Homogeneous Differential Equations of the Form y' = P/Q
A specific class of first-order differential equations takes the form $$ y' = \frac{P(x,y)}{Q(x,y)} $$ where \( P \) and \( Q \) are homogeneous polynomials of the same degree.
When this type of equation can't be solved using separation of variables or the method of integrating factors, an effective approach is to introduce an auxiliary variable.
$$ t = \frac{y}{x} $$
We then express the unknown function \( y \) and its derivative \( y' \) in terms of \( t \) and \( x \):
$$ y = t \cdot x $$
$$ y' = D_x[t \cdot x] = t'x + t $$
Substituting these expressions into the original equation yields:
$$ y' = \frac{P(x,y)}{Q(x,y)} $$
$$ t'x + t = \frac{P(x,tx)}{Q(x,tx)} $$
We now attempt to solve this transformed equation using the method of separation of variables in terms of \( t \) and \( x \).
Once we find the general solution in terms of \( t \), we revert back to the original variable \( y \).
A Worked Example
Consider the following differential equation, which is an example of a homogeneous equation:
$$ xyy' = x^2 + y^2 $$
We first isolate \( y' \) by dividing both sides by \( xy \):
$$ \frac{xyy'}{xy} = \frac{x^2 + y^2}{xy} $$
$$ y' = \frac{x^2}{xy} + \frac{y^2}{xy} $$
$$ y' = \frac{x}{y} + \frac{y}{x} $$
Here, \( y' \) is expressed as a function of a ratio between two homogeneous polynomials in \( x \) and \( y \).
To proceed, we introduce the auxiliary variable \( t = y/x \):
$$ y' = \frac{x}{y} + \frac{y}{x} $$
$$ y' = \frac{1}{t} + t $$
Now we express \( y \) in terms of \( t \) and \( x \):
$$ y = t \cdot x $$
and differentiate both sides with respect to \( x \):
$$ D_x[y] = D_x[t \cdot x] $$
$$ y' = t'x + t $$
Substituting \( y' = t'x + t \) into the equation gives:
$$ y' = \frac{1}{t} + t $$
$$ t'x + t = \frac{1}{t} + t $$
Subtracting \( t \) from both sides simplifies the expression:
$$ t'x = \frac{1}{t} $$
We rewrite \( t' \) using differential notation:
$$ \frac{dt}{dx} \cdot x = \frac{1}{t} $$
At this point, the equation can be tackled using separation of variables.
We rearrange the terms to isolate \( t \) and \( x \):
$$ t \cdot dt = \frac{1}{x} \cdot dx $$
Now we integrate both sides:
$$ \int t \cdot dt = \int \frac{1}{x} \cdot dx $$
The integral on the right yields the antiderivative \( \log(x) + c \):
$$ \int t \cdot dt = \log(x) + c $$
The integral on the left gives \( t^2 / 2 \):
$$ \frac{t^2}{2} = \log(x) + c $$
$$ t^2 = 2 \cdot \log(x) + c $$
We substitute back \( t = y/x \) to return to the original variable:
$$ \left(\frac{y}{x}\right)^2 = 2 \cdot \log(x) + c $$
$$ \frac{y^2}{x^2} = 2 \cdot \log(x) + c $$
$$ y^2 = x^2 \cdot [2 \log(x) + c] $$
$$ y^2 = 2x^2 \cdot \log(x) + c \cdot x^2 $$
Taking the square root of both sides gives:
$$ \sqrt{y^2} = \sqrt{2x^2 \cdot \log(x) + c \cdot x^2} $$
$$ y = \sqrt{2x^2 \cdot \log(x) + c \cdot x^2} $$
This expression represents the general solution to the original differential equation.
And so on.
