一阶线性微分方程：积分因子法

First-Order Linear ODE

专题: General / 综合
难度: L4
来源: QuantQuestion

个人笔记

题目详情

解微分方程（ $x>0$ ）：

\frac{dy}{dx}+\frac{y}{x}=\frac{1}{x^2},\quad y(1)=1.

英文原题

A standard first-order linear ODE:
$> \frac{dy}{dx} \;+\; P(x)\,y \;=\; Q(x). >$
Use an integrating factor
$> I(x) \;=\; \exp\!\Bigl(\,\int P(x)\,dx\Bigr). >$
Then multiply both sides by $I(x)$ :
$> I(x)\,\frac{dy}{dx} + I(x)\,P(x)\,y > \;=\; > \bigl(I(x)\,y\bigr)' > \;=\; > I(x)\,Q(x). >$
Integrate and solve for $y$ :
$> I(x)\,y > \;=\; > \int I(x)\,Q(x)\,dx, > \quad > y > \;=\; > e^{-\int P(x)\,dx} > \Bigl[ > \int > e^{\,\int P(x)\,dx}\,Q(x)\,dx > + C > \Bigr]. >$

Bernoulli equation:
$\frac{dy}{dx} \;+\; P(x)\,y \;=\; Q(x)\,y^n$
can be turned into a linear ODE by the transformation $u = y^{1-n},$ etc.

Question: Solve
$\frac{dy}{dx} \;+\; \frac{y}{x} \;=\; \frac{1}{x^2},\quad y(1)=1,\;x>0.$

解析

积分因子 $I(x)=e^{\int (1/x)dx}=x$ 。

乘上 $x$ 得 $(xy)'=1/x$ ，积分：

xy=\ln x+C\Rightarrow y=\frac{\ln x+C}{x}.

用 $y(1)=1$ 得 $C=1$ ，所以

y(x)=\frac{\ln x+1}{x}.

英文解析

We identify $P(x)=\frac{1}{x},\; Q(x)=\frac{1}{x^2}.$
The integrating factor is
$I(x) = \exp\!\Bigl(\int \tfrac{1}{x}\,dx\Bigr) = \exp(\ln x) = x.$
Then
$x\,\frac{dy}{dx} + y \;=\; \frac{1}{x}. \quad\Longrightarrow\quad \bigl(x\,y\bigr)' \;=\; 1.$
Integrate:
$x\,y \;=\; x + C \;\Longrightarrow\; y \;=\; 1 + \frac{C}{x}.$

However, from the direct formula:
$y = \frac{1}{I(x)} \int I(x)\,Q(x)\,dx = \frac{1}{x} \int x \,\frac{1}{x^2}\,dx = \frac{1}{x} \int \frac{1}{x}\,dx = \frac{1}{x} \Bigl(\ln x + C\Bigr).$
Using $y(1)=1$ implies $1 = 1\cdot(\ln 1 + C)=C.$ So
$y = \frac{\ln x + 1}{x}.$

Hence the solution:
$y(x) = \frac{\ln x + 1}{x}.$