利用反函数导数公式的三阶非线性ODE

青青子衿

\[ y'''=\frac{3\left(y''\right)^2+x\!\cdot\!\left(y'\right)^5}{y'} \]
\begin{align*}
\frac{{\rm\,d}x}{{\rm\,d}y}&=\frac{1}{y'}\\
\frac{{\rm\,d}^2x}{{\rm\,d}y^2}&=\frac{{\rm\,d}}{{\rm\,d}y}\left(\frac{{\rm\,d}x}{{\rm\,d}y}\right)
=\frac{{\rm\,d}}{{\rm\,d}y}\left(\frac{1}{y'}\right)
=\frac{{\rm\,d}}{{\rm\,d}x}\left(\frac{1}{y'}\right)\frac{{\rm\,d}x}{{\rm\,d}y}\\
&=-\frac{y''}{\left(y'\right)^2}\frac{{\rm\,d}x}{{\rm\,d}y}=-\frac{y''}{\left(y'\right)^3}\\
\frac{{\rm\,d}^3x}{{\rm\,d}y^3}&=\frac{{\rm\,d}}{{\rm\,d}y}\left(\frac{{\rm\,d}^2x}{{\rm\,d}y^2}\right)
=\frac{{\rm\,d}}{{\rm\,d}y}\left(-\frac{y''}{\left(y'\right)^3}\right)
=\frac{{\rm\,d}}{{\rm\,d}x}\left(-\frac{y''}{\left(y'\right)^3}\right)\frac{{\rm\,d}x}{{\rm\,d}y}\\
&=-\frac{y'''\!\cdot\!\left(y'\right)^3-3\left(y'\right)^2y''\!\cdot\!y''}{\left(y'\right)^6}\frac{{\rm\,d}x}{{\rm\,d}y}=-\frac{y'''\!\cdot\!\left(y'\right)^3-3\left(y'\right)^2\left(y''\right)^2}{\left(y'\right)^7}\\
&=\frac{3\left(y'\right)^2\left(y''\right)^2-y'''\!\cdot\!\left(y'\right)^3}{\left(y'\right)^7}=\frac{3\left(y''\right)^2-y'y'''}{\left(y'\right)^5}
\end{align*}

青青子衿

回复 1# 青青子衿
\begin{align*}
&&\frac{{\rm\,d}^3x}{{\rm\,d}y^3}&=\frac{\color{red}{3\left(y''\right)^2-y'y'''}}{\left(y'\right)^5}=\frac{\color{red}{-x\!\cdot\!\left(y'\right)^5}}{\left(y'\right)^5}=-x\\
&\Rightarrow&\frac{{\rm\,d}^3x}{{\rm\,d}y^3}+x&=0\\
&\Rightarrow&x'''+x&=0\\
\end{align*}
\begin{align*}
x'''+x&=0\\
x&=C_1e^{-y}+C_2e^{\frac{y}{2}}\cos\left(\frac{\sqrt{3}}{2}y\right)+C_3e^{\frac{y}{2}}\sin\left(\frac{\sqrt{3}}{2}y\right)
\end{align*}

\begin{align*}
&&\frac{{\rm\,d}^3x}{{\rm\,d}y^3}&=\frac{\color{red}{3\left(y''\right)^2-y'y'''}}{\left(y'\right)^5}=\frac{\color{red}{-x\!\cdot\!\left(y'\right)^{\color{blue}{4}\,}}}{\left(y'\right)^5}=-\frac{x}{y'}\\
&\Rightarrow&\frac{{\rm\,d}^3x}{{\rm\,d}y^3}+\frac{x}{y'}&=0\\
&\Rightarrow&\frac{{\rm\,d}^3x}{{\rm\,d}y^3}+x\frac{{\rm\,d}x}{{\rm\,d}y}&=0\\
&\Rightarrow&x'''+x\cdot x'&=0\\
\end{align*}
\begin{align*}
x'''+x\cdot x'&=0\\
x&=\,\,???
\end{align*}

青青子衿

回复 1# 青青子衿
更一般：
\[ y'y'''-3\left(y''\right)^2+a\left(y'\right)^2y''-b\left(y'\right)^4-cx\!\cdot\!\left(y'\right)^5=0 \]
\[ \xrightarrow[\frac{{\rm\,d}^3x}{{\rm\,d}y^3}=\frac{3\left(y''\right)^2-y'y'''}{\left(y'\right)^5}]
{\frac{{\rm\,d}x}{{\rm\,d}y}=\frac{1}{y'}\quad\frac{{\rm\,d}^2x}{{\rm\,d}y^2}=-\frac{y''}{\left(y'\right)^3}} \]
\[ \frac{{\rm\,d}^3x}{{\rm\,d}y^3}+a\frac{{\rm\,d}^2x}{{\rm\,d}y^2}+b\frac{{\rm\,d}x}{{\rm\,d}y}+cx=0 \]

\[ f'(x)y'y'''-3f'(x)\left(y''\right)^2+3f''(x)y'y''+af'(x)\left(y'\right)^2y''=af''(x)\left(y'\right)^3+bf'(x)\left(y'\right)^4+f'''(x)\left(y'\right)^2+cf(x)\left(y'\right)^5 \]
\[\xrightarrow[]{u(y)=f(x)}\]
\[\frac{{\rm\,d}^3u(y)}{{\rm\,d}y^3}+a\frac{{\rm\,d}^2u(y)}{{\rm\,d}y^2}+b\frac{{\rm\,d}u(y)}{{\rm\,d}y}+cu(y)=0\]

hbghlyj

青青子衿 发表于 2018-11-25 03:38
\begin{align*}
x'''+x\cdot x'&=0\\
x&=\,\,???
\end{align*}

DSolve[x'''[y] + x[y] x'[y] == 0, x[y], y]
\[x(y)=-(-2)^{2/3} \sqrt[3]{3} \wp \left(\frac{\sqrt[3]{-\frac{1}{3}} (y+c_1)}{2^{2/3}};2 (-2)^{2/3} \sqrt[3]{3} c_1,c_2\right)\]

Czhang271828

hbghlyj 发表于 2023-1-10 21:49
$x'''+xx'=0$.

剩下的交给椭圆函数

\begin{align*}
&x'''+xx'=0\\
\implies &2x''+x^2=C_0\\
\implies &6x'x''+3x'x^2=3C_0x'\\
\implies &3x'^2+x^3=3C_0x+C_1\\
\implies &\dfrac{\mathrm dx}{\mathrm dy}=\pm \sqrt{C_0x+\frac{x^3-C_1}{3}}.
\end{align*}