Homogeneous

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

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Contents

Definition
Methods Of Solution
The General Form Of A Homogeneous Linear Equation
The Use Of The D Operator To Solve Homogeneous Equations
Equations Which Can Be Reduced To The Homogeneous Form
Exceptional Case
Page Comments

Definition

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

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

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

Example:

Example - Simple example

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 $\inline y = vx$ on both sides of the equation:

Example:

Example - Rational Example

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

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

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

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

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

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

Hence

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

(2)

And

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

(3)

The Use Of The D Operator To Solve Homogeneous Equations

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

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

Then from equation (2)

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

And from equation (3)

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

(4)

Example:

Example - A complex example

Problem

Solve the following Differential equation:

$x^2\;\frac{d^2y}{dx^2} + 7x\,\frac{dy}{dx} + 9\,y = 18\,x^3$

Workings

By putting $x = e^t\;$ and using the D factor then the equation reduces to:-

$D(D - 1)y + 7D\,y + 9y = 18\,e^{3t}$

$(D^2 + 6D + 9)y = 18\,e^{3t}$

Therefore

$(D + 3)^2y = 18\,e^{3t}$

Therefore

$y = (A + B\,t)\;e^{-3t} + \frac{1}{(D + 3)^2}\times18\,e^{3t}$

$= (A + B\,t)\;e^{-3t} + \frac{18}{36}\,e^{3t}$

Solution

Therefore

$y = (A + B\,ln\,x)\;\frac{1}{x^3} + \frac{1}{2}\,x^{3}$

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 $\inline + 4$ and $\inline - 10$

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

Example:

Example - Rational Example

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

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 $\inline x = -\,Z - 3\,\ln\,(Z - 6) + K$ Thus $\inline 3\ln\,(3y - 4x - 6) = 3x - 3y + K$