Homogeneous Differential Equations - Math Is Fun

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?

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, we have a function of yx.

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 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.

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?

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 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?

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 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:

9419, 9420, 9421, 9422, 9423, 9424, 9425, 9426, 9427, 9428 Homogeneous Functions Differential Equation Calculus Index