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.