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# 基础练习
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## 1. 求下列方程的通解
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### (1)
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$$
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\frac{dy}{dx} - \frac{y+1}{x+1} = x+1
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$$
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**参考解答**：
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令 $X = x + 1$，$Y = y + 1$，则 $dy = dY$，$dx = dX$。
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方程可化为：
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$$
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\frac{dY}{dX} - \frac{Y}{X} = X,
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$$
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由一阶非齐次线性微分方程通解公式有：
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$$
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Y = e^{\int \frac{dX}{X}} \left[ \int X \cdot e^{-\int \frac{dX}{X}} dX + C' \right] = X \left( \int dX + C' \right) = X(X + C').
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$$
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将 $X = x + 1$，$Y = y + 1$ 代入上面通解得原方程通解为：
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$$
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y = (x+1)(x+C) - 1.
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$$
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---
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### (2)
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$$
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y'y'' = 3(y')^2
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$$
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**参考解答**：
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**方法1**：令 $z = y'$，方程化为：
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$$
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z z'' = 3(z')^2,
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$$
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再令 $z' = p(z)$，则
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$$
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z'' = p \frac{dp}{dz},
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$$
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方程化为：
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$$
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z p \frac{dp}{dz} = 3p^2 \quad\Rightarrow\quad \frac{1}{p} dp = \frac{3}{z} dz.
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$$
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积分得：
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$$
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\ln|p| = 3\ln|z| + \ln|C_1|,
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$$
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整理得：
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$$
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z' = C_1 z^3 \quad\Rightarrow\quad \frac{dz}{dx} = C_1 z^3 \quad\Rightarrow\quad \frac{dz}{z^3} = C_1 dx,
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$$
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积分得：
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$$
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-\frac{1}{2z^2} = C_1 x + C_2 \quad\Rightarrow\quad z = \pm \frac{1}{\sqrt{C_1 x + C_2}} \quad\Rightarrow\quad dy = \pm \frac{1}{\sqrt{C_1 x + C_2}} dx,
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$$
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再积分得方程通解为：
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$$
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y = \pm \frac{2\sqrt{C_1 x + C_2}}{C_1} + C_3,
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$$
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整理得方程通解可化为：
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$$
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x = C_1 y^2 + C_2 y + C_3.
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$$
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**方法2**：交换因变量与自变量地位：将
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$$
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y'(x) = \frac{1}{x'(y)},\quad y''(x) = -\frac{x''(y)}{[x'(y)]^3},\quad y'''(x) = \frac{3[x''(y)]^2 - x'(y)x'''(y)}{[x'(y)]^5},
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$$
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代入原方程化为：
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$$
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\frac{1}{x'(y)} \cdot \frac{3[x''(y)]^2 - x'(y)x'''(y)}{[x'(y)]^5} = 0,
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$$
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即有：
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$$
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x''(y) = 0 \quad \text{或} \quad x'(y) = 0 \; (\text{包含于第一个等式中}),
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$$
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对 $x''(y) = 0$ 连续积分得原方程通解为：
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$$
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x(y) = C_1 y^2 + C_2 y + C_3.
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$$
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---
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### (3)
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$$
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x y y'' + x (y')^2 = 3 y y'
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$$
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**参考解答**：
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**方法1**：方程可变形为：$x(yy')' = 3yy'$，令 $yy' = u$，方程化为：
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$$
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\frac{du}{dx} = \frac{3u}{x},
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$$
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分离变量得：
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$$
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\frac{du}{u} = \frac{3dx}{x},
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$$
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积分得：
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$$
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\ln|u| = 3\ln|x| + \ln|C_1| \quad\Rightarrow\quad yy' = C_1 x^3,
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$$
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再分离变量得：
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$$
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y dy = C_1 x^3 dx,
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$$
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积分整理得方程通解为：
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$$
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y^2 = \frac{C_1}{2} x^4 + C_2.
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$$
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**方法2**：令 $u = y^2$，则有
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$$
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\frac{du}{dx} = 2yy',\quad \frac{d^2u}{dx^2} = 2(y')^2 + 2yy'',
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$$
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代入原方程，方程化为：
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$$
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x u'' = 3u',
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$$
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令 $p = u'$，则有
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$$
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x \frac{dp}{dx} = 3p,
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$$
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用分离变量积分得：
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$$
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\frac{dp}{p} = \frac{3dx}{x},
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$$
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积分得：
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$$
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\ln p = 3\ln|x| + \ln|C_1| \quad\Rightarrow\quad \frac{du}{dx} = C_1 x^3,
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$$
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再分离变量得：
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$$
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du = C_1 x^3 dx,
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$$
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积分整理得方程通解为：
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$$
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y^2 = \frac{C_1}{4} x^4 + C_2.
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$$
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**方法3**：分析方程三项的结构，由于
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$$
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(x y y')' = y y' + x (y')^2 + x y y'',
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$$
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故原方程可以改写为：
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$$
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(x y y')' = 4 y y',
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$$
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于是令 $x y y' = u(x)$，$y y' = \frac{u}{x}$，
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方程可化为可分离变量方程：
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$$
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\frac{du}{dx} = 4 \frac{u}{x} \quad\Rightarrow\quad \frac{du}{u} = 4 \frac{dx}{x},
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$$
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积分得：
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$$
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u = C_1 x^4 = x y y',
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$$
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再分离变量有：
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$$
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C_1 x^3 dx = y dy,
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$$
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积分得：
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$$
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\frac{C_1}{4} x^4 = \frac{1}{2} y^2 + C_2,
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$$
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于是方程通解为：
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$$
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y^2 = C_3 x^4 + C_4.
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$$
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---
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### (4)
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$$
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(2x-1)^2 y'' - 4(2x-1) y' + 8y = 8x
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$$
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**参考解答**：
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令 $2x - 1 = t$，则
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$$
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\frac{dy}{dx} = \frac{dy}{dt} \cdot \frac{dt}{dx} = 2 \frac{dy}{dt},\quad \frac{d^2y}{dx^2} = 4 \frac{d^2y}{dt^2}.
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$$
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代入原方程整理得欧拉方程：
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$$
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t^2 \frac{d^2y}{dt^2} - 2t \frac{dy}{dt} + 2y = t + 1.
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$$
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由欧拉方程的求解方法，令 $t = e^u$，$u = \ln t$，于是
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$$
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\frac{dy}{dt} = \frac{dy}{du} \cdot \frac{du}{dt} = \frac{1}{t} \frac{dy}{du},
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$$
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$$
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\frac{d^2y}{dt^2} = -\frac{1}{t^2} \frac{dy}{du} + \frac{1}{t} \frac{d^2y}{du^2} \cdot \frac{du}{dt} = \frac{1}{t^2} \left( \frac{d^2y}{du^2} - \frac{dy}{du} \right),
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$$
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化为关于 $u$ 变量的线性微分方程：
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$$
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\frac{d^2y}{du^2} - 3 \frac{dy}{du} + 2y = e^u + 1.
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$$
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其特征方程为 $r^2 - 3r + 2 = 0$，得特征根为 $r_1 = 1$，$r_2 = 2$。
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对应齐次方程的通解为：
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$$
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Y = C_1 e^u + C_2 e^{2u}.
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$$
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设非齐次线性微分方程有特解为：
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$$
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y^* = A u e^u + B,
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$$
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代入非齐次线性微分方程比较系数得：
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$$
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A = -1,\quad B = \frac{1}{2}.
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$$
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于是方程通解为：
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$$
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y = Y + y^* = C_1 e^u + C_2 e^{2u} - u e^u + \frac{1}{2}.
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$$
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代回 $u = \ln t = \ln(2x-1)$，得
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$$
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y = C_1 (2x-1) + C_2 (2x-1)^2 - (2x-1) \ln(2x-1) + \frac{1}{2}.
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$$
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---
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### (5) 常系数非齐次线性方程（待定系数法）
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$$
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y'' - 2y' + y = x e^x
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$$
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**参考解答**：
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先解齐次方程 $y'' - 2y' + y = 0$，特征方程 $r^2 - 2r + 1 = 0$，得重根 $r = 1$，齐次通解为
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$$
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y_h = (C_1 + C_2 x) e^x.
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$$
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非齐次项 $f(x) = x e^x$，而 $e^x$ 是齐次解的一部分（且 $x e^x$ 也是齐次解，因为重根），故应设特解形式为
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$$
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y^* = x^2 (A x + B) e^x = (A x^3 + B x^2) e^x.
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$$
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为简化计算，通常设 $y^* = x^2 (A x + B) e^x$。计算导数：
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$$
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y^* = e^x (A x^3 + B x^2),
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$$
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$$
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y'^* = e^x (A x^3 + B x^2) + e^x (3A x^2 + 2B x) = e^x [A x^3 + (B+3A) x^2 + 2B x],
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$$
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$$
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y''^* = e^x [A x^3 + (B+3A) x^2 + 2B x] + e^x [3A x^2 + 2(B+3A)x + 2B] = e^x [A x^3 + (B+6A) x^2 + (2B+6A)x + 2B].
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$$
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代入原方程 $y'' - 2y' + y = x e^x$，两边约去 $e^x$，得
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$$
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\big[A x^3 + (B+6A) x^2 + (2B+6A)x + 2B\big] - 2\big[A x^3 + (B+3A) x^2 + 2B x\big] + \big(A x^3 + B x^2\big) = x.
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$$
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合并同类项：
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$x^3$ 项：$A - 2A + A = 0$；
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$x^2$ 项：$(B+6A) - 2(B+3A) + B = B+6A -2B -6A + B = 0$；
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$x$ 项：$(2B+6A) - 4B + 0 = 6A - 2B$；
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常数项：$2B - 0 + 0 = 2B$。
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比较系数得
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$$
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\begin{cases}
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6A - 2B = 1, \\
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2B = 0,
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\end{cases}
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$$
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解得 $B=0$，$A = \frac{1}{6}$。所以特解为
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$$
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y^* = \frac{1}{6} x^3 e^x.
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$$
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原方程通解为
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$$
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y = y_h + y^* = (C_1 + C_2 x) e^x + \frac{1}{6} x^3 e^x.
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$$
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---
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已知二阶线性齐次微分方程
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$$
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(x-1) y'' - x y' + y = 0 \qquad (x > 1)
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$$
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的一个特解为 $y_1 = e^x$，求该方程的通解。
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**参考解答**：
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先将原方程化为标准形式 $y'' + p(x) y' + q(x) y = 0$。原方程为
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$$
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(x-1) y'' - x y' + y = 0,
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$$
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两边除以 $x-1$（$x>1$）：
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$$
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y'' - \frac{x}{x-1} y' + \frac{1}{x-1} y = 0.
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$$
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因此
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$$
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p(x) = -\frac{x}{x-1}.
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$$
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计算积分
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$$
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\int p(x) dx = -\int \frac{x}{x-1} dx = -\int \left(1 + \frac{1}{x-1}\right) dx = -\left[ x + \ln(x-1) \right] = -x - \ln(x-1).
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$$
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所以
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$$
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e^{-\int p(x) dx} = e^{x + \ln(x-1)} = (x-1) e^x.
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$$
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已知 $y_1 = e^x$，则 $y_1^2 = e^{2x}$。由刘维尔公式，另一线性无关解为
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$$
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y_2 = y_1 \int \frac{e^{-\int p(x) dx}}{y_1^2} dx = e^x \int \frac{(x-1) e^x}{e^{2x}} dx = e^x \int (x-1) e^{-x} dx.
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$$
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计算积分：
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$$
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\int (x-1) e^{-x} dx = \int x e^{-x} dx - \int e^{-x} dx.
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$$
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分别计算：
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$$
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\int x e^{-x} dx = -x e^{-x} - e^{-x} + C = -e^{-x}(x+1) + C,
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$$
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$$
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\int e^{-x} dx = -e^{-x} + C.
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$$
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所以
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$$
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\int (x-1) e^{-x} dx = -e^{-x}(x+1) - (-e^{-x}) + C = -e^{-x}(x+1) + e^{-x} + C = -x e^{-x} + C.
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$$
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取 $C=0$ 得一个特解
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$$
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y_2 = e^x \cdot (-x e^{-x}) = -x.
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$$
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常数因子可吸收，故可取 $y_2 = x$。因此两个线性无关的特解为 $y_1 = e^x$，$y_2 = x$。
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原方程的通解为
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$$
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y = C_1 e^x + C_2 x.
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$$
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**注**：直接验证可知 $y = x$ 也满足原方程，与 $e^x$ 线性无关，因此通解如上。
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$$
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y'' + 2y' + 2y = e^{-x} \cos x
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$$
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**参考解答**：
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先解齐次方程 $y'' + 2y' + 2y = 0$。特征方程为 $r^2 + 2r + 2 = 0$，解得 $r = -1 \pm i$。故齐次通解为
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$$
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y_h = e^{-x} (C_1 \cos x + C_2 \sin x).
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$$
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非齐次项 $f(x) = e^{-x} \cos x$，而 $e^{-x} \cos x$ 恰好是齐次解的一部分（对应复根 $-1 \pm i$），因此应设特解形式为
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$$
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y^* = x e^{-x} (A \cos x + B \sin x).
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$$
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计算导数（可简化计算，利用复数法或直接求导）。采用待定系数法，设
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$$
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y^* = x e^{-x} (A \cos x + B \sin x) = e^{-x} \left( A x \cos x + B x \sin x \right).
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$$
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先计算 $y'^*$ 和 $y''^*$。令 $u(x) = A x \cos x + B x \sin x$，则 $y^* = e^{-x} u$，于是
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$$
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y'^* = e^{-x} (u' - u), \quad y''^* = e^{-x} (u'' - 2u' + u).
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$$
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代入原方程 $y'' + 2y' + 2y = e^{-x} \cos x$，两边约去 $e^{-x}$，得
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$$
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(u'' - 2u' + u) + 2(u' - u) + 2u = \cos x,
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$$
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即
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$$
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u'' + 0 \cdot u' + u = \cos x.
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$$
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所以只需解 $u'' + u = \cos x$。
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计算 $u$ 的各阶导数：
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$$
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u = A x \cos x + B x \sin x,
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$$
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$$
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u' = A \cos x - A x \sin x + B \sin x + B x \cos x = (A + B x) \cos x + (B - A x) \sin x,
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$$
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$$
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u'' = -A \sin x - A \sin x - A x \cos x + B \cos x + B \cos x - B x \sin x = (-2A - B x) \sin x + (2B - A x) \cos x.
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$$
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注意，更简便的方法是：由于 $u'' + u$ 应等于 $\cos x$，而 $u$ 中 $x \cos x$ 和 $x \sin x$ 项的导数会与自身抵消后留下 $\cos x$ 和 $\sin x$ 项。直接代入求系数：
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由 $u'' + u$ 的计算，实际上对于 $u = x (A \cos x + B \sin x)$，有 $u'' + u = 2(-A \sin x + B \cos x)$（这是一个常用结果，可通过复数推导）。验证：令 $z = x e^{ix}$，则 $z'' + z = 2i e^{ix}$，实部对应 $\cos$，虚部对应 $\sin$。因此
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$$
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u'' + u = 2(-A \sin x + B \cos x).
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$$
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令其等于 $\cos x$，得
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$$
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2B \cos x - 2A \sin x = \cos x,
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$$
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比较系数得
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$$
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2B = 1, \quad -2A = 0 \quad\Rightarrow\quad A = 0,\ B = \frac{1}{2}.
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$$
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因此
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$$
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u = \frac{1}{2} x \sin x,
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$$
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所以
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$$
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y^* = e^{-x} u = \frac{1}{2} x e^{-x} \sin x.
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$$
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原方程的通解为
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$$
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y = y_h + y^* = e^{-x} (C_1 \cos x + C_2 \sin x) + \frac{1}{2} x e^{-x} \sin x.
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$$
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