# Ordinary Differential Equations/Homogeneous x and y

Not to be confused with homogeneous equations, an equation homogeneous in x and y of degree n is an equation of the form

F(x,y,y')=0

Such that

$a^nF(x,y,y')=F(ax,ay,y')$.

Then the equation can take the form

$x^nF(1,\frac{y}{x},y')=0$

Which is essentially another in the form

$x^nF(\frac{y}{x},y')=0$.

If we can solve this equation for y', then we can easily use the substitution method mentioned earlier to solve this equation. Suppose, however, that it is more easily solved for $\frac{y}{x}$,

$\frac{y}{x}=f(y')$

So that

$y=xf(y')$.

We can differentiate this to get

y'=f(y')+xf'(y')y

Then re-arranging things,

$\frac{dx}{x}=\frac{f'(y')}{y'-f(y')}dy'$

So that upon integrating,

$ln(x)=\int \frac{f'(y')}{y'-f(y')}dy'+C$

We get

$x=Ce^{\int \frac{f'(y')}{y'-f(y')}dy'}$

Thus, if we can eliminate y' between two simultaneous equations

$y=xf(y')$

and

$x=Ce^{\int \frac{f'(y')}{y'-f(y')}dy'}$,

then we can obtain the general solution..

## Homogeneous Ordinary Differential EquationsEdit

A function P is homogeneous of order $\alpha$ if $a^\alpha P(x,y)=P(ax,ay)$. A homogeneous ordinary differential equation is an equation of the form P(x,y)dx+Q(x,y)dy=0 where P and Q are homogeneous of the same order.

The first usage of the following method for solving homogeneous ordinary differential equations was by Leibniz in 1691. Using the substitution y=vx or x=vy, we can make turn the equation into a separable equation.

$\frac{dy}{dx}=F \left (\frac{y}{x} \right)$
$v(x,y)=\frac{y}{x}$
$y=vx \,$

Now we need to find v':

$\frac{dy}{dx}=v+\frac{dv}{dx}x$

Plug back into the original equation

$v+x\frac{dv}{dx}=F(v)$
$\frac{dv}{dx}=\frac{F(v)-v}{x}$
Solve for v(x), then plug into the equation of v to get y
$y(x)=xv(x) \,$

Again, don't memorize the equation. Remember the general method, and apply it.

### Example 2Edit

$\frac{dy}{dx}=5\frac{y}{x}+3\frac{x}{y}$

Let's use $v=\frac{y}{x}$. Solve for y'(x,v,v')

$y=vx \,$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Now plug into the original equation

$v+x\frac{dv}{dx}=5v+\frac{3}{v}$
$x\frac{dv}{dx}=4v+\frac{3}{v}$
$v\frac{dv}{dx}=\frac{(4v^2+3)}{x}$

Solve for v

$\frac{vdv}{4v^2+3}=\frac{dx}{x}$
$\int \frac{vdv}{4v^2+3}=\int \frac{dx}{x}$
$\frac{1}{8}\ln(4v^2+3)=\ln(x)$
$4v^2+3=e^{8 \ln(x)} \,$
$4v^2+3=e^{\ln(x^8)} \,$
$4v^2+3=x^8 \,$
$v^2=\frac{x^8-3}{4}$

Plug into the definition of v to get y.

$y=vx \,$
$y^2=v^2x^2 \,$
$y^2=\frac{x^{10}-3x^2}{4}$

We leave it in $y^2$ form, since solving for y would lose information.

Note that there should be a constant of integration in the general solution. Adding it is left as an exercise.

### Example 3Edit

$\frac{dy}{dx}=\frac{x}{\sin(\frac{y}{x})}+\frac{y}{x}$

Lets use $v=\frac{y}{x}$ again. Solve for $y'(x,v,v')$

$y=vx \,$
$\frac{dy}{dx}=v+x\frac{dv}{dx}$

Now plug into the original equation

$v+x\frac{dv}{dx}=\frac{x}{\sin(v)}+v$
$x\frac{dv}{dx}=\frac{x}{\sin(v)}$
$\sin(v)dv=dx$

Solve for v:

$\int \sin(v)dv=\int dx$
$-\cos v=x+C \,$
$v=\arccos(-x+C) \,$

Use the definition of v to solve for y.

$y=vx \,$
$y=\arccos(-x+C)x \,$

### An equation that is a function of a quotient of linear expressionsEdit

Given the equation $dy+f(\frac{a_1x+b_1y+c_1}{a_2x+b_2y+c_2})dx=0$,

We can make the substitution x=x'+h and y=y'+k where h and k satisfy the system of linear equations:

$a_1h+b_1k+c_1=0$
$a_2h+b_2k+c_2=0$

Which turns it into a homogeneous equation of degree 0:

$dy+f(\frac{a_1x'+b_1y'}{a_2x'+b_2y'})dx=0$