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

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

then we say the differential equation is homogeneous. Now put

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

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

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

and

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

This is separable. Thus integrate both sides

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

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 $M(x,y)$ is the homogeneous function of degree $n$.

Given

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

if $M(x,y)$ and $N(x,y)$ 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

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

Let $y = vx$. Then $y' = v'x + v$ and

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

Simplifying

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

or

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

Integrate both sides

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

Multiplying 2 to both sides to get

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

Now take $e$ as base to get

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

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

Finally replace $v = y/x$ to obtain the general solution

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

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

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

is $(0,1)$. we move the axis so that $(0,1)$ is the origin. Let $X = 0, Y = y-1$. Then $x = X + 0, y = Y + 1$ and

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

Thus we have

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

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

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

Since $Y = y-1, X = x$, we have

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

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

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

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

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

Integrate both sides

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

Then the general solution is

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