General Definition
The concept of homogeneous/isobaric remains the same when it comes to higher-order ODEs just like the first-order ones, as described in this tutorial. Again, we assign a unit of \( \alpha \) to \( x \) and \( dx \), \( \alpha^m \) to \(y \) and \( dy\), and see if every term in the concerned ODE has the same dimensionality. Subsequently, we can apply the previous substitution method \( z = y/x^m \) and then further make a change of the independent variable \( x = e^t \). This new independent variable \( t \) will then not appear explicitly in the equation.
Example
Solve the ODE
\begin{align}
x^3\frac{d^2y}{dx^2} + (2x^2+xy)\frac{dy}{dx} -(2xy+y^2) = 0 \tag{1}
\end{align}
The form of the bracketed terms suggests that this equation may be isobaric with \( m = 1 \), which is indeed true because each additive term has dimensionality \( m+1 \), \( m+1 \), \( 2m \), \( m+1 \), \( 2m \) and are equal to \( 2 \) in this case. Now we let \( y = zx \) and hence \( dy/dx = z + x dz/dx \), \( d^2y/dx^2 = 2 dz/dx + x d^2z/dx^2 \). Plugging these into (1) gives
\begin{align}
x^3(2 \frac{dz}{dx} + x \frac{d^2z}{dx^2}) + (2x^2+zx^2)(z + x\frac{dz}{dx}) -(2x^2z+x^2z^2) &= 0 \\
x^4 \frac{d^2z}{dx^2} + (4x^3 + zx^3)\frac{dz}{dx} &= 0 \\
x\frac{d^2z}{dx^2} + (4+ z)\frac{dz}{dx} &= 0 \tag{2}
\end{align}
Now further let \( x = e^t \), and
\begin{align}
\frac{dz}{dx} &= \frac{dz}{dt} \frac{dt}{dx} \\
&= \frac{dz}{dt} (\frac{1}{dx/dt}) = e^{-t}\frac{dz}{dt} \tag{3} \\
\frac{d^2z}{dx^2} &= \frac{d}{dx} (\frac{dz}{dx}) \\
&= \frac{dt}{dx}\frac{d}{dt}(e^{-t}\frac{dz}{dt}) \\
&= e^{-t} (e^{-t} \frac{d^2z}{dt^2} -e^{-t}\frac{dz}{dt}) \tag{4}
\end{align}
Substitution into (2) produces
\begin{align}
e^t (e^{-t} (e^{-t} \frac{d^2z}{dt^2} -e^{-t}\frac{dz}{dt})) + (4+ z)(e^{-t}\frac{dz}{dt}) &= 0 \\
\frac{d^2z}{dt^2} + (3+z)\frac{dz}{dt} &= 0 \\
\frac{d}{dt}(\frac{dz}{dt}) + \frac{d}{dt}(\frac{1}{2}z^2) + 3\frac{dz}{dt} &= 0 \tag{5}
\end{align}
by rewriting as differentials. Then
\begin{align}
\frac{dz}{dt} + \frac{1}{2}z^2 + 3z &= c_1 \\
\int \frac{dz}{c_1 -3z -\frac{1}{2}z^2} &= \int dt \\
\int \frac{dz}{(z+3)^2 -2c_1 -9} &= -\int \frac{1}{2} dt \\
-\frac{1}{\sqrt{2c_1+9}} \tanh^{-1} (\frac{z+3}{\sqrt{2c_1+9}}) &= -\frac{1}{2}t + c_2 \tag{6}
\end{align}
A reverse transformation then finally leads to
\begin{align}
\frac{z+3}{\sqrt{2c_1+9}} &= \tanh (\frac{1}{2}\sqrt{2c_1+9}t -c_2\sqrt{2c_1+9}) \\
z &= \sqrt{2c_1+9}\tanh(\frac{1}{2}\sqrt{2c_1+9}t -c_2\sqrt{2c_1+9}) -3 \\
y &= d_1x\tanh(\frac{1}{2}d_1 \ln x -c_2d_1) -3x \tag{7}
\end{align}
where we have replaced \( \sqrt{2c_1 + 9} \) by \( d_1 \).
Exercise
Derive the general solution to
\begin{align}
xy\frac{d^2y}{dx^2} -x(\frac{dy}{dx})^2 + (y-x) \frac{dy}{dx} + y = 0 \tag{8}
\end{align}
Answer
It is not hard to see that the ODE is isobaric with \( m = 1 \). Again, we can let \( y = zx \) and \( dy/dx = z + x dz/dx \), \( d^2y/dx^2 = 2 dz/dx + x d^2z/dx^2 \) to transform it into
\begin{align}
x^2z(2 \frac{dz}{dx} + x \frac{d^2z}{dx^2}) -x(z + x\frac{dz}{dx})^2 + (xz-x) (z + x\frac{dz}{dx}) + xz &= 0 \\
xz\frac{d^2z}{dx^2} + (z-1)\frac{dz}{dx} -x(\frac{dz}{dx})^2 &= 0
\end{align}
Substituting \( x = e^t \) and (3), (4) leads to
\begin{align}
e^tz(e^{-t} (e^{-t} \frac{d^2z}{dt^2} -e^{-t}\frac{dz}{dt})) + (z-1)(e^{-t}\frac{dz}{dt}) -e^t(e^{-t}\frac{dz}{dt})^2 &= 0 \\
z\frac{d^2z}{dt^2} -\frac{dz}{dt} -(\frac{dz}{dt})^2 &= 0
\end{align}
By this tutorial, we can further make a change of variable of \( p = dz/dt \), and we have
\begin{align}
zp\frac{dp}{dz} -p -p^2 &= 0 \\
z\frac{dp}{dz} &= 1+p \\
\int \frac{dp}{1+p} &= \int \frac{1}{z}dz \\
\ln |1+p| &= \ln |z| + c_1 \\
1+p &= \pm e^{c_1} z = Az
\end{align}
where we set \( A = \pm e^{c_1} \). Then
\begin{align}
1 + \frac{dz}{dt} &= Az \\
\frac{dz}{dt} -Az &= -1 \\
e^{-At}\frac{dz}{dt} -Ae^{-At}z &= -e^{-At} \\
\frac{d}{dt}(e^{-At}z) &= -e^{-At} \\
e^{-At}z &= \int -e^{-At} dt = \frac{1}{A} e^{-At} + B \\
z &= \frac{1}{A} + Be^{At}
\end{align}
and hence \( y = x(1/A + Bx^A) \).
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