Homogeneous Differential Equations

A homogeneous differential equation is a type of ordinary differential equation (ODE) that can be expressed in the form:

$$\frac{dy}{dx} = F\left(\frac{y}{x}\right)$$

where $F$ is a function of the ratio $\frac{y}{x}$. This means that the equation is homogeneous of degree zero, as it can be rewritten in terms of the variable $v = \frac{y}{x}$.

To solve a homogeneous differential equation, we can use the substitution $v = \frac{y}{x}$, which implies that $y = vx$. Differentiating both sides with respect to $x$, we get:

$$\frac{dy}{dx} = v + x\frac{dv}{dx}$$

Substituting this back into the original equation gives us:

$$v + x\frac{dv}{dx} = F(v)$$

Rearranging this, we can express it as:

$$x\frac{dv}{dx} = F(v) - v$$

This is a separable differential equation, which can be solved by separating the variables:

$$\frac{1}{F(v) - v}~dv = \frac{1}{x}~dx$$

Integrating both sides, we have:

$$\int \frac{1}{F(v) - v}~dv = \int \frac{1}{x}~dx$$

The integral on the right side is straightforward and yields:

$$\int \frac{1}{x}~dx = \ln|x| + C$$

where $C$ is the constant of integration. The integral on the left side will depend on the specific form of the function $F(v)$. Once we evaluate the integral on the left, we can express the solution in terms of $v$ and then substitute back to find $y$ in terms of $x$.

Example

Consider the homogeneous differential equation:

$$2xy \frac{dy}{dx}=y^2-x^2 \quad \quad x>0$$

Show a general solution, where $c$ is a positive constant, for this equation is:

$$x^2+y^2=cx$$

First, we can rewrite the given equation in the form of a homogeneous differential equation. We have:

$$\frac{dy}{dx} =\frac{y^2 - x^2}{2xy}$$

Then, recognizing the fraction can be separated into two:

$$=\frac{1}{2}v-\frac{1}{2v}$$

Then equating the derivative substitution (for $y=vx$) to the above gives us:

$$v + x\frac{dv}{dx} = \frac{1}{2}v-\frac{1}{2v}$$

Rearranging this, we can express it as:

$$x\frac{dv}{dx} = -\frac{1}{2}v-\frac{1}{2v}$$

This is a separable differential equation, which can be solved by separating the variables:

$$\frac{1}{x}~dx = \frac{1}{-\frac{1}{2}v-\frac{1}{2v}}~dv$$

Integrating both sides, we have:

$$\int \frac{1}{x}~dx = \int \frac{1}{-\frac{1}{2}v-\frac{1}{2v}}~dv$$

The integral on the left side is straightforward and yields:

$$\int \frac{1}{x}~dx = \ln|x| + C$$

The integral on the right side can be simplified as follows:

$$-\int \frac{2v}{1+v^2}~dv$$

Using u-substitution, let $u = 1 + v^2$, then $du = 2v~dv$. This gives us:

$$-\int \frac{1}{u}~du = -\ln|u| + C = -\ln|1 + v^2| + C$$

Equating the two integrals, we have:

$$\ln|x| + C = -\ln|1 + v^2|$$

Rearranging this, we can express it as:

$$\ln|x| + \ln|1 + v^2| = C$$

Using the properties of logarithms, we can combine the left side:

$$\ln|x(1 + v^2)| = C$$

Exponentiating both sides gives us:

$$x(1 + v^2) = e^C$$

Let $c = e^C$, which is a positive constant. Then we have:

$$x(1 + v^2) = c$$

Substituting back $v = \frac{y}{x}$, we get:

$$x\left(1 + \left(\frac{y}{x}\right)^2\right) = c$$

This simplifies to:

$$x^2 + y^2 = cx$$