# 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

${\displaystyle a^{n}F(x,y,y')=F(ax,ay,y')}$.

Then the equation can take the form

${\displaystyle x^{n}F(1,{\frac {y}{x}},y')=0}$

Which is essentially another in the form

${\displaystyle x^{n}F({\frac {y}{x}},y')=0}$.

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 ${\displaystyle {\frac {y}{x}}}$,

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

So that

${\displaystyle y=xf(y')}$.

We can differentiate this to get

${\displaystyle y'=f(y')+xf'(y'){\frac {dy'}{dx}}}$

Then re-arranging things,

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

So that upon integrating,

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

We get

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

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

${\displaystyle y=xf(y')}$

and

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

then we can obtain the general solution..

## Homogeneous Ordinary Differential Equations

A function P is homogeneous of order ${\displaystyle \alpha }$ if ${\displaystyle a^{\alpha }P(x,y)=P(ax,ay)}$. 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.

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

Now we need to find v':

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

Plug back into the original equation

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

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

### Example 2

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

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

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

Now plug into the original equation

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

Solve for v

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

Plug into the definition of v to get y.

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

We leave it in ${\displaystyle y^{2}}$ 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 3

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

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

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

Now plug into the original equation

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

Solve for v:

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

Use the definition of v to solve for y.

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

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

Given the equation ${\displaystyle dy+f({\frac {a_{1}x+b_{1}y+c_{1}}{a_{2}x+b_{2}y+c_{2}}})dx=0}$,

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

${\displaystyle a_{1}h+b_{1}k+c_{1}=0}$
${\displaystyle a_{2}h+b_{2}k+c_{2}=0}$

Which turns it into a homogeneous equation of degree 0:

${\displaystyle dy+f({\frac {a_{1}x'+b_{1}y'}{a_{2}x'+b_{2}y'}})dx=0}$