# Homogeneous Differential Equations

A Differential Equation is an equation with a function and one or more of its derivatives:

Example: an equation with the function y and its derivative dy dx

Here we look at a special method for solving "Homogeneous Differential Equations"

## Homogeneous Differential Equations

A first order Differential Equation is Homogeneous when it can be in this form:

dy dx = F( y x )

We can solve it using Separation of Variables but first we create a new variable v = y x

v = y x   which is also   y = vx
And dy dx = d (vx) dx = v dx dx + x dv dx (by the Product Rule)
Which can be simplified to dy dx = v + x dv dx

Using y = vx and dy dx = v + x dv dx we can solve the Differential Equation.

An example will show how it is all done:

### Example: Solve dy dx = x2 + y2xy

Can we get it in F( y x ) style?

Separate terms: x2 xy + y2 xy
Simplify: x y + y x
Reciprocal of first term:( y x )−1 + y x

Yes, we have a function of (y/x).

So let's go:

Start with: dy dx = ( y x )−1 + y x
y = vx and dydx = v + x dvdx:v + x dv dx = v−1 + v
Subtract v from both sides:x dv dx = v−1

Now use Separation of Variables:

Separate the variables:v dv = 1 x dx
Integrate: v2 2 = ln(x) + C
Then we make C = ln(k): v2 2 = ln(x) + ln(k)
Combine ln: v2 2 = ln(kx)
Simplify:v = ±√(2 ln(kx))

Now substitute back v = y x

Substitute v = y x : y x = ±√(2 ln(kx))
Simplify:y = ±x √(2 ln(kx))

And we have the solution.

The positive portion looks like this:

Another example:

### Example: Solve dy dx = y(x−y) x2

Can we get it in F( y x ) style?

Separate terms: xy x2 y2 x2
Simplify: y x − ( y x )2

Yes! So let's go:

Start with: dy dx = y x − ( y x )2
y = vx and dy dx = v + x dvdx v + x dv dx = v − v2
Subtract v from both sides:x dv dx = −v2

Now use Separation of Variables:

Separate the variables: 1 v2 dv = 1 x dx
Put the integral sign in front: 1 v2 dv = 1 x dx
Integrate: 1 v = ln(x) + C
Then we make C = ln(k): 1 v = ln(x) + ln(k)
Combine ln: 1 v = ln(kx)
Simplify:v = 1 ln(kx)

Now substitute back v = y x

Substitute v = y x : y x = 1 ln(kx)
Simplify:y = x ln(kx)

And we have the solution.

Here are some sample k values:

And one last example:

### Example: Solve dy dx = x−y x+y

Can we get it in F( y x ) style?

Divide through by x: x/x−y/x x/x+y/x
Simplify: 1−y/x 1+y/x

Yes! So let's go:

y = vx and dy dx = v + x dvdx v + x dv dx = 1−v 1+v
Subtract v from both sides:x dv dx = 1−v 1+v − v
Then:x dv dx = 1−v 1+v v+v2 1+v
Simplify:x dv dx = 1−2v−v2 1+v

Now use Separation of Variables:

Separate the variables: 1+v 1−2v−v2 dv = 1 x dx
Put the integral sign in front: 1+v 1−2v−v2 dv = 1 x dx
Integrate: 1 2 ln(1−2v−v2) = ln(x) + C
Then we make C = ln(k): 1 2 ln(1−2v−v2) = ln(x) + ln(k)
Combine ln:(1−2v−v2)−½ = kx
Square and Reciprocal:1−2v−v2 = 1 k2x2

Now substitute back v = y x

Substitute v = y x :1−2( y x )−( y x )2 = 1 k2x2
Multiply through by x2:x2−2xy−y2 = 1 k2

We are nearly there ... it is nice to separate out y though!
We can try to factor x2−2xy−y2 but we must do some rearranging first:

Change signs:y2+2xy−x2 = − 1 k2
Replace − 1 k2 by c:y2+2xy−x2 = c
Add 2x2 to both sides:y2+2xy+x2 = 2x2+c
Factor:(y+x)2 = 2x2+c
Square root:y+x = ±√(2x2+c)
Subtract x from both sides:y = ±√(2x2+c) − x

And we have the solution.

The positive portion looks like this: