Homogenous differential equations – a first type

Up a level : Differential Equations
Previous page : Homogenous functions
Next page : Linear differential equations

Say we have an equation of the form

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

This we will solve by doing a variable substitution. Let

$y = vx$

That means that

$\frac{{dy}}{{dx}} = \frac{{d(vx)}}{{dx}} = \frac{{dv}}{{dx}}x + v\frac{{dx}}{{dx}} = \frac{{dv}}{{dx}}x + v$

This gives us that our original equation will become

$\frac{{dv}}{{dx}}x + v = f\left( v \right)$

$\frac{{dv}}{{dx}}x = f\left( v \right) - v$

and

$\frac{{dv}}{{f\left( v \right) - v}} = \frac{{dx}}{x}$

Now we can integrate both sides separately. In practice, it will often end up way easier than what it looks like from the above.

Example 1

We have already solved equations of this form, but let us try to solve one of them using this method to see that we get the same result. Let us look at

$\frac{{dy}}{{dx}} = - \frac{x}{y}$

The substitutions above give us

$\frac{{dv}}{{dx}}x + v = - \frac{x}{{vx}} = - \frac{1}{v}$

This gives us

$\frac{{dv}}{{dx}}x = - \frac{1}{v} - v$

$\frac{{dv}}{{ - \frac{1}{v} - v}} = - \frac{{vdv}}{{1 + {v^2}}} = \frac{{dx}}{x}$

We can now do the substitution

$\left| {\begin{array}{*{20}{c}} {w = 1 + {v^2}} \\ {dw = 2vdv} \\ {\frac{{dw}}{2} = vdv} \end{array}} \right.$

to get

$- \frac{{dw}}{{2w}} = \frac{{dx}}{x}$

or, after integrations

$- \frac{1}{2}\ln |w| = \ln |x| + {C_1}$

$\ln |w| = - 2\ln |x| + {C_2}$

and thus

$w = C{x^{ - 2}}$

or, given that y=vx

$1 + \frac{{{y^2}}}{{{x^2}}} = C{x^{ - 2}}$

Then we multiply through by x² to get

${x^2} + {y^2} = C$

I.e. the same solution as before, but in a way harder way. So let us look at an example where this method helps.

Example 2

Say we have

$xyy' + {x^2} + {y^2} = 0$

We can rewrite this as

$y' = - \frac{{{x^2} + {y^2}}}{{xy}} = - \frac{x}{y} - 1/\left( {\frac{x}{y}} \right)$

So it is an equation of the desired form.

Next, we do the substitutions, and I suggest you do it on the original equation. We get

$xvx\left( {\frac{{dv}}{{dx}}x + v} \right) + {x^2} + {(vx)^2} = 0$

$v{x^2}\left( {\frac{{dv}}{{dx}}x + v} \right) + {x^2} + {v^2}{x^2} = 0$

or,

${x^2}\left( {v\left( {\frac{{dv}}{{dx}}x + v} \right) + 1 + {v^2}} \right) = 0$

By the null factor law either x=0 or

$v\left( {\frac{{dv}}{{dx}}x + v} \right) + 1 + {v^2} = 0$

This gives us

$\begin{gathered} \frac{{dv}}{{dx}}x = \frac{{ - 1 - {v^2}}}{v} - v = \frac{{ - 1 - {v^2}}}{v} - \frac{{{v^2}}}{v} \hfill \\ \quad \quad = \frac{{ - 1 - 2{v^2}}}{v} \hfill \\ \end{gathered}$