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

Last modified on 12 April 2012, at 23:18