利用高阶全微分方程的三阶非线性ODE

青青子衿

\[ y'''=\frac{3\left(y''\right)^2+\left(y'\right)^4}{y'} \]
\[ y'''=\frac{3x^3\!\cdot\!\left(y''\right)^2+\left(y'\right)^4}{x^3y'} \]

青青子衿

回复 1# 青青子衿
\begin{align*}
y'''=\frac{3\left(y''\right)^2+\left(y'\right)^4}{y'}
\quad\Rightarrow\quad
y'y'''-3\left(y''\right)^2=\left(y'\right)^4
\end{align*}
\begin{align*}
\left[\frac{y''}{\left(y'\right)^3}\right]'&=\frac{\left(y'\right)^3y'''-3\left(y'\right)^2\left(y''\right)^2}{\left(y'\right)^6}
=\frac{\color{red}{y'y'''-3\left(y''\right)^2}\,}{\left(y'\right)^4}=1\\
\end{align*}
\begin{align*}
\frac{y''}{\left(y'\right)^3}&=x+C_{0}\\
-\frac{1}{2}\left[\frac{1}{\left(y'\right)^2}\right]'&=x+C_{0}\\
\left[\frac{1}{\left(y'\right)^2}\right]'&=-2x-2C_{0}\\
\frac{1}{\left(y'\right)^2}&=-x^2+C_{1}x+C_{2}\\
\left(y'\right)^2&=\frac{1}{C_{2}+C_{1}x-x^2}\\
y'&=\pm\frac{1}{\sqrt{C_{2}+C_{1}x-x^2}\,}\\
y&=\pm\int\frac{{\rm\,d}x}{\sqrt{C_{2}+C_{1}x-x^2}\,}+C_3\\
\end{align*}