Homogeneous Substitution

If a first order equation isn’t separable or linear, then one of three types of substitutions might solve the equation.

The Essentials

The second kind of substitution is a homogeneous substitution. A continuous function \( f(x, y) \) is homogeneous if \( f(tx, ty) = t^nf(x, y) \) where n is an integer greater than zero. The standard form for a homogeneous function is:

\[ \frac{dy}{dx} = \frac{M(x, y)}{N(x, y)} \]

The two substitutions that can be used for homogeneous equations are:

\[ u = \frac{x}{y} \]

(1)

\[ u = \frac{y}{x} \]

(2)

Either could be used, but the first is usually easier when the M is simpler and the second when N is simpler. The left hand side must also be expressed in terms of u.

Example

Solve the equation:

\[ 6xy \frac{dy}{dx} = 3x^2 + 6y^2 \]

First, we put the equation in standard form:

\[ \frac{dy}{dx} = \frac{x^2 + 2y^2}{2xy} \]

Then, we write our our substitution and solve for \( \frac{dy}{dx} \):

\[ u = \frac{y}{x} \] \[ y = u{x} \] \[ \frac{dy}{dx} = x \frac{du}{dx} + u \]

Next, we substitute u in for the x’s and y’s and we substitute the \( \frac{dy}{dx} \) term:

\[ x \frac{du}{dx} + u = \frac{x^2(1 + 2u^2)}{2x^2u} \] \[ x \frac{du}{dx} + u = \frac{1 + 2u^2}{2u} \] \[ x \frac{du}{dx} + u = \frac{1}{2u} + u \] \[ x \frac{du}{dx} = \frac{1}{2u} \] \[ \int 2u \cdot du = \int \frac{1}{x} \cdot dx \] \[ u^2 = \ln |x| + C \]

Finally, we undo the substitution and solve for y(x):

\[ \frac{y^2}{x^2} = \ln |x| + C \] \[ y^2 = x^2 (\ln |x| + C) \] \[ y = \pm x \sqrt{\ln|x| + C} \]