Homogeneous Differential Equations

Using substitution, we often can transform a differential equation into simpler form. Here we consider how to transform the homogeneous differential equation into separable differential equation.

If a 1st order differential equation $y' = f(x,y)$ is put into the form

$\displaystyle y^{\prime} = f(\frac{y}{x}),$

then we say the differential equation is homogeneous. Now put

$\displaystyle v = \frac{y}{x}$

Then

and $y^{\prime} = v^{\prime}x + v$ . Substitute these into the above equation, we have

$\displaystyle v^{\prime}x + v = f(v)$

and

$\displaystyle \frac{dv}{f(v) - v} = \frac{1}{x} dx$

This is separable. Thus integrate both sides

$\displaystyle \int \frac{dv}{f(v) - v} = \log{x} + c$

Finally, substitute $v = \frac{y}{x}$ to get the general solution.

If a function $M(x,y)$ satisfies $M(tx,ty) = t^{n}M(x,y)$ , then we say is the homogeneous function of degree $n$ .

Given

$\displaystyle M(x,y)dx + N(x,y)dy = 0, \ y^{\prime} = -\frac{M(x,y)}{N(x,y)}$

and

are the homogeneous functions of the same degree, then the differential equation is homogeneous.

Example 1..6 Find the general solution of $y^{\prime} = \frac{x - y}{x + y}$ .

SOLUTION $M(x,y),N(x,y)$ are the homogeneous functions of the same degree 1. Divide the numerator and the denominator by $x$ . Then

$\displaystyle y^{\prime} = \frac{1 - (y/x)}{1 + (y/x)} = f(\frac{y}{x})$

Let

. Then

and

$\displaystyle v^{\prime}x + v = \frac{1 - v}{1 + v}$

Simplifying

$\displaystyle v^{\prime}x = \frac{1 - v}{1 + v} - v = \frac{1 - v - v - v^{2}}{1 + v}$

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

Integrate both sides

$\displaystyle \frac{1}{2}\log{\vert v^{2} + 2v - 1\vert} = -\log{x} + c,$

Multiplying 2 to both sides to get

$\displaystyle \log\vert v^{2} + 2v - 1\vert = -\log{x^2} + c.$

Now take

as base to get

$\displaystyle e^{\log\vert v^{2} + 2v - 1\vert} = e^{-\log{x^2}}c$

$\displaystyle (v^{2} + 2v - 1)x^{2} = c.$

Finally replace $v = y/x$

to obtain the general solution

$\displaystyle y^{2} + 2xy - x^{2} = c \ensuremath{\ \blacksquare}$

Example 1..7 Find the general solution to $y^{\prime} = \frac{x - y + 1}{x + y - 1}$ .

SOLUTION This is not homogeneou. But once we can get rid of the constant term, it becomes homogeneous. The intersetion of two lines

$\displaystyle x - y + 1 = 0, x + y - 1 = 0$

. we move the axis so that $(0,1)$

is the origin. Let $X = 0, Y = y-1$

. Then

and

$\displaystyle \frac{dy}{dx} = \frac{dy}{dY}\frac{dY}{dX}\frac{dX}{dx} = \frac{dY}{dX}$

Thus we have

$\displaystyle \frac{dY}{dX} = \frac{X - Y}{X + Y}$

This differential equation is solved in example 1.6. Thus the general solution is

$\displaystyle Y^{2} + 2XY - X^{2} = c .$

Since

, we have

$\displaystyle (y - 1)^{2} + 2(y - 1)x - x^{2} = c\ensuremath{\ \blacksquare}$

Example 1..8 Find the general solution of $y^{\prime} = \frac{x + y - 1}{x + y + 1}$ .

SOLUTION Let $u = x + y$ . Then $y = u - x$ and $y^{\prime} = u^{\prime} - 1$ . Put this back into the original differential equation to obtain

$\displaystyle u^{\prime} - 1 = \frac{u - 1}{u + 1},$

$\displaystyle u^{\prime} = \frac{u - 1}{u + 1} + 1 = \frac{u - 1 + u + 1}{u + 1} = \frac{2u}{u + 1},$

$\displaystyle \frac{u + 1}{u}du = 2dx.$

Integrate both sides

$\displaystyle u + \log{\vert u\vert} = 2x + c .$

Then the general solution is

$\displaystyle y - x + \log{\vert x + y\vert} = c \ensuremath{\ \blacksquare}$

Subsections

Exercise