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

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

Which is essentially another in the form

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

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&#39;:


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

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.


 * $$y=vx \,$$


 * $$y^2=v^2x^2 \,$$


 * $$y^2=\frac{x^{10}-3x^2}{4}$$

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

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

$$a_1h+b_1k+c_1=0$$

$$a_2h+b_2k+c_2=0$$

Which turns it into a homogeneous equation of degree 0:

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