Site Overlay

微分方程 dy/dx=(x+y)/(x-y) 和 y^{‘}\cos (x)+y\sin (x)=1

微分方程 dy/dx=(x+y)/(x-y)

第一种方法:

$y=2x$

$$
\begin{align}
&Reslove:\frac{dy}{dx} = \frac{x+y}{x-y}\\
y' &=\dfrac{1+\dfrac{y}{x}}{1-\dfrac{y}{x}}\\
\text{Let }u(x)=\frac{y(x)}{x}\\
&=\frac{1+u}{1-u}\\
u(x)=\frac{y(x)}{x}\Rightarrow y=ux
\Rightarrow y' = u'x + u\\
u'x+u&=\frac{1+u}{1-u}\\
u'x&=\frac{1+u}{1-u}-u\\
&=\frac{1+u}{1-u}-\frac{u-u^2}{1-u}\\
&=\frac{1+u^2}{1-u}\\
\text{Seperate}\\
\frac{du}{dx}x&=\frac{1+u^2}{1-u}\\
\frac{dx}{xdu}&=\frac{1-u}{1+u^2}\\
\Rightarrow \frac{dx}{x}&=\frac{1-u}{1+u^2}du\\
\int \frac{dx}{x}&=\int \frac{1-u}{1+u^2}du\\
\ln x &=\int \frac{1}{1+u^2}du-\int \frac{u}{1+u^2}du\\
\text{As for:}\int \frac{u}{1+u^2}du\\
\text{let }t=u^2+1\\
\Rightarrow dt = 2udu\\
\int \frac{u}{1+u^2}du=\int\frac{dt}{2t}=\frac{1}{2}ln|x^2+1|+c\\
&=\arctan u-\frac{1}{2}ln|x^2+1|+c\\
\text{result: }\\
ln|x| &= \arctan \frac{y}{x}-\frac{1}{2}ln|x^2+1|+c\\
\end{align}
$$

ODE 解方程 y^{'}\cos (x)+y\sin (x)=1

$$
\begin{aligned}
&\quad y^{\prime} \cos x+y \sin x=1 \\
&\Rightarrow y^{\prime}+y \frac{\sin x}{\cos x}=\frac{1}{\cos x} \\
&\Rightarrow y^{\prime}+y \tan x=\sec x \\
\end{aligned}
$$
$$
\begin{aligned}
&y=\frac{\int e^{\int \tan x d x} \cdot \sec x d x+c}{e^{\int \tan x d x}} \\
e^{\int \tan x d x}=e^{-\ln |\cos x|} =e^{\ln \left|\frac{1}{\cos x}\right|}=\left|\frac{1}{\cos x}\right|=|\sec x| \\
&=\frac{\int \sec ^{2} x d x+c}{\sec x} \\
&=\frac{\tan x d x+c}{\sec x} \\
\end{aligned}
$$

发表评论

电子邮件地址不会被公开。 必填项已用*标注