Homogeneous Differential Equations

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TYPE - 2: 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

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

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:

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

For example,

\begin{align}&\qquad {f_1}\left( {x,y} \right) = {x^3} + {y^3}\\ & \qquad \qquad \quad \;= {x^3}\left( {1 + {{\left( {\frac{y}{x}} \right)}^3}} \right) = {y^3}\left( {1 + {{\left( {\frac{x}{y}} \right)}^3}} \right)\end{align}

Having seen homogeneous functions we define homogeneous DEs as follows :

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

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

\begin{align}\frac{{dy}}{{dx}} = - \frac{{M\left( {x,y} \right)}}{{N\left( {x,y} \right)}} = - \frac{{{x^n}M\left( {\frac{y}{x}} \right)}}{{{x^n}N\left( {\frac{y}{x}} \right)}} = - \frac{{M\left( {\frac{y}{x}} \right)}}{{N\left( {\frac{y}{x}} \right)}} = P\left( {\frac{y}{x}} \right)\end{align}

(The function $$P\left( t \right)$$ stands for\begin{align}\frac{{ - M\left( t \right)}}{{N\left( t \right)}}\end{align})

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

\begin{align} &\qquad \qquad y = vx\\& \Rightarrow \quad \; \;\frac{{dy}}{{dx}} = v + x\frac{{dv}}{{dx}}\end{align}

Thus, \begin{align}\frac{{dy}}{{dx}} = P\left( {\frac{y}{x}} \right)\end{align} transforms to

\begin{align}& \qquad \qquad v + x\frac{{dv}}{{dx}} = P\left( v \right)\\& \Rightarrow \quad \;\;\;\frac{{dv}}{{P\left( v \right) - v}} = \frac{{dx}}{x}\end{align}

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

Let us see some examples of solving homogeneous DEs.

Example – 8

Solve the DE \begin{align}\frac{{dy}}{{dx}} = \frac{{2x - y}}{{x + y}}.\end{align}

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

\begin{align}\frac{{2x - y}}{{x + y}} = \frac{{x \cdot \left( {2 - \frac{y}{x}} \right)}}{{x \cdot \left( {1 + \frac{y}{x}} \right)}} = \frac{{2 - \frac{y}{x}}}{{1 + \frac{y}{x}}}\end{align}

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

\begin{align} &\qquad\qquad v + x\frac{{dv}}{{dx}} = \frac{{2 - v}}{{1 + v}}\\&\Rightarrow \quad x\frac{{dv}}{{dx}} = \frac{{2 - v}}{{1 + v}} - v\\ &\qquad\qquad\;\;\; = \frac{{2 - 2v - {v^2}}}{{1 + v}}\\ &\qquad\qquad\;\;\; = \frac{{3 - {{\left( {1 + v} \right)}^2}}}{{1 + v}}\\&\Rightarrow \quad \frac{{\left( {1 + v} \right)}}{{3 - {{\left( {1 + v} \right)}^2}}}dv = \frac{{dx}}{x}\end{align}

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

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

Integrating, we have

\begin{align} &\qquad \int {\frac{t}{{3 - {t^2}}}dt} = \int {\frac{{dx}}{x}} \\&\Rightarrow \quad - \frac{1}{2}\ln \left| {3 - {t^2}} \right| = \ln x + \ln {C_1}\\& \Rightarrow \quad \ln \left( {{x^2}\left( {3 - {t^2}} \right)} \right) = {C_2}\\& \Rightarrow \quad {x^2}\left( {3 - {t^2}} \right) = C\\ &\Rightarrow \quad {x^2}\left( {3 - {{\left( {1 + v} \right)}^2}} \right) = C\\ &\Rightarrow \quad {x^2}\left( {2 - 2v - {v^2}} \right) = C\end{align}

Substituting $$\frac{y}{x}$$ for $$v,$$ we finally obtain the required general solution to the DE:

$2{x^2} - 2xy - {y^2} = C.$