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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

Step-5

Substituting \dfrac{y}{x} for v, we finally obtain the required general solution to the DE:
2{x^2} - 2xy - {y^2} = C
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