Homogeneous Differential Equation is of a prime importance in physical applications of mathematics due to its simple structure and useful solution. In this section, we will discuss the homogeneous differential equation of the first order. Since they feature homogeneous functions in one or the other form, it is crucial that we understand what are homogeneous functions first. So let’s dive straight into it!

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## Homogeneous Functions

A homogeneous function is one that exhibits multiplicative scaling behavior i.e. if all of its arguments are multiplied by a factor, then the value of the function is multiplied by some power of that factor. Mathematically, we can say that a function in two variables f(x,y) is a homogeneous function of degree *n* if –

\(f(\alpha{x},\alpha{y}) = \alpha^nf(x,y)\)

where α is a real number. For example, if given f(x,y,z) = x^{2} + y^{2} + z^{2} + xy + yz + zx. We can note that f(αx,αy,αz) = (αx)^{2}+(αy)^{2}+(αz)^{2}+αx.αy+αy.αz+αz.αx=α^{2}f(x,y,z). Thus, given f(x,y,z) is a homogeneous function of degree 2.

**Browse more Topics under Differential Equations**

- Order and Degree of Differential Equations
- Linear Differential Equations
- General and Particular Solutions of a Differential Equation
- Formation of differential Equation whose General Solution is Given
- Differential Equations with Variables Separable

## Homogeneous Differential Equation of the First Order

A first-order differential equation, that may be easily expressed as $${\frac{dy}{dx} = f(x,y)}$$ is said to be a homogeneous differential equation if the function on the right-hand side is homogeneous in nature, of degree = 0. This implies that for any real number α –

\(f(\alpha{x},\alpha{y}) = \alpha^0f(x,y)\)

\( = f(x,y)\)

An alternate form of representation of the differential equation can be obtained by rewriting the homogeneous function on the right-hand side in terms of two homogeneous functions M(x,y) and N(x,y) of the same degree –

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

\(N(x,y)dy = -M(x,y)dx\)

\(M(x,y)dx + N(x,y)dy = 0\)

Another representation is possible, which is a bit non-trivial to prove and is only valid for the case of the first order. It is shown below –

$${\frac{dy}{dx} = f(\frac{x}{y})}$$

Now that you’ve learnt to identify the homogeneous differential equations, let us look at the general method for solving such equations.

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## Methods of Solving a Homogeneous Differential Equation

- Introduce a new dependent variable \(v = \frac{y}{x}\). It would involve substituting \(y = vx\) in the differential equation.
- The differential equation now becomes – $${x\frac{dv}{dx} + v = f(\frac{1}{v})}$$
- Solve the above differential equation by the variables separation method.
- Substitute the expression obtained for v back in \(y = vx\) to obtain the general solution to the differential equation.

## Equations Reducible to the Homogeneous Form

The first-order differential equations of the form – $${\frac{dy}{dx} = f(\frac{ax + by + c}{kx + ly + m})}$$ where a, b, c, k, l, m are constants is reducible to Homogeneous Form. First, let us note how does it even comes close to the homogeneous form. On rewriting the equation –

\(\frac{dy}{dx} = f(\frac{a + b\frac{y}{x} + \frac{c}{x}}{k + l\frac{y}{x} + \frac{m}{x}})\)

This clearly is the Homogeneous Form itself, if c = m = 0. In order to do just that, we shift the coordinates i.e. introduce new variables in the system –

X = x – α, Y = y – β

where α and β are some constants whose values we’ll determine from the condition c = m = 0. Following this process, the equation will reduce to homogeneous form which then can be directly solved by the substitution \(Y = vX\). Although, in the end, we must back-substitute Y in the equation Y = y – β to get the final solution.

Now take a look at the solved examples below to clear your methods!

## Solved Examples For You

### Example 1: Solve

\((2x^3 + y^3)dx – 3xy^2dy = 0\)

**Solution: **The given differential equation is a homogeneous differential equation of the first order since it has the form \(M(x,y)dx + N(x,y)dy = 0\), where M(x,y) and N(x,y) are homogeneous functions of the same degree = 3 in this case.

Here, \(M(x,y) = 2x^3 + y^3\) and \(N(x,y) = – 3xy^2\). To solve, we first rearrange the differential equation in the following format –

\(\frac{dy}{dx} = \frac{2x^3 + y^3}{3xy^2}\) …… (1)

To see that the equation is homogeneous, we can also see that the right-side can be converted to a function of the form \(f(\frac{y}{x})\) as

\(\frac{2x^3}{3xy^2} + \frac{y^3}{3xy^2} = \frac{2}{3(\frac{y}{x})^2} + \frac{1}{3}(\frac{y}{x})\)

Anyway, to solve now we proceed with the substitution \(y = vx\) where \(x\) is the new dependent variable. Then the differential equation in the form 1) gets converted into

\(v + x\frac{dv}{dx} = \frac{2x^3 + y^3}{3xy^2}\)

\(=\frac{2x^3 + (vx)^3}{3x(vx)^2}\)

\(=\frac{2 + v^3}{3v^2}\)

#### Then the differential equation can be converted to variables-separable form as

\(x\frac{dv}{dx} = \frac{2 + v^3}{3v^2} – v\)

\(= \frac{2 – 2v^3}{3v^2}\)

\(\frac{3v^2}{v^3 – 1}\frac{dv}{dx} = -\frac{2}{x}\)

On integrating with respect to x, the general solution is found out to be

\(\int{\frac{3v^2}{v^3 – 1}\frac{dv}{dx}dx} = \int{-\frac{2}{x}dx}\)

\(ln[mod(v^3 – 1)] = -2lnx + c\)

\(ln[mod(v^3 – 1)x^2] = c\)

\([mod(v^3 – 1)]x^2 = e^c = C\)

This is the implicit form of the general solution for v(x). Now let us find y(x) by back-substituting \(v = \frac{y}{x}\).

\([mod((\frac{y}{x})^3 – 1)]x^2 = C\)

\((\frac{y}{x})^3 – 1)x^2 = \pm{C}\)

\(y^3 – x^3 = \pm{Cx}\)

which is the final solution. It is actually a one-parameter family of curves which satisfies the homogeneous differential equation.

### Example 2: Solve the equation

\((2x – y +1)dx – (x – 2y – 1)dy = 0\)

**Solution: **The given differential equation can be rearranged to get

\(\frac{dy}{dx} = \frac{(2x – y +1)}{(x – 2y – 1)}\)

This is clearly the form that is reducible to the homogeneous differential equation form. Let us introduce two new variables

X = x – α, Y = y – β

where α, β are unknown constants. Thus, in our equation, we must proceed with the substitution

x = X + α, y = Y + β

dx = dX, dy = dY

#### We can then get the resultant differential equation as

\(\frac{dY}{dX} = \frac{(2(X + α) – (Y + β) + 1)}{((X + α) – 2(Y + β) – 1)}\)

\(\frac{dY}{dX} = \frac{(2X – Y + (2α – β + 1))}{(X – 2Y + (α – 2β – 1))}\)

In order for the differential equation to be homogeneous, the terms (2α – β + 1) and (α – 2β – 1) must be identically equal to zero. Thus we have two simultaneous linear equations in two unknowns (α and β) as

2α – β + 1 = 0

α – 2β – 1 = 0

These can be easily solved to get α = -1, and β = -1. On using these values, we will get the resultant differential equation as

\(\frac{dY}{dX} = \frac{(2X – Y)}{(X – 2Y)}\)

#### This is clearly Homogeneous in nature and we can proceed by substituting Y = vX

\(v + X\frac{dv}{dX} = \frac{2X – (vX)}{X – 2(vX)}\)

\(= \frac{2 – v}{1 – 2v}\)

\(X\frac{dv}{dX} = \frac{2 – v}{1 – 2v} – v\)

\( = \frac{2 – 2v + 2v^2}{1 – 2v}\)

\( = \frac{2(v^2 – v + 1)}{1 – 2v}\)

\(\frac{1 – 2v}{2(v^2 – v + 1)}\frac{dv}{dX} = \frac{1}{X}\)

Integrate with respect to X

\(\int{\frac{1 – 2v}{2(v^2 – v + 1)}\frac{dv}{dX}dX} = \int{\frac{1}{X}dX}\)

\(ln(v^2 – v + 1)^{-\frac{1}{2}}= lnX + c\)

\(ln(\frac{(v^2 – v + 1)^{-\frac{1}{2}}}{X}) = c\)

\((v^2 – v + 1)^{\frac{1}{2}}X = \frac{1}{e^c} = C\)

\((v^2 – v + 1)X^2 = C’\)

Now let us back-substitute to get Y

\((\frac{Y}{X})^2 – \frac{Y}{X} + 1)X^2 = C’\)

Finally, we must remember to change the variables X and Y back to the original variables x and y. We can then get the general solution, in the implicit form, as

\(((\frac{(y + 1)}{(x + 1)})^2 – \frac{(y + 1)}{(x + 1)} + 1)(x + 1)^2 = C’\)

This concludes our discussion on this very important topic. The examples that we have discussed above, if understood properly, are sufficient to enable you to solve any problem involving homogeneous differential equations now.

**Solved Question for You**

**Question 1: What are Homogeneous Functions?**

**Answer:** A homogeneous function refers to one which demonstrates multiplicative scaling behaviour. In other words, if we multiply all of its arguments by a factor, after that we multiply the value of the function by some power of that factor.

**Question 2: What makes a system linear?**

**Answer:** A linear system refers to a mathematical model of a system which is on the basis of the use of a linear operator. Linear systems usually demonstrate features and properties which are much simpler than the nonlinear case.

** Question 3: What is non-homogeneous?**

**Answer:** Homogeneous means made up of various kinds of people or things. For instance, there are non-homogeneous neighbourhoods. Then, we have the non-homogenous atmosphere of the earth as well as the non-homogenous allocation of particles.

**Question 4: What makes a function homogeneous?**

**Answer:** A multiplicative scaling behaviour makes a function homogenous. For instance, a homogeneous real-valued function of two variables x and y is a real-valued function which satisfies the condition.