问题描述
已知，有二阶欧拉方程：
$$
x^{2} y^{\prime \prime} + a x y^{\prime} + b y = f(x)
$$
计算过程
首先，若 $x$ $>$ $0$, 则令：
$$
x = e^{t}
$$
即：
$$
t = \ln x
$$
于是：
$$
y^{\prime} =
$$
$$
\frac{\mathrm{d} y}{\mathrm{d} x} =
$$
$$
\frac{\mathrm{d} y}{\mathrm{d} t} \times \frac{\mathrm{d} t}{\mathrm{d} x} =
$$
$$
\frac{\mathrm{d} y}{\mathrm{d} t} \times \frac{1}{x}.
$$
进而：
$$
y^{\prime \prime} =
$$
$$
\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}} =
$$
$$
\bigg( \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{1}{x} \bigg)^{\prime}_{x} =
$$
$$
\frac{\mathrm{d} (\frac{\mathrm{d} y}{\mathrm{d} t})}{\mathrm{d} t} \cdot \frac{\mathrm{d} t}{\mathrm{d} x} \cdot \frac{1}{x} + \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \big(\frac{-1}{x^{2}} \big) =
$$
$$
\frac{\mathrm{d} (\frac{\mathrm{d} y}{\mathrm{d} t})}{\mathrm{d} t} \cdot \frac{1}{x} \cdot \frac{1}{x} + \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \big(\frac{-1}{x^{2}} \big) =
$$
$$
\frac{\mathrm{d ^{2}} y}{\mathrm{d} t^{2}} \cdot \frac{1}{x} \cdot \frac{1}{x} + \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \big(\frac{-1}{x^{2}} \big) \Rightarrow
$$
$$
\frac{\mathrm{d^{2}} y}{\mathrm{d} x^{2}} =
\bigg(\frac{\mathrm{d ^{2}} y}{\mathrm{d} t^{2}} – \frac{\mathrm{d} y}{\mathrm{d} t}\bigg) \cdot \frac{1}{x^{2}}.
$$
综上：
$$
x^{2} y^{\prime \prime} + a x y^{\prime} + b y = f(x) \Rightarrow
$$
$$
e^{2t} \bigg(\frac{\mathrm{d ^{2}} y}{\mathrm{d} t^{2}} – \frac{\mathrm{d} y}{\mathrm{d} t}\bigg) \cdot \frac{1}{x^{2}} + a x \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{1}{x} + b y = f(x) \Rightarrow
$$
$$
e^{2t} \bigg(\frac{\mathrm{d ^{2}} y}{\mathrm{d} t^{2}} – \frac{\mathrm{d} y}{\mathrm{d} t}\bigg) \cdot \frac{1}{e^{2 t}} + a x \frac{\mathrm{d} y}{\mathrm{d} t} \cdot \frac{1}{x} + b y = f(x) \Rightarrow
$$
$$
\bigg(\frac{\mathrm{d ^{2}} y}{\mathrm{d} t^{2}} – \frac{\mathrm{d} y}{\mathrm{d} t}\bigg) \cdot + a \frac{\mathrm{d} y}{\mathrm{d} t} + b y = f(x) \Rightarrow
$$
于是，可得一个以 $t$ 为自变量，$y(t)$ 为末知函数的二阶线性常系数微分方程：
$$
\frac{\mathrm{d}^{2} y}{\mathrm{~d} t^{2}}+(a-1) \frac{\mathrm{d} y}{\mathrm{~d} t}+b y=f(\mathrm{e}^{t}).
$$
接着，通过计算二阶线性常系数微分方程的方法计算出以 $t$ 为自变量的函数 $y(t)$ 后，用 $t$ $=$ $\ln x$ 进行反代，即可得出以 $x$ 为自变量的函数 $y(x)$.