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$$F(x,y,y^{\prime })=0$$Such that

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

Then the equation can take the form

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

Which is essentially another in the form

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

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

$\frac{y}{x}=f(y^{\prime })$

So that

$y=xf(y^{\prime })$.

We can differentiate this to get

$y^{\prime }=f(y^{\prime })+x{f}^{\prime }(y^{\prime })\frac{d{y}^{\prime }}{dx}$

Then re-arranging things,

$\frac{dx}{x}=\frac{f^{\prime }(y^{\prime })}{y^{\prime }-f(y^{\prime })}d{y}^{\prime }$

So that upon integrating,

$ln(x)=\int \frac{f^{\prime }(y^{\prime })}{y^{\prime }-f(y^{\prime })}d{y}^{\prime }+C$

We get

$x=C{e}^{\int \frac{f^{\prime }(y^{\prime })}{y^{\prime }-f(y^{\prime })}d{y}^{\prime }}$

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

$y=xf(y^{\prime })$

and

$x=C{e}^{\int \frac{f^{\prime }(y^{\prime })}{y^{\prime }-f(y^{\prime })}d{y}^{\prime }}$,

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 $y^{\prime }(x,v,v^{\prime })$

$y=vx$

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

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 $y^{\prime }(x,v,v^{\prime })$

$y=vx$

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

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 $dy+f\left(\frac{a_{1}x+b_{1}y+c_1}{a_{2}x+b_{2}y+c_2}\right)dx=0$,

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:

$dy+f\left(\frac{a_1{x}^{\prime }+b_1{y}^{\prime }}{a_2{x}^{\prime }+b_2{y}^{\prime }}\right)dx=0$