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

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The equation $\displaystyle P\;\frac{dy}{dx}\;=\;Q$ is said to be homogeneous if P and Q are homogeneous functions of x and y of the same degree.

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

(1)

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

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

(2)

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

(3)

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

Example 2

$\frac{dy}{dx}\;=\;\frac{y^3}{x^2}$

(4)

$Becomes\;\;\;\;\frac{dy}{dx}\;=\;x\;v^2$

(5)

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 3

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

(6)

Putting y - vx

Since y is a function of x so is v

$\frac{dy}{dx}\;=\;v\;+\;x\,\frac{dv}{dx}$

(7)

$\therefore\;\;\;\;v\;+\;x\,\frac{dv}{dx}\;=\;\frac{x^2\;+\;y^2}{2x^2}\;=\;\frac{1\;+\;v^2}{2}$

(8)

$\therefore\;\;\;\;2x\,dv\;=\;(1\;+\;v^2\;-\;2v)\,dx$

(9)

Separating the variables

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

(10)

Integrating

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

(11)

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

(12)

Substituting (12) in equation (10)

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

(13)

The General Form of a Homogeneous Linear Equation.

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

(14)

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

$If\;\;\;x\;=\;e^t$

(15)

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

(16)

$\therefore\;\;\;\;\frac{dy}{dt}\;=\;\frac{dy}{dx}\;.\;\frac{dx}{dt}\;=\;x\,\frac{dy}{dx}$

(17)

Also

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

(18)

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

(19)

Hence

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

(20)

And

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

(21)

The Use of the D operator to solve Homogeneous Equations.

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

(22)

Then from equation (27)

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

(23)

And from equation (21)

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

(24)

Example 4

Solve the following Differential equation:-

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

(25)

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

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

(26)

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

(27)

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

(28)

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

(29)

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

(30)

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

(31)

Example 5

Solve the following Differential Equation:-

$x^2\;\frac{d^2y}{dx^2}\;-\;2x\frac{dy}{dx}\;+\;2y\;=\;24\,x^4$

(32)

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

$D(D\;-\;1)\;-\;2Dy\;+\;2y\;=\;24\,e^{4t}$

(33)

$\therefore\;\;\;\;\;(D^2\;-\;3D\;+\;2)y\;=\;24\,e^{4t}$

(34)

$\therefore\;\;\;\;\;(D\;-\;2)(D\;-\;1)y\;=\;24\,e^{4t}$

(35)

$Thus\;\;\;\;y\;=\;\left(A\,e^t\;+\;B\,e^{2t} \right)\;+\;\frac{24\,e^{4t}}{(D\;-\;2)(D\;-\;1)}$

(36)

$\therefore\;\;\;\;y\;=\;\left(A\,e^t\;+\;B\,e^{2t} \right)\;+\;4\,e^{4t}$

(37)

$Or\;\;\;\;y\;=\;\left(A\,x\;+\;B\,x^{2} \right)\;+\;4\,x^{4}$

(38)

Equations which can be reduced to the Homogeneous Form.

Consider the following equation:-

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

(39)

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 6

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

(40)

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