How to Solve Homogeneous Equations? - Calculus

In this post, we'll learn how to solve homogeneous differential equations (DE):

What is a Homogeneous Function?

To properly understand homogeneous DEs, we'll first start by defining what homogeneous functions are.

Say we have a function $f (x, y)$ and a random variable $t$ . Let's evaluate it with $x = t x$ and $y = t y$ . The function $f$ would then be $f (t x, t y)$ . If this expression is the same as $t^{n} \cdot f (t x, t y)$ , then we say that function $f$ is homogeneous.

To illustrate, let's look at $f (x, y) = 3 x y + 5 x^{2}$ . We can check if the function is homogeneous:

First, let $x = t x$ and $y = t y$ . Evaluate the expression using these values.

$f (x, y) = 3 x y + 5 x^{2}$
$f (t x, t y) = 3 (t x) (t y) + 5 (t x)^{2}$

Next, manipulate the function so that $t$ can be factored out. Match the form $t^{n} f (x, y)$

$f (t x, t y) = 3 x y t^{2} + 5 x^{2} t^{2}$
$f (t x, t y) = t^{2} (3 x y + 5 x^{2})$

The function is homogeneous if we reach the form $t^{n} f (x, y)$ . In the example, $t^{n} f (x, y) = t^{2} (3 x y + 5 x^{2})$

\) where $n$ is 2. This value, 2, is the degree of the homogeneous function.

Homogeneous Differential Equations

If we have a DE of the form: $M (x, y) d x + N (x, y) d y = 0$ and the functions $M (x, y)$ and $N (x, y)$ are homogeneous with the same degree, then we have a homogeneous differential equation.

When we have a homogeneous DE, we can convert the form into one that is solvable using separation of variables. We can summarize the concept of this method using these steps:

We create a placeholder expression relating $x$ and $y$ with a random variable.
Substitute the placeholder expression to the homogeneous DE, replacing either $x$ or $y$ .
Solve the DE with the random variable using separation of variables.
Return the replaced $x$ or $y$ using the placeholder expression.

Example

Consider the differential equation: $x y d x + 2 x^{2} d y = 0$ . Let's try to solve it.

Investigate DE

First, we would investigate if this DE observes the form: $M (x, y) d x + N (x, y) d y = 0$ . Let's identify the components of the equation:

$M (x, y) = x y$
$N (x, y) = 2 x^{2}$

We then check if $M (x, y)$ and $N (x, y)$ are homogeneous functions of the same degree:

$M (x, y) = x y$
$M (t x, t y) = (t x) (t y)$
$M (t x, t y) = t^{2} (x y)$

$M (x, y) = x y$ is a homogeneous function of degree 2.

$N (x, y) = 2 x^{2}$
$N (t x, t y) = 2 (t x)^{2}$
$N (t x, t y) = t^{2} (2 x^{2})$

$N (x, y) = 2 x^{2}$ is a homogeneous function of degree 2.

Because both functions are of the same degree, we can use this method to solve the DE.

Apply Method

To begin the homogeneous DE solution, we relate $x$ and $y$ with another variable - let's say $a$ . We create a simple placeholder expression: $y = a x$ . In preparation for the later stage, we need to find the differential of this expression: $d y = a d x + x d a$ .

Afterward, we substitute the placeholder expression and its derivatives into the DE.

$x y d x + 2 x^{2} d y = 0$
$x (a x) d x + 2 x^{2} (a d x + x d a) = 0$

Next, we simplify the expression to reach a DE of the form $M (x) + N (y) y^{'} = 0$ . To achieve this, we must divide the whole equation with a value common to all terms.

$\frac{x^{2} a d x + 2 x^{2} a d x + 2 x^{3} d a = 0}{x^{2}}$
$a d x + 2 a d x + 2 x d a = 0$
$3 a d x + 2 x d a = 0$

We can apply separation of variables afterward to reach a solution. We won't show the process, but the general solution will be:

$\frac{3 d x}{x} = \frac{- 2 d a}{a}$
$3 \ln (x) = - 2 \ln (a) + C_{1}$

For solutions that involve logarithms, it's better to express them in algebraic terms. We can do this by letting $C_{1}$ be $\ln C$ and applying logarithm rules to reach an expression free from it. We can do this since $C_{1}$ is a constant expression encompassing a family of different solutions.

$3 \ln (x) = - 2 \ln (a) + C_{1}$
$3 \ln (x) = - 2 \ln (a) + \ln (C)$
$\ln (x)^{3} = \ln (a)^{- 2} + \ln (C)$
$e^{\ln (x)^{3} =} e^{\ln (\frac{C}{a^{2}})}$
$x^{3} = \frac{C}{a^{2}}$

From the placeholder expression $a = y / x$ , we return the original variable $y$ and simplify.

$x^{3} = \frac{C}{a^{2}}$
$x^{3} = \frac{C}{{(\frac{y}{x})}^{2}}$
$x = \frac{C}{y^{2}}$

The final result is the general solution. We can verify this by evaluating the original DE with the answer.

Summary

Homogeneous differential equations are another approach to solving differential equations analytically.

Say we have a function $f (x, y)$ and a random variable $t$ . Let's evaluate it with $x = t x$ and $y = t y$ . The function $f$ would then be $f (t x, t y)$ . If this expression is the same as $t^{n} \cdot f (t x, t y)$ , then we say that function $f$ is homogeneous. The variable $n$ is the degree of the homogeneous function.

If we have a DE of the form: $M (x, y) d x + N (x, y) d y = 0$ and the functions $M (x, y)$ and $N (x, y)$ are homogeneous with the same degree $n$ , then we have a homogeneous differential equation.

When we have a homogeneous DE, we can convert the form into one that is solvable using separation of variables.

To use this method, we create a placeholder expression relating $x$ and $y$ with a random variable.

Next, we substitute the placeholder expression to the homogeneous DE, replacing either $x$ or $y$ .

Solve the DE with the random variable using separation of variables.