Homogeneous

A homogeneous polynomial is a polynomial whose monomials with nonzero coefficients all have the same total degree.
For example : $x^7+xy^6+y^7$ is homogeneous polynomial .

The D operator is a linear operator defined as : $D \equiv \frac{d}{dt}$ . For example : $D(2x+1)=2$

The solution of homogeneous differential equations including the use of the D operator

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Definition

The equation $Py' = Q$ is said to be homogeneous if P and Q are homogeneous functions of $x$ and $y$ of the same degree.
For example :

If $\displaystyle \frac{dy}{dx} = f\left(\frac{y}{x} \right)$

We can test to see whether this first order equation is homogeneous by substituting $\displaystyle y = v\,x$ . If the result is in the form $f(v)$ i.e. all the $x$ 's are canceled then the test is satisfied and the equation is Homogeneous.

Example 1

Problem

$\frac{dy}{dx} = \frac{x^2 + y^2}{2\,x^2}$

Workings

Becomes $\frac{dy}{dx} = \frac{1 + v^2}{2}$

Solution

There are no terms in $x$ on the right hand side and the equation is Homogereous.

So the original equation is not homogeneous.

Methods of Solution

A solution can be found by putting $y = vx$ on both sides of the equation:

Example 1

Problem

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

Workings

Putting $y - vx$

Since y is a function of x so is v

$\frac{dy}{dx} = v + x\,\frac{dv}{dx}$ Therefore $v + x\,\frac{dv}{dx} = \frac{x^2 + y^2}{2x^2} = \frac{1 + v^2}{2}$ Therefore $2x\,dv = (1 + v^2 - 2v)\,dx$

Separating the variables

$\frac{2\,dv}{(v - 1)^2} = \frac{dx}{x}$

Integrating

$\frac{-\,2}{(v - 1)} = \ln x + C$

(1)

But $v = \frac{y}{x}$ so

$\frac{- 2}{v - 1} = \frac{-\,2}{\frac{y}{x} - 1} = \frac{2x}{x - y}$

(2)

Solution

Substituting equation (#2) in equation (#1)

$2x = (x - y)(\ln x + C)$

The General Form of a Homogeneous Linear Equation

$p_0\,x^n\,\frac{d^{n}y}{dx^{n}} + p_1\,x^{n-1}\,\frac{d^{n-1}}{dx^{n-1}} + ....+\;p_{n-2}\,x^2\;\frac{d^2y}{dx^2}{ + p_{n-1}\,x\,\,\frac{dy}{dx}} + p_n\,y = fx$

The method to solve this is to put $\displaystyle\; x = e^t$ and the equation then reduces to a linear type with constant coefficients.

If $x = e^t$
Then $\displaystyle \frac{dy}{dt} = x$
Therefore $\displaystyle \frac{dy}{dt} = \frac{dy}{dx}\;.\;\frac{dx}{dt} = x\,\frac{dy}{dx}$

Also

$\frac{d^2y}{dt^2} = \frac{d}{dx}\left(x\,\frac{dy}{dx} \right)\frac{dx}{dt} = x\,\left(\frac{dy}{dx} + x\,\frac{d^2y}{dx^2} \right)$
Therefore $\displaystyle x^2\,\frac{d^2y}{dx^2} + x\,\frac{dy}{dt}= \frac{d^2y}{dt^2}$

Hence

$x\;\frac{dy}{dx} = \frac{dy}{dt}$

(3)

And

$x^2\,\frac{d^2y}{dx^2} = \frac{d^2y}{dt^2} - \frac{dy}{dt}$

(4)

The Use of the D operator to solve Homogeneous Equations

If $\displaystyle x\;=z\;e^t$ and $D = \frac{d}{dt}$

Then from equation (#1)

$x\;\frac{dy}{dx} = Dy$

And from equation (#2)

$x^2\;\frac{d^2y}{dx^2} = D(D - 1)\,y$

(5)

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Equations which can be reduced to the Homogeneous Form

Consider the following equation:

$\frac{dy}{dx} = \frac{2x + 3y + 4}{4x + 5y - 10}$

The equation is not Homogeneous due to the constant terms $+ 4$ and $- 10$

However if we shift the origin to the point of intersection of the straight lines $\displaystyle 2x + 3y + 4 = 0$ and $4x + 5y - 10 = 0$ , then the constant terms in the differential equation will disappear.

Example 1

Problem

$\frac{dy}{dx} = \frac{2x + 9y - 20}{6x + 2y - 10}$

Workings

The lines $\displaystyle 2x + 9y - 20 = 0$ and $6x + 2y - 10 = 0$ meet at the point (1, 2). We therefore make the following substitutions:

$x = X + 1$ $y = Y + 2$

The equation now becomes:

$\frac{dY}{dX} = \frac{2(X + 1) + 9(Y + 2) - 20}{6(X + 1) + 2(y + 2) - 10} = \frac{2X + 9Y}{6X + 2Y}$

Solution

This is homogeneous and can be solved by putting Y = v X. The solution is given by: $(2x - y)^2 = C(x + 2y - 5)$

Exceptional case

If the two straight lines are parallel, then there is no finite point of intersection and we proceed as follows:

Let $\displaystyle \frac{dy}{dx} = \frac{3y - 4x - 2}{3y - 4x - 3}$ Put $Z = 3y - 4x$ Then $\displaystyle \frac{dZ}{dx} = 3\,\frac{dy}{dx} - 4$

Thus the equation becomes:

$\frac{1}{3}\left(\frac{dZ}{dx} + 4} \right) = \frac{Z - 2}{Z - 3}$ Therefore $\frac{dZ}{dx} = \left( \frac{3Z - 6}{Z - 3} \right) - 4 = \frac{-\,Z + 6}{Z - 3}$ Therefore $dx = -\,\left(\frac{Z - 3}{Z - 6} \right)\,dZ = -\,\left(\frac{Z - 6 + 3}{Z - 6} \right)\,dZ$ $= \left(-\,1 - \frac{3}{z - 6} \right)\,dZ$ Therefore $x = -\,Z - 3\,\ln\,(Z - 6) + K$ Thus $3\ln\,(3y - 4x - 6) = 3x - 3y + K$