Section03_可降阶的高阶微分方程（不要求）

$y^{(n)} = f(x)$

$f(x,y',y'') = 0$ 缺 $y$

解法

1. 令 $p = y'$
2. $f(x, p, \frac{dp}{dx}) = 0$
3. $p = \phi(x,C_{1})$
4. $y = \int \phi(x,C_{1})\cdot dx + C_{2}$

例题

1. $\displaystyle xy''+2y'=x^{2}$ $$\begin{array}{ll} & \text{令} p = y' \\ \therefore & x \frac{dp}{dx} + 2p = x^{2}\\ & \frac{dp}{dx} + \frac{2}{x}p = x \\ \therefore & p = [\int_{}^{}xe^{\int_{}^{}\frac{2}{x}\cdot dx}\cdot dx + C_{1}]e^{-\int_{}^{}\frac{2}{x}\cdot dx} \\ & p = [\frac{1}{4} x^{4} + C_{1}] x^{-2} = \frac{1}{4}x^{2} + C_{1} x^{-2} \\ \therefore & y = \int_{}^{} (\frac{1}{4}x^{2} + C_{1} x^{-2})\cdot dx + C_{2} \\ & y = \frac{1}{12}x^{3} - C_{1}x^{-1} + C_{2} \\ \end{array}$$

$f(y,y',y'')=0$ 缺$x$

解法

1. 令 $y'=p$
2. $y'' = \frac{dp}{dx} = \frac{dy}{dx}\frac{dp}{dy} = p\frac{dp}{dy}$
3. $f(y,p,p\frac{dp}{dy})$
4. $\frac{dy}{dx} = p = \phi(y,C_{1})$
5. $\displaystyle\int \frac{dy}{\phi(y,C_{1})} =\int dx + C_{2}$

例题

1. $\displaystyle yy''-(y')^{2} = 0, y(0)=y'(0)=1$ 求特解 $$\begin{array}{ll} & \text{令} p = y' \\ \therefore & py \frac{dp}{dy} - p^{2} = 0 \\ \because & y(0)=y'(0) = 1 = p \\ \therefore & \frac{dp}{dy} - \frac{p}{y} = 0 \\ \therefore & p = C_{1}e^{- \int_{}^{}- \frac{1}{y}\cdot dy} \\ & p = C_{1}e^{\ln y} = C_{1} y \\ \therefore & \frac{dy}{dx} = C_{1}y \\ & \frac{dy}{y} = C_{1}dx \\ \therefore & \int_{}^{}\frac{dy}{y} = \int_{}^{}C_{1}dx + C_{2}\\ \therefore & \ln y = C_{1}x + C_{2} \\ \therefore & y = C_{2}e^{C_{1}x} \\ \because & y(0) = y'(0)=1 \\ \therefore & C_{2} = 1; C_{1}C_{2} = 1 \\ \therefore & y = e^{x} \\ \end{array}$$