# 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

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

Then the equation can take the form

$x^{n}F(1,{\frac {y}{x}},y')=0$ Which is essentially another in the form

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

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

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

$y=xf(y')$ .

We can differentiate this to get

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

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 2

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

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.

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

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 3

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

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 expressions

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

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

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

Which turns it into a homogeneous equation of degree 0:

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