# 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   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 + y2 xy

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

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

Yes! So let's go:

 Start with: dy dx = ( y x )-1 + y x y = vx and dy dx = v + x dv dx 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 Put the integral sign in front: ∫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.

Another example:

### Example: Solve dy dx = y(x−y) x2

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

 Start with: y(x−y) x2 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 dv dx 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.

And one last example:

### Example: Solve dy dx = x−y x+y

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

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

Yes! So let's go:

 Start with: dy dx = 1−y/x 1+y/x y = vx and dy dx = v + x dv dx 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.