Definition of Degree-k Homogeneous
A function \( f(x,y) \) is known as being homogeneous of degree \(k\) if
\begin{equation}
f(tx, ty) = t^k f(x,y) \tag{1}
\end{equation}
for all \( t \). This means that whenever \( x \) and \( y \) are replaced by \( tx \) and \( ty \), the resulting function becomes \( t^k\) times the original function. Here we show some examples:
- \( x/y, \sin(2x/y), \ln(y/x), \sqrt{x^2+y^2}/x \), or functions generally in the form of \( g(x/y) \), are homogeneous of degree \( 0 \);
- \( x, y, 2x+3y, \sqrt{x^2 + 5y^2}, x^2/\pi y \) are homogeneous of degree \( 1 \);
- \( x^2 + 3xy, \sqrt{x^3y}, (4x+y)^2 \) are homogeneous of degree \( 2 \).
Basically, we can check if a function \( f(x,y) \) is homogeneous of degree \(k\) by thinking of \(x, y\) as having the same unit \( \alpha \) and see whether this implies \( f(x,y) \) has the consistent unit of \( \alpha^k \).
A first-order ODE in the form of
\begin{equation}
\frac{dy}{dx} = f(x,y) \tag{2}
\end{equation}
is then called homogeneous (notice this is different from the type of homogeneous in this tutorial) if \( f(x,y) \) is homogeneous of degree \( 0 \). Or equivalently,
\begin{equation}
M(x,y) dx + N(x,y) dy = 0 \Rightarrow \frac{dy}{dx} = -\frac{M(x,y)}{N(x,y)} \tag{3}
\end{equation}
is homogeneous if both \( M(x,y) \) and \( N(x,y) \) are homogeneous functions of the same degree.
Substitution Method
The strategy to solve (2) is to apply the substitution \( z = y/x \), transforming it into a separable ODE. By the definition of (1), we have
\begin{equation}
f(tx, ty) = t^0 f(x,y) = f(x,y) \tag{4}
\end{equation}
By letting \( t = 1/x \), we have
\begin{equation}
f(x,y) = f((1/x)x, (1/x)y) = f(1, y/x) = f(1, z) \equiv h(z) \tag{5}
\end{equation}
where we are removing the redundant dependence on the constant \( 1 \) from the function \( f(1,z) \) to write it as \( h(z) \). Moreover, due to the substitution \( y = zx \), by Product Rule from elementary Calculus:
\begin{equation}
\frac{dy}{dx} = z + x\frac{dz}{dx} \tag{6}
\end{equation}
Plugging these into (2) gives
\begin{align}
z + x\frac{dz}{dx} &= h(z) \\
\frac{dz}{dx} &= \frac{h(z) -z}{x} \tag{7}
\end{align}
which is now a separable equation in \(x\) and \(z\). It can be solved by a simple integration (see this tutorial):
\begin{align}
\int\frac{dz}{h(z) -z} &= \int\frac{dx}{x} \tag{8}
\end{align}
and the final answer can be retrieved by putting back \( z = y/x \).
Example
Our example is to solve
\begin{equation}
x\frac{dy}{dx} = 3x + 4y \tag{9}
\end{equation}
By moving the \( x \) factor from L.H.S. to R.H.S., it is not hard to see that (9) is a homogeneous ODE:
\begin{equation}
\frac{dy}{dx} = \frac{3x + 4y}{x} = 3 + 4\frac{y}{x} \tag{10}
\end{equation}
Using the substitution method \( z = y/x \) and with (6), we have
\begin{align}
z + x\frac{dz}{dx} &= 3 + 4z \\
x\frac{dz}{dx} &= 3+3z = 3(1+z) \tag{11}
\end{align}
Now, this is a separable ODE, easily solved as follows:
\begin{align}
\int \frac{dz}{1+z} &= \int \frac{3}{x}dx \\
\ln |1+z| &= 3 \ln |x| + C = \ln |x|^3 + C\\
|1 + z| &= e^{(\ln |x|^3 + C)} = e^C|x|^3 \\
1 + z &= Ax^3 \tag{12}
\end{align}
where we have assigned \( A = \pm e^C \). Putting back \( z = y/x \) leads to
\begin{align}
1 + \frac{y}{x} &= Ax^3 \\
y &= Ax^4 -x \tag{13}
\end{align}
Exercise
Argue that an ODE in the general form of
\begin{equation}
\frac{dy}{dx} = f\left(\frac{ax + by + p}{cx + dy + q}\right) \tag{14}
\end{equation}
can be transformed into a homogeneous ODE via the substitutions
\begin{align}
\left\{\begin{aligned}
\tilde{x} &= x -h \\
\tilde{y} &= y -k
\end{aligned}\right. \tag{15}
\end{align}
by choosing the values for the constants \(h, k\) appropriately, if \( ad \neq bc \). Hence solve
\begin{equation}
\frac{dy}{dx} = \frac{x + 2y -1}{2x -y + 3} \tag{16}
\end{equation}
Answers
This amounts to enforcing, for
\begin{align}
\frac{ax + by + p}{cx + dy + q} &= \frac{a(\tilde{x} + h) + b(\tilde{y} + k) + p}{c(\tilde{x} + h) + d(\tilde{y} + k) + q} \\
&= \frac{a\tilde{x}+ b\tilde{y} + (ah + bk + p)}{c\tilde{x} + d\tilde{y} + (ch + dk + q)}
\end{align}
we have
\begin{align}
\left\{\begin{aligned}
ah + bk + p &= 0 \\
ch + dk + q &= 0
\end{aligned}\right.
\end{align}
so that the argument becomes homogeneous in \( \tilde{x}, \tilde{y} \):
\begin{align}
f\left(\frac{ax + by + p}{cx + dy + q}\right) &= f\left(\frac{a\tilde{x}+ b\tilde{y} + (ah + bk + p)}{c\tilde{x} + d\tilde{y} + (ch + dk + q)}\right) \\
&= f\left(\frac{a\tilde{x}+ b\tilde{y}}{c\tilde{x} + d\tilde{y}}\right) = f\left(\frac{a + b\tilde{y}/\tilde{x}}{c + d\tilde{y}/\tilde{x}}\right)
\end{align}
For (16), we need to satisfy
\begin{align}
\left\{\begin{aligned}
h + 2k -1 &= 0 \\
2h -k + 3 &= 0
\end{aligned}\right.
\end{align}
whose solution can be seen to be \( h=-1, k=1 \). With this substitution, (16) becomes
\begin{equation}
\frac{d\tilde{y}}{d\tilde{x}} = \frac{\tilde{x} + 2\tilde{y}}{2\tilde{x} -\tilde{y}} = \frac{1 + 2\tilde{y}/\tilde{x}}{2 -\tilde{y}/\tilde{x}}
\end{equation}
By a further substitution \( \tilde{z} = \tilde{y}/\tilde{x} \) and according to (6), it is changed into and solved as
\begin{align}
\tilde{z} + \tilde{x}\frac{d\tilde{z}}{d\tilde{x}} &= \frac{1 + 2\tilde{z}}{2 -\tilde{z}} \\
\tilde{x}\frac{d\tilde{z}}{d\tilde{x}} &= \frac{1 + 2\tilde{z} -\tilde{z}(2 -\tilde{z})}{2 -\tilde{z}} = \frac{1 + \tilde{z}^2}{2 -\tilde{z}} \\
\int\frac{2-\tilde{z}}{1 + \tilde{z}^2}d\tilde{z} &= \int \frac{d\tilde{x}}{\tilde{x}} \\
\int\frac{2}{1 + \tilde{z}^2}d\tilde{z} -\int\frac{\tilde{z}}{1 + \tilde{z}^2}d\tilde{z} &= \ln |\tilde{x}| + C \\
2 \tan^{-1}(\tilde{z}) -\frac{1}{2}\int\frac{d(1 + \tilde{z}^2)}{1 + \tilde{z}^2} &= \ln |\tilde{x}| + C \\
2 \tan^{-1}(\tilde{z}) -\frac{1}{2}\ln|1 + \tilde{z}^2| &= \ln |\tilde{x}| + C \\
\end{align}
\begin{align}
2 \tan^{-1}(\tilde{y}/\tilde{x}) -\frac{1}{2}\ln(1 + (\tilde{y}/\tilde{x})^2) -\ln |\tilde{x}| &= C \\
2 \tan^{-1}(\frac{y-1}{x+1}) -\frac{1}{2}\ln(1 + (\frac{y-1}{x+1})^2) -\ln |x+1| &= C
\end{align}
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