# 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** \frac{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:

\frac{dy}{dx} = F( \frac{y}{x} )

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

*which is also*y = vx

Using **y = vx** and** \frac{dy}{dx} = v + x\frac{dv}{dx}** we can solve the Differential Equation.

An example will show how it is all done:

### Example: Solve \frac{dy}{dx} = \frac{x^{2} + y^{2}}{xy}

Can we get it in F( \frac{y}{x} ) style?

^{-1}+ \frac{y}{x}

Yes, we have a function of \frac{y}{x}.

So let's go:

^{-1}+ \frac{y}{x}

**y = vx**and

**\frac{dy}{dx} = v + x\frac{dv}{dx}**:v + x\frac{dv}{dx} = v

^{-1}+ v

^{-1}

Now use Separation of Variables:

**C = ln(k)**:\frac{v^{2}}{2} = ln(x) + ln(k)

Now substitute back v = \frac{y}{x}

And we have the solution.

The positive portion looks like this:

Another example:

### Example: Solve \frac{dy}{dx} = \frac{y(x−y)}{x^{2}}

Can we get it in F( \frac{y}{x} ) style?

^{2}

Yes! So let's go:

^{2}

**y = vx**and \frac{dy}{dx}

**= v + x\frac{dv}{dx}**v + x\frac{dv}{dx} = v − v

^{2}

^{2}

Now use Separation of Variables:

**C = ln(k)**:\frac{1}{v} = ln(x) + ln(k)

Now substitute back v = \frac{y}{x}

And we have the solution.

Here are some sample k values:

And one last example:

### Example: Solve \frac{dy}{dx} = \frac{x−y}{x+y}

Can we get it in F( \frac{y}{x} ) style?

Yes! So let's go:

**y = vx**and \frac{dy}{dx}

**= v + x\frac{dv}{dx}**v + x\frac{dv}{dx} = \frac{1−v}{1+v}

Now use Separation of Variables:

^{2}) = ln(x) + C

**C = ln(k)**:−\frac{1}{2} ln(1−2v−v

^{2}) = ln(x) + ln(k)

^{2})

^{-½}= kx

^{2}= \frac{1}{k^{2}x^{2}}

Now substitute back v = \frac{y}{x}

^{2}= \frac{1}{k^{2}x^{2}}

**x**:x

^{2}^{2}−2xy−y

^{2}= \frac{1}{k^{2}}

We are nearly there ... it is nice to separate out y though!

We can try to factor x^{2}−2xy−y^{2} but we must do some rearranging first:

^{2}+2xy−x

^{2}= − \frac{1}{k^{2}}

^{2}+2xy−x

^{2}= c

^{2}to both sides:y

^{2}+2xy+x

^{2 }= 2x

^{2}+c

^{2}= 2x

^{2}+c

^{2}+c)

^{2}+c) − x

And we have the solution.

The positive portion looks like this: