}{dt} + ay = b$
Rearranging to make $\frac{dy}{dt}$ the subject:
$\frac{dy}{dt} = b - ay \quad \text{(shown)}$

1. Separate the variables to solve the differential equation for $y$.

$\frac{dy}{dt} = b - ay$
$\int \frac{1}{b-ay} \, dy = \int 1 \, dt$

2. Integrate both sides with respect to their variables.

$-\frac{1}{a} \ln |b - ay| = t + C$

3. Make $y$ the subject of the formula.

$\ln |b - ay| = -at - aC$
$b - ay = \pm e^{-at - aC}$
$b - ay = A e^{-at} \quad \text{where } A = \pm e^{-aC}$
$y = \frac{b}{a} - \frac{A}{a} e^{-at}$
Let $B = -\frac{A}{a}$.
$y = \frac{b}{a} + B e^{-at}$

4. Substitute $y = \frac{dx}{dt}$ and set up the integral to find $x$.

$\frac{dx}{dt} = \frac{b}{a} + B e^{-at}$
$x = \int \left( \frac{b}{a} + B e^{-at} \right) \, dt$

5. Evaluate the integral to find the general solution for $x$.

$x = \frac{b}{a}t - \frac{B}{a} e^{-at} + D$
where $B$ and $D$ are arbitrary constants.