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Homogeneous

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

Definition

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

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

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

Example:
Example - Simple example
Problem

$\frac{dy}{dx}&space;=&space;\frac{x^2&space;+&space;y^2}{2\,x^2}$
Workings
Becomes
$\frac{dy}{dx}&space;=&space;\frac{1&space;+&space;v^2}{2}$
Solution
There are no terms in $\inline&space;x$ 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 $\inline&space;y&space;=&space;vx$ on both sides of the equation:
Example:
Example - Rational Example
Problem
$\frac{dy}{dx}&space;=&space;\frac{x^2&space;+&space;y^2}{2x^2}$
Workings
Putting $\inline&space;y&space;-&space;vx$

Since y is a function of x so is v

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

Separating the variables

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

Integrating

$\frac{-\,2}{(v&space;-&space;1)}&space;=&space;\ln&space;x&space;+&space;C$
But $\inline&space;v&space;=&space;\frac{y}{x}$ so
$\frac{-&space;2}{v&space;-&space;1}&space;=&space;\frac{-\,2}{\frac{y}{x}&space;-&space;1}&space;=&space;\frac{2x}{x&space;-&space;y}$
Solution
Substituting equation (2) in equation (1)

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

The General Form Of A Homogeneous Linear Equation

A homogeneous polynomial is a polynomial whose monomials with nonzero coefficients all have the same total degree.
For example : $\inline&space;x^7+xy^6+y^7$ is homogeneous polynomial .
$p_0\,x^n\,\frac{d^{n}y}{dx^{n}}&space;+&space;p_1\,x^{n-1}\,\frac{d^{n-1}}{dx^{n-1}}&space;+&space;....+\;p_{n-2}\,x^2\;\frac{d^2y}{dx^2}{&space;+&space;p_{n-1}\,x\,\,\frac{dy}{dx}}&space;+&space;p_n\,y&space;=&space;fx$

The method to solve this is to put $\inline&space;\displaystyle\;&space;x&space;=&space;e^t$ and the equation then reduces to a linear type with constant coefficients.
If $\inline&space;x&space;=&space;e^t$
Then $\inline&space;\displaystyle&space;\frac{dy}{dt}&space;=&space;x$
Therefore $\inline&space;\displaystyle&space;\frac{dy}{dt}&space;=&space;\frac{dy}{dx}\;.\;\frac{dx}{dt}&space;=&space;x\,\frac{dy}{dx}$

Also

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

Therefore $\inline&space;\displaystyle&space;x^2\,\frac{d^2y}{dx^2}&space;+&space;x\,\frac{dy}{dt}=&space;\frac{d^2y}{dt^2}$

Hence
$x\;\frac{dy}{dx}&space;=&space;\frac{dy}{dt}$
And
$x^2\,\frac{d^2y}{dx^2}&space;=&space;\frac{d^2y}{dt^2}&space;-&space;\frac{dy}{dt}$

The Use Of The D Operator To Solve Homogeneous Equations

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

The D operator is a linear operator defined as : $\inline&space;D&space;\equiv&space;\frac{d}{dt}$ . For example : $\inline&space;D(2x+1)=2$
Then from equation (1)

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

And from equation (2)

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

Example:
Example - A complex example
Problem
Solve the following Differential equation:

$x^2\;\frac{d^2y}{dx^2}&space;+&space;7x\,\frac{dy}{dx}&space;+&space;9\,y&space;=&space;18\,x^3$
Workings
By putting $\inline&space;&space;x&space;=&space;e^t\;$ and using the D factor then the equation reduces to:-

$D(D&space;-&space;1)y&space;+&space;7D\,y&space;+&space;9y&space;=&space;18\,e^{3t}$
Or
$(D^2&space;+&space;6D&space;+&space;9)y&space;=&space;18\,e^{3t}$
Therefore
$(D&space;+&space;3)^2y&space;=&space;18\,e^{3t}$
Therefore
$y&space;=&space;(A&space;+&space;B\,t)\;e^{-3t}&space;+&space;\frac{1}{(D&space;+&space;3)^2}\times18\,e^{3t}$

$=&space;(A&space;+&space;B\,t)\;e^{-3t}&space;+&space;\frac{18}{36}\,e^{3t}$
Solution
Therefore
$y&space;=&space;(A&space;+&space;B\,ln\,x)\;\frac{1}{x^3}&space;+&space;\frac{1}{2}\,x^{3}$

Equations Which Can Be Reduced To The Homogeneous Form

Consider the following equation:

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

The equation is not Homogeneous due to the constant terms $\inline&space;+&space;4$ and $\inline&space;-&space;10$

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

Example:
Example - Rational Example
Problem
$\frac{dy}{dx}&space;=&space;\frac{2x&space;+&space;9y&space;-&space;20}{6x&space;+&space;2y&space;-&space;10}$
Workings
The lines $\inline&space;\displaystyle&space;2x&space;+&space;9y&space;-&space;20&space;=&space;0$ and $\inline&space;6x&space;+&space;2y&space;-&space;10&space;=&space;0$ meet at the point (1, 2). We therefore make the following substitutions:

$x&space;=&space;X&space;+&space;1$
$y&space;=&space;Y&space;+&space;2$

The equation now becomes:

$\frac{dY}{dX}&space;=&space;\frac{2(X&space;+&space;1)&space;+&space;9(Y&space;+&space;2)&space;-&space;20}{6(X&space;+&space;1)&space;+&space;2(y&space;+&space;2)&space;-&space;10}&space;=&space;\frac{2X&space;+&space;9Y}{6X&space;+&space;2Y}$
Solution
This is homogeneous and can be solved by putting Y = v X. The solution is given by:
$(2x&space;-&space;y)^2&space;=&space;C(x&space;+&space;2y&space;-&space;5)$

Exceptional Case

If the two straight lines are parallel, then there is no finite point of intersection and we proceed as follows:
Let $\inline&space;\displaystyle&space;\frac{dy}{dx}&space;=&space;\frac{3y&space;-&space;4x&space;-&space;2}{3y&space;-&space;4x&space;-&space;3}$ Put $\inline&space;Z&space;=&space;3y&space;-&space;4x$ Then $\inline&space;\displaystyle&space;\frac{dZ}{dx}&space;=&space;3\,\frac{dy}{dx}&space;-&space;4$

Thus the equation becomes:

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