同济高数 第7章 第7-4-1题

[AI解答]

[AI解答]
以下为各小题的详细求解过程，均为一阶线性微分方程，使用通解公式
$$
y = e^{-\int P(x)\,dx} \left( \int Q(x) e^{\int P(x)\,dx} dx + C \right)
$$
或相应变量代换。

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(1)
方程：$\displaystyle \frac{dy}{dx} + y = e^{-x}$
这里 $P(x)=1,\ Q(x)=e^{-x}$。
积分因子：$\displaystyle \mu = e^{\int 1\,dx} = e^{x}$
则
$$
y = e^{-x} \left( \int e^{-x} \cdot e^{x} dx + C \right) = e^{-x} \left( \int 1\,dx + C \right) = e^{-x}(x + C)
$$
通解：
$$
\boxed{y = (x + C)e^{-x}}
$$

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(2)
方程：$\displaystyle x y' + y = x^{2} + 3x + 2$
化为标准形式：$\displaystyle y' + \frac{1}{x}y = x + 3 + \frac{2}{x}$
积分因子：$\displaystyle \mu = e^{\int \frac{1}{x}dx} = x$
则
$$
y = \frac{1}{x} \left( \int x\left(x + 3 + \frac{2}{x}\right) dx + C \right) = \frac{1}{x} \left( \int (x^{2} + 3x + 2) dx + C \right)
$$
$$
= \frac{1}{x} \left( \frac{x^{3}}{3} + \frac{3x^{2}}{2} + 2x + C \right)
$$
通解：
$$
\boxed{y = \frac{x^{2}}{3} + \frac{3x}{2} + 2 + \frac{C}{x}}
$$

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(3)
方程：$\displaystyle y' + y \cos x = e^{-\sin x}$
积分因子：$\displaystyle \mu = e^{\int \cos x\,dx} = e^{\sin x}$
则
$$
y = e^{-\sin x} \left( \int e^{-\sin x} \cdot e^{\sin x} dx + C \right) = e^{-\sin x} (x + C)
$$
通解：
$$
\boxed{y = (x + C)e^{-\sin x}}
$$

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(4)
方程：$\displaystyle y' + y \tan x = \sin 2x$
积分因子：$\displaystyle \mu = e^{\int \tan x\,dx} = e^{-\ln|\cos x|} = \sec x$
则
$$
y = \cos x \left( \int \sec x \cdot \sin 2x\,dx + C \right)
$$
注意 $\sin 2x = 2\sin x \cos x$，所以
$$
\sec x \cdot \sin 2x = \frac{1}{\cos x} \cdot 2\sin x \cos x = 2\sin x
$$
积分得 $\int 2\sin x\,dx = -2\cos x$，于是
$$
y = \cos x (-2\cos x + C) = -2\cos^{2}x + C\cos x
$$
通解：
$$
\boxed{y = C\cos x - 2\cos^{2}x}
$$

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(5)
方程：$\displaystyle (x^{2}-1)y' + 2xy - \cos x = 0$
化为标准形式：$\displaystyle y' + \frac{2x}{x^{2}-1}y = \frac{\cos x}{x^{2}-1}$
积分因子：$\displaystyle \mu = e^{\int \frac{2x}{x^{2}-1}dx} = e^{\ln|x^{2}-1|} = x^{2}-1$
则
$$
y = \frac{1}{x^{2}-1} \left( \int \cos x\,dx + C \right) = \frac{\sin x + C}{x^{2}-1}
$$
通解：
$$
\boxed{y = \frac{\sin x + C}{x^{2}-1}}
$$

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(6)
方程：$\displaystyle \frac{d\rho}{d\theta} + 3\rho = 2$
积分因子：$\displaystyle \mu = e^{\int 3\,d\theta} = e^{3\theta}$
则
$$
\rho = e^{-3\theta} \left( \int 2 e^{3\theta} d\theta + C \right) = e^{-3\theta} \left( \frac{2}{3}e^{3\theta} + C \right) = \frac{2}{3} + Ce^{-3\theta}
$$
通解：
$$
\boxed{\rho = \frac{2}{3} + Ce^{-3\theta}}
$$

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(7)
方程：$\displaystyle y' + 2xy = 4x$
积分因子：$\displaystyle \mu = e^{\int 2x\,dx} = e^{x^{2}}$
则
$$
y = e^{-x^{2}} \left( \int 4x e^{x^{2}} dx + C \right) = e^{-x^{2}} \left( 2e^{x^{2}} + C \right) = 2 + Ce^{-x^{2}}
$$
通解：
$$
\boxed{y = 2 + Ce^{-x^{2}}}
$$

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(8)
方程：$\displaystyle y\ln y\,dx + (x - \ln y) dy = 0$
改写为：$\displaystyle \frac{dx}{dy} + \frac{1}{y\ln y}x = \frac{1}{y}$
积分因子：$\displaystyle \mu = e^{\int \frac{1}{y\ln y}dy} = e^{\ln|\ln y|} = \ln y$
则
$$
x = \frac{1}{\ln y} \left( \int \ln y \cdot \frac{1}{y} dy + C \right) = \frac{1}{\ln y} \left( \int \frac{\ln y}{y} dy + C \right)
$$
令 $t = \ln y$，则 $dt = \frac{1}{y}dy$，积分 $\int t\,dt = \frac{t^{2}}{2} = \frac{(\ln y)^{2}}{2}$
所以
$$
x = \frac{1}{\ln y} \left( \frac{(\ln y)^{2}}{2} + C \right) = \frac{\ln y}{2} + \frac{C}{\ln y}
$$
通解：
$$
\boxed{x = \frac{\ln y}{2} + \frac{C}{\ln y}}
$$

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(9)
方程：$\displaystyle (x-2)\frac{dy}{dx} = y + 2(x-2)^{3}$
化为标准形式：$\displaystyle y' - \frac{1}{x-2}y = 2(x-2)^{2}$
积分因子：$\displaystyle \mu = e^{-\int \frac{1}{x-2}dx} = e^{-\ln|x-2|} = \frac{1}{x-2}$
则
$$
y = (x-2) \left( \int \frac{1}{x-2} \cdot 2(x-2)^{2} dx + C \right) = (x-2) \left( \int 2(x-2) dx + C \right)
$$
$$
= (x-2) \left( (x-2)^{2} + C \right) = (x-2)^{3} + C(x-2)
$$
通解：
$$
\boxed{y = (x-2)^{3} + C(x-2)}
$$

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(10)
方程：$\displaystyle (y^{2} - 6x)\frac{dy}{dx} + 2y = 0$
改写为：$\displaystyle \frac{dx}{dy} - \frac{3}{y}x = -\frac{y}{2}$
积分因子：$\displaystyle \mu = e^{-\int \frac{3}{y}dy} = e^{-3\ln|y|} = y^{-3}$
则
$$
x = y^{3} \left( \int y^{-3} \cdot \left(-\frac{y}{2}\right) dy + C \right) = y^{3} \left( -\frac{1}{2} \int y^{-2} dy + C \right)
$$
$$
= y^{3} \left( \frac{1}{2y} + C \right) = \frac{y^{2}}{2} + C y^{3}
$$
通解：
$$
\boxed{x = \frac{y^{2}}{