ONE MORE STEP TOWARD BETTER TOMORROW : CHAPTER 7

Tuesday, 5 August 2014

CHAPTER 7 - Homogeneous DEs

By definition, a homogeneous function $f\left( {x,y} \right)$ of degree $n$ satisfies the property

$f\left( {\lambda x,\lambda y} \right) = {\lambda ^n}f\left( {x,y} \right)$

For example, the functions

	${f_1}\left( {x,y} \right) = {x^3} + {y^3}$
	${f_2}\left( {x,y} \right) = {x^2} + xy + {y^2}$
	${f_3}\left( {x,y} \right) = {x^3}{e^{x/y}} + x{y^2}$

are all homogeneous functions, of degrees three, two and three respectively (verify this assertion).

Observe that any homogeneous function $f\left( {x,y} \right)$ of degree $n$ can be equivalently written as follows:

$f\left( {x,y} \right) = {x^n}f\left( {\dfrac{y}{x}} \right) = {y^n}f\left( {\dfrac{x}{y}} \right)$

For example,

	${f_1}\left( {x,y} \right) = {x^3} + {y^3}$
	$= {x^3}\left( {1 + {{\left( {\dfrac{y}{x}} \right)}^3}} \right) = {y^3}\left( {1 + {{\left( {\dfrac{x}{y}} \right)}^3}} \right)$

Having seen homogeneous functions we define homogeneous DEs as follows :

Any DE of the form $M(x, y) dx +N(x, y) dy = 0$ or $\dfrac{{dy}}{{dx}} = - \dfrac{{M\left( {x,y} \right)}}{{N\left( {x,y} \right)}}$ is called homogeneous if $M(x, y)$ and $N(x, y)$ are homogeneous functions of the same degree.

What is so special about homogeneous DEs ? Well, it turns out that they are extremely simple to solve. To see how, we express both $M(x, y)$ and $N(x, y)$ as, say ${x^n}M\left( {\dfrac{y}{x}} \right)$ and ${x^n}N\left( {\dfrac{y}{x}} \right)$ . This can be done since $M(x, y)$ and $N(x, y)$ are both homogeneous functions of degree $n$ . Doing this reduces our DE to

$\dfrac{{dy}}{{dx}} = - \dfrac{{M\left( {x,y} \right)}}{{N\left( {x,y} \right)}} = - \dfrac{{{x^n}M\left( {\dfrac{y}{x}} \right)}}{{{x^n}N\left( {\dfrac{y}{x}} \right)}} = - \dfrac{{M\left( {\dfrac{y}{x}} \right)}}{{N\left( {\dfrac{y}{x}} \right)}} = P\left( {\dfrac{y}{x}} \right)$

The function $P(t)$ stands for $\dfrac{{ - M\left( t \right)}}{{N\left( t \right)}}$

Now, the simple substitution $y=vx$ reduces this DE to a VS form :

	$y=vx$
	$\Rightarrow \dfrac{{dy}}{{dx}} = v + x\dfrac{{dv}}{{dx}}$

Thus, $\dfrac{{dy}}{{dx}} = P\left( {\dfrac{y}{x}} \right)$ transforms to

	$v + x\dfrac{{dv}}{{dx}} = P\left( v \right)$
	$\Rightarrow \dfrac{{dv}}{{P\left( v \right) - v}} = \dfrac{{dx}}{x}$

This can now be integrated directly since it is in VS form.

Let us see some examples of solving homogeneous DEs

Example: 1

Solve the DE $\dfrac{{dy}}{{dx}} = \dfrac{{2x - y}}{{x + y}}$

Solution: 1

Step-1

This is obviously a homogeneous DE of degree one since the RHS can be written as

$\dfrac{{2x - y}}{{x + y}} = \dfrac{{x \cdot \left( {2 - \dfrac{y}{x}} \right)}}{{x \cdot \left( {1 + \dfrac{y}{x}} \right)}} = \dfrac{{2 - \dfrac{y}{x}}}{{1 + \dfrac{y}{x}}}$

Step-2

Using the substitution $y=vx$ reduces this DE to

	$v + x\dfrac{{dv}}{{dx}} = \dfrac{{2 - v}}{{1 + v}}$
	$\Rightarrow x\dfrac{{dv}}{{dx}} = \dfrac{{2 - v}}{{1 + v}} - v$
	$= \dfrac{{2 - 2v - {v^2}}}{{1 + v}}$
	$= \dfrac{{3 - {{\left( {1 + v} \right)}^2}}}{{1 + v}}$
	$\Rightarrow \dfrac{{\left( {1 + v} \right)}}{{3 - {{\left( {1 + v} \right)}^2}}}dv = \dfrac{{dx}}{x}$

Step-3

Using $t=1+v$ above, we have

$\dfrac{t}{{3 - {t^2}}}dt = \dfrac{{dx}}{x}$

Step-4

Integrating, we have

	$\int {\dfrac{t}{{3 - {t^2}}}dt} = \int {\dfrac{{dx}}{x}}$
	$\Rightarrow -\dfrac{1}{2}\ln \left\| {3 - {t^2}} \right\| = \ln x + \ln {C_1}$
	$\Rightarrow \ln \left( {{x^2}\left( {3 - {t^2}} \right)} \right) = {C_2}$
	$\Rightarrow {x^2}\left( {3 - {t^2}} \right) = C$
	$\Rightarrow {x^2}\left( {3 - {{\left( {1 + v} \right)}^2}} \right) = C$
	$\Rightarrow {x^2}\left( {2 - 2v - {v^2}} \right) = C$