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利用首次积分法求解两道一阶非线性常微分方程组

本帖最后由 青青子衿 于 2018-11-30 13:14 编辑

利用首次积分法(First Integral)求解对称形式的常微分方程组:
\[ \frac{{\rm\,d}x}{-x+y+z}=\frac{{\rm\,d}y}{x-y+z}=\frac{{\rm\,d}z}{x+y-z}  \]
\[ \frac{{\rm\,d}x}{-x^2+y^2+z^2}=\frac{{\rm\,d}y}{x^2-y^2+z^2}=\frac{{\rm\,d}z}{x^2+y^2-z^2}  \]
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本帖最后由 青青子衿 于 2018-11-30 13:11 编辑

回复 1# 青青子衿
\begin{align*}  
&&\frac{{\rm\,d}x}{-x+y+z}&=\frac{{\rm\,d}y}{x-y+z}\\  
&\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{-2\left(x-y\right)}&=\frac{{\rm\,d}\left(x+y+z\right)}{x+y+z}\\
&\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{\left(x-y\right)}&=-\frac{2{\rm\,d}\left(x+y+z\right)}{\left(x+y+z\right)}\\   
&\Rightarrow&\ln\left|x-y\right|&=-2\ln\left|x+y+z\right|+\ln|C_1|\\  
&\Rightarrow&\left(x-y\right)\left(x+y+z\right)^2&=C_1\\  
\end{align*}
同理
In like manner
\begin{align*}
&&\frac{{\rm\,d}y}{x-y+z}&=\frac{{\rm\,d}z}{x+y-z}\\
&\Rightarrow&\frac{{\rm\,d}\left(y-z\right)}{-2\left(y-z\right)}&=\frac{{\rm\,d}\left(x+y+z\right)}{x+y+z}\\
&\Rightarrow&\frac{{\rm\,d}\left(y-z\right)}{\left(y-z\right)}&=-\frac{2{\rm\,d}\left(x+y+z\right)}{\left(x+y+z\right)}\\
&\Rightarrow&\ln\left|y-z\right|&=-2\ln\left|x+y+z\right|+\ln|C_2|\\
&\Rightarrow&\left(y-z\right)\left(x+y+z\right)^2&=C_2\\
\end{align*}

\begin{align*}
\left\{\begin{array}{r}  
\begin{split}  
\frac{{\rm\,d}x}{-x+y+z}&=\frac{{\rm\,d}y}{x-y+z}\\  
\frac{{\rm\,d}y}{x-y+z}&=\frac{{\rm\,d}z}{x+y-z}\\  
\end{split}  
\end{array}\right.
\Rightarrow
\left\{\begin{array}{r}  
\begin{split}  
\left(x-y\right)\left(x+y+z\right)^2&=C_1\\  
\left(y-z\right)\left(x+y+z\right)^2&=C_2\\  
\end{split}  
\end{array}\right.
\end{align*}

\begin{align*}
\color{red}{
\left\{\begin{array}{r}   
\begin{split}   
\frac{{\rm\,d}x}{-x+y+a}&=\frac{{\rm\,d}y}{x-y+a}\\   
\frac{{\rm\,d}y}{b-y+z}&=\frac{{\rm\,d}z}{b+y-z}\\   
\end{split}   
\end{array}\right.  
\Rightarrow  
\left\{\begin{array}{r}   
\begin{split}   
x-y&=C_1e^{-\frac{x+y}{z}}\\   
y-z&=C_2e^{-\frac{y+z}{x}}\\   
\end{split}   
\end{array}\right. }
\end{align*}

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本帖最后由 青青子衿 于 2018-11-30 22:32 编辑

回复 1# 青青子衿

\begin{align*}
\color{red}{
\left\{\begin{array}{r}  
\begin{split}  
\frac{{\rm\,d}x}{-x^2+y^2+a^2}&=\frac{{\rm\,d}y}{x^2-y^2+a^2}\\  
\frac{{\rm\,d}y}{b^2-y^2+z^2}&=\frac{{\rm\,d}z}{b^2+y^2-z^2}\\  
\end{split}  
\end{array}\right.
\Rightarrow
\left\{\begin{array}{r}  
\begin{split}  
x-y&=C_1e^{-\frac{\left(x+y\right)^2}{2a^2}}\\  
y-z&=C_2e^{-\frac{\left(y+z\right)^2}{2b^2}}\\  
\end{split}  
\end{array}\right. }
\end{align*}
另外:
\begin{align*}
&&\frac{{\rm\,d}x}{-x^3+y^3+a^3}&=\frac{{\rm\,d}y}{x^3-y^3+a^3}\\
&\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{-2\left(x^3-y^3\right)}&=\frac{{\rm\,d}\left(x+y\right)}{2a^3}\\
&\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{{\rm\,d}\left(x+y\right)}&=\frac{-2\left(x^3-y^3\right)}{2a^3}\\
&\Rightarrow&\frac{{\rm\,d}\left(x-y\right)}{{\rm\,d}\left(x+y\right)}&=\frac{-\left(x-y\right)\left(x^2+xy+y^2\right)}{a^2}\\
&\Rightarrow&\frac{\left(x-y\right){\rm\,d}\left(x-y\right)}{{\rm\,d}\left(x+y\right)}&=\frac{-\left(x-y\right)^2\left(3\left(x+y\right)^2+\left(x-y\right)^2\right)}{a^2}\\
&\Rightarrow&\frac{{\rm\,d}\left(\left(x-y\right)^2\right)}{{\rm\,d}\left(x+y\right)}&=-\frac{3\left(x+y\right)^2}{a^2}\left(x-y\right)^2-\frac{1}{a^2}\left(x-y\right)^4\\
&\Rightarrow&\frac{{\rm\,d}u}{{\rm\,d}v}&=-\frac{3v^2}{a^2}u-\frac{1}{a^2}u^2\\
&\Rightarrow&\\
&\Rightarrow&\\
\end{align*}

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