Wednesday, June 14, 2017

Unizor - Ordinary Differential Equations - Major Types

Notes to a video lecture on

Ordinary Differential Equations
Major Types of Equations

In this lecture we will only consider first order ordinary differential equations for a function of one argument y(x) (no higher order derivatives). The general form of these equations is
F(x, y, dy/dx) = 0

We will consider three major types of these differential equations with known approaches to integration:
separable equations,
homogeneous equations,
linear non-homogeneous equations.

Separable Ordinary Differential Equations

A few examples we were working with in the introductory lecture to ordinary differential equations are separable in a sense that the original differential equation, that can be generally expressed as F(x, y, dy/dx) = 0, can be transformed into
f(y)·dy = g(x)·dx
that can be separately integrated, using the techniques of calculating indefinite integrals, and, hopefully, resolved for y.
Even if it will not be possible to resolve it for y, the result of integration will be a simpler formula G(x,y)=0 (it will also include a constant as a result of integration, which can be found if some initial condition on a function y(x) is imposed).
In any case, whether the result of integration can or cannot be resolved for y, it's still a significantly better than original equation that includes a derivative.


y' + x·y = 0
Let's use the Leibniz notation for derivatives to facilitate the separation of function from its argument and resolve the equation for a derivative.
dy/dx = −x·y
Separate x and y:
dy/y = −x·dx
Now we can apply an indefinite integral to both sides to solve the equation.

Homogeneous Ordinary Differential Equations

Homogeneous equations can be defined using the following criterion.
Replace all occurrences of x with λ·x and all occurrences of y with λ·y. Do not change anything with derivative dy/dx. If, as a result, all λ's cancel each other out, the equation is homogeneous.
For example, consider the following equation:
y' + x/y + x²/y² = 0
Substitute x with λ·x and y with λ·y:
y' + (λ·x)/(λ·y) + (λ·x)²/(λ·y)² = 0
Obviously, we can reduce both ratios, getting exactly the same equation as before.
Now we will use the above example to explain the method of solving homogeneous equations.
Let's introduce a new function z(x)=y(x)/x, which results in y(x)=z(x)·x and express the initial equation in terms of xz and z'.
The expression for a derivative y' is:
y' = (z·x)' = z'·x+z
So, our equation looks like
z'·x + z + x/(z·x) + x²/(z·x)² = 0
Simplifying by reducing the ratios by x and , we get
z'·x + z + 1/z + 1/z²= 0
This equation can be solved using the method of separation.
z'·x = −(z + 1/z + 1/z²)
dz/(z+1/z+1/z²) = −dx/x
Now we can apply an indefinite integral to both sides to solve the equation for z(x) and then multiply it by x to get y(x).

Linear Non-Homogeneous Ordinary Differential Equations

Standard form of this type of differential equations is
f(x)·y' + g(x)·y + h(x) = 0
As the first step, we can divide all members of this equation by f(x) (assuming it's not identically equal to 0), getting a simpler equation
y'+u(x)·y+v(x) = 0
The suggested solution lies in the substitution y(x)=p(x)·q(x), where p(x) and q(x) are unknown (for now) functions.
Express y'(x) in terms of p(x) and q(x):
y' = p·q'+q·p'
Substitute this into our equation:
p·q'+q·p'+u·p·q+v = 0
Let's simplify this
p(q'+u·q)+q·p'+v = 0
If there are such functions p(x) and q(x) that satisfy conditions
(1) q'+u·q = 0 and
(2) q·p'+v = 0
our job would be finished.
Let's try to find such functions.
From the equation (1) in our pair of equations we derive
q'/q = −u,
which can be converted into
dq(x)/q(x) = −u(x)·dx
that can be solved by integrating:
ln(q(x)) = −u(x)·dx
q(x) = e−∫u(x)·dx
Once q(x) is found, we solve the equation (2) for p(x):
p'(x) = −v(x)/q(x),
which can be integrated to find
p(x) = −v(x)/g(x) dx
and, consequently, y(x)=p(x)·q(x) can be fully determined.
Let's consider an example.
y' + x·y + x² = 0
If y(x)=p(x)·q(x), our equation looks like this:
p'·q+p·q'+x·p·q+x² = 0
(q'+x·q)·p+(q·p'+x²) = 0
Now we have to solve the following equation to nullify the first term:
q'+x·q = 0
(which is solvable through separation)
and substitute the resulting function q(x) into
q·p'+x² = 0
to solve it for p(x)
(which is a simple integration).

No comments: