Is $\frac {dy} {dx} = \frac {2x} {3y}$ a homogeneous differential equation?

Question

I have a differential equation $\dfrac {dy} {dx} = \dfrac {2x} {3y}$ whose solutions are $y = \pm \sqrt {\dfrac 2 3}x $ which when I back-substitute I get $LHS=RHS$.

From the definition on wikipedia here

A linear differential equation is called homogeneous if the following condition is satisfied: If $\phi(x)$ is a solution, so is $c \phi(x)$, where c is an arbitrary (non-zero) constant

So taking the above definition $y = 5 \sqrt {\dfrac 2 3}x $ should work too. But if I backsubstitute this in original differential equation, I get $LHS = 5 \sqrt {\dfrac 2 3}$ and $RHS = \dfrac 1 5 \sqrt {\dfrac 2 3}$ which are not equal.

So is this still a homogeneous linear equation by definition? Am I making a mistake somewhere? Please help me understand why there are two types of homogeneity defined here. What does it mean to be homogeneous?

Answer 1

Note that the solution should come out as $$\frac {3y^2}2=x^2+C$$or $$y^2=\frac {2x^2}3+D$$

(where $D =\frac {2C}3$ is a constant) whence $$y=\pm\sqrt{\left(\frac {2x^2}3+D\right)}$$

The constant of integration "gets in the way" here - it is added under the square root, rather than being a factor. You have ignored the constant of integration in your answer, which has misled you as to the nature of the solution.

If you had $ \dfrac {dy} {dx} =\dfrac {2y} {3x} $ that would be homogeneous. You'd get $y^3=Dx^2$ where the constant of integration comes out as a factor - so an arbitrary multiple will automatically work. So you can tell from the solution that the original equation is homogeneous. In your case clearing fractions gives you a term in $ y\dfrac {dy} {dx} $ which cannot be part of the form that your linked article says a homogeneous equation must have.

Answer 2

This is a first order homogeneous ODE.

$$\Large \frac{dy}{dx}=\frac{2x}{3y}= \frac{2}{3\left(\frac{y}{x}\right)}.$$

Now let $$\Large v=\frac{y}{x} \iff \underbrace{y=vx\implies \frac{dy}{dx}=v+x\frac{dv}{dx}}_{\text{Product rule}}.$$

Then we have $$\Large v+x \frac{dv}{dx}=\frac{2}{3v}$$ which is separable.

---

Another way of describing a (first order) homogeneous ODE is that it is of the form $$\frac{dy}{dx}=f\left(\frac{y}{x}\right),$$ where $f$ is some suitable function.