Definition of Isobaric ODEs
Isobaric ODEs are the generalization of homogeneous ODEs discussed in the last tutorial. Given a function \( f(x,y) \), it is isobaric if
\begin{equation}
f(sx, s^m y) = s^{m-1}f(x,y) \tag{1}
\end{equation}
We can check the isobaric property by assigning a unit of \( \alpha \) to \( x \) and \( dx \), but \( \alpha^m \) to \( y \) and \( dy \), then see whether the first-order ODE
\begin{equation}
\frac{dy}{dx} = f(x,y) \tag{2}
\end{equation}
can have a consistent unit of \( \alpha^{m-1} \) on both sides. Note that a homogeneous ODE is then the special case when \(m=1\).
For example,
\begin{equation}
\frac{dy}{dx} = \frac{y}{x} -\frac{y^2}{x^3} \tag{3}
\end{equation}
can be identified as an isobaric ODE: the \( y/x \) term has a unit of \( \alpha^{m-1} \), and the \( y^2/x^3 \) term has a unit of \( \alpha^{2m-3} \), so the only possibility is \( m = 2 \) if they have to be consistent. Indeed, when \( m = 2 \), both sides have a unit of \( \alpha^1 \), and this confirms our suspicion.
Substitution Method
The solution method is very similar to the last one, using a slightly modified substitution of \( z = y/x^m \) (i.e. \( y = x^mz \)). Then as in the derivation from the last tutorial:
\begin{align}
f(x,y) &= f(x, x^mz) \\
&= x^{m-1}f(1, z) & & \text{(By (1))} \\
& \equiv x^{m-1} h(z) \tag{4}
\end{align}
Moreover, similarly by the Product Rule in elementary Calculus:
\begin{align}
\frac{dy}{dx} = x^m\frac{dz}{dx} + mx^{m-1}z \tag{5}
\end{align}
Plugging in (4) and (5) into (2) then produces a separable ODE:
\begin{align}
x^m\frac{dz}{dx} + mx^{m-1}z &= x^{m-1} h(z) \\
x\frac{dz}{dx} + mz &= h(z) \\
\frac{dz}{dx} &= \frac{h(z)-mz}{x} \tag{6}
\end{align}
Finally, integrate this equation and do a backward substitution \( z = y/x^m \) to get the final answer just as before.
Example
We will demonstrate how to solve (3). The required substitution is \( z = y/x^2 \) and thus
\begin{align}
\frac{dy}{dx} = x^2\frac{dz}{dx} + 2xz \tag{7}
\end{align}
(3) then can be proceeded as
\begin{align}
x^2\frac{dz}{dx} + 2xz &= xz -xz^2 \\
x\frac{dz}{dx} &= -z -z^2 \\
-\frac{dz}{z + z^2} &= \frac{dx}{x} \tag{8} \\
\end{align}
By utilizing partial fractions:
\begin{align}
\int (\frac{1}{z+1} -\frac{1}{z}) dz &= \int \frac{dx}{x} \\
\ln |\frac{z+1}{z}| &= \ln |x| + C \\
\frac{z+1}{z} &= \pm e^C x \\
1 + \frac{1}{z} &= Ax \\
\frac{x^2}{y} &= Ax -1 \\
y &= \frac{x^2}{Ax -1} \tag{9}
\end{align}
where we set \( A = \pm e^C \) as usual.
Exercise
Solve
\begin{equation}
\frac{dy}{dx} = \frac{2}{y} -\frac{2x}{y^3} \tag{10}
\end{equation}
Answer
It can be checked that the ODE is isobaric with \( m = 1/2 \). Setting \( y = x^{1/2}z \), then (5) becomes
\begin{align}
\frac{dy}{dx} = x^\frac{1}{2}\frac{dz}{dx} + \frac{1}{2}x^{-\frac{1}{2}}z
\end{align}
and the ODE is transformed into
\begin{align}
x^\frac{1}{2}\frac{dz}{dx} + \frac{1}{2}x^{-\frac{1}{2}}z &= \frac{2}{zx^{\frac{1}{2}}} -\frac{2x}{(zx^{\frac{1}{2}})^3} \\
x^\frac{1}{2}\frac{dz}{dx} + \frac{1}{2}x^{-\frac{1}{2}}z &= 2z^{-1}x^{-\frac{1}{2}} – 2z^{-3}x^{-\frac{1}{2}} \\
x\frac{dz}{dx} + \frac{1}{2}z &= \frac{2}{z} -\frac{2}{z^3} \\
\frac{dz}{ -\frac{1}{2}z +\frac{2}{z} -\frac{2}{z^3}} &= \frac{dx}{x} \\
-\int \frac{2z^3 dz}{z^4 -4z^2 + 4} &= \int \frac{dx}{x} \\
-\int \frac{z^2 d(z^2-2)}{(z^2 -2)^2} &= \int \frac{dx}{x} \\
-\int \frac{d(z^2-2)}{(z^2 -2)} -\int \frac{2d(z^2-2)}{(z^2 -2)^2} &= \ln |x| + C \\
– \ln|z^2 -2| +\frac{2}{z^2 -2} &= \ln |x| + C
\end{align}
Putting back \( z = y/x^{1/2} \) yields
\begin{align}
– \ln|\frac{y^2}{x}-2| + \frac{2}{\frac{y^2}{x}-2} &= \ln |x| + C \\
\frac{2x}{y^2-2x} &= \ln |y^2 -2x| + C
\end{align}
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