Independent Variable Absent
In the case of the independent variable \( x \) missing, the order of the ODE can be lowered by one as well if we let \( p = dy/dx \), and hence by Chain Rule and Product Rule:
\begin{align}
\frac{d^2y}{dx^2} &= \frac{dp}{dx} \\
&= \frac{dp}{dy} \frac{dy}{dx} = p\frac{dp}{dy} \tag{1} \\
\frac{d^3y}{dx^3} &= \frac{d}{dx} (\frac{d^2y}{dx^2}) \\
&= \frac{d}{dx}(p\frac{dp}{dy}) \\
&= \frac{dy}{dx}\frac{d}{dy}(p\frac{dp}{dy}) \\
&= p (\frac{dp}{dy}\frac{dp}{dy} + p \frac{d^2p}{dy^2}) \\
&= p (\frac{dp}{dy})^2 + p^2 \frac{d^2p}{dy^2} \tag{2}
\end{align}
where the independent variable now becomes \( y \) and the rest goes on. (If instead the dependent variable \( y \) is absent, just again substitute \( p = dy/dx \), which will be much more direct.)
Example
Solve the ODE
\begin{align}
(\frac{dy}{dx})^2 -2y\frac{d^2y}{dx^2} &= k \tag{3}
\end{align}
where \( k \) is any constant.
Substitute \( p = dy/dx \) and (1) into (3) yields
\begin{align}
p^2 -2y(p\frac{dp}{dy}) &= k \\
p -2y\frac{dp}{dy} &= \frac{k}{p} \\
\frac{dp}{dy} -\frac{p}{2y} &= -\frac{k}{2py} \tag{4}
\end{align}
This is a Bernoulli ODE where \( n = -1 \) and so we let \( v = p^{1-(-1)} = p^2 \), then \( dp/dy = (dp/dv) (dv/dy) = (1/(2p)) (dv/dy)\), so (4) becomes
\begin{align}
\frac{1}{2p} \frac{dv}{dy} -\frac{p}{2y} &= -\frac{k}{2py} \\
\frac{dv}{dy} -\frac{p^2}{y} &= -\frac{k}{y} \\
\frac{dv}{dy} -\frac{v}{y} &= -\frac{k}{y} \\
(\frac{1}{y}\frac{dv}{dy} -\frac{v}{y^2}) &= -\frac{k}{y^2} \\
\frac{d}{dy}(\frac{v}{y}) &= -\frac{k}{y^2} \\
\frac{v}{y} &= -\int \frac{k}{y^2} dy \\
\frac{v}{y} &= \frac{k}{y} + c_1 \\
v &= k + c_1 y \Rightarrow p = \pm \sqrt{k + c_1 y} \tag{5}
\end{align}
So
\begin{align}
\frac{dy}{dx} &= \pm \sqrt{k + c_1 y} \\
\pm \int \frac{1}{\sqrt{k + c_1 y}} dy &= \int dx \\
\pm\frac{2}{c_1} \sqrt{k+c_1 y} &= x + c_2 \\
\frac{4}{c_1^2} (k+c_1 y) &= (x + c_2)^2 \\
k + c_1 y &= \frac{c_1^2}{4} (x + c_2)^2 \\
y &= \frac{c_1}{4} (x + c_2)^2 -\frac{k}{c_1} \tag{6}
\end{align}
Exercise
Find the general solution of
\begin{align}
\frac{dy}{dx} -y\frac{d^2y}{dx^2} = 1 \tag{7}
\end{align}
Answer
Plugging \( p = dy/dx \) and (1) into (7):
\begin{align}
p -yp\frac{dp}{dy} &= 1 \\
y\frac{dp}{dy} &= \frac{p-1}{p} \\
\int \frac{p}{p-1}dp = \int (1 + \frac{1}{p-1})dp &= \int \frac{1}{y} dy \\
p + \ln |p-1| &= \ln |y| + c_1 \\
(p -1)e^p &= Ay
\end{align}
According to this tutorial, we can differentiate both sides with respect to \( x \) to get
\begin{align}
Ay’ &= \frac{\partial}{\partial y’}((y’ -1)e^{y’})\frac{dy’}{dx} \\
Ay’ &= y’e^{y’} \frac{dy’}{dx} \\
A &= e^{y’} \frac{dy’}{dx} \\
\int e^{y’} dy’ &= \int A dx \\
e^{y’} &= Ax + B \\
y’ &= \ln (Ax + B)
\end{align}
Substitute this back into the equation relating \( p = y’ \) and \( y \) at the start then produces
\begin{align}
(\ln (Ax + B) -1)e^{\ln (Ax + B)} &= Ay \\
y &= (\ln (Ax + B) -1)(x + B/A)
\end{align}
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