## Tuesday, June 13, 2017

### Unizor - Ordinary Differential Equations - Introduction

Notes to a video lecture on http://www.unizor.com

Ordinary Differential Equations
Introduction

Ordinary differential equations are equations, where derivatives of some function participate in the equation.
Assuming that y(x) is some unknown function, a differential equation, in its general form, looks like this:
F(x, y, y', y'',...) = 0
where F(...) is some function of many arguments.
The goal is to find the function y(x) that satisfies this equation.

y'(x) = 2x
We can easily guess that, if a derivative of a function equals to 2x, the function must be y(x)=x²+C, where C - any constant.

On the other hand, we can represent this equation in the form
dy/dx = 2x
and transform it into
dy = 2x·dx
This is a relationship between two infinitesimals that signifies that these infinitesimals are equal in a sense that the difference between them is an infinitesimal of a higher order than themselves.
Now we can apply an operation of integration to both getting the following
1·dy =  2x·dx

Integration results in the following equality
y + C1 = x² + C2,
where C1 and C2 are any constants, and therefore, can be combined into one, getting
y = x² + C
This method of integration is a little more "scientific" than straight guessing that we employed above, though, by itself, might be difficult since it involves the operation of integration.

Notice the presence of any constant in the result. This is typical for differential equations and is similar to indefinite integrals.

Arguably, the method of separation of argument x and function y into different sides of an equation with subsequent integration is the most effective way to solve differential equations. Those equations that allow solution of this type are called separable differential equations

Let's consider a few more examples.

Example 1

x²·y'(x) = y(x)
Let's represent y'(x) as a ratio of differentials dy/dx, our equation will look like
x²·dy/dx = y(x)
Now we can separate argument x and function y into different sides of an equation
dy/y = dx/x²
Integrate both sides
dy/y = dx/x²
which results in
ln(y) = −1/x + C
(where C is any constant) or, since we have to find an expression for y in terms of x, we can use this equality as exponents and raise e into it, getting
y = C·e−1/x

Let's check this result.
y'(x) = C·e−1/x·(1/x²)
x²·y'(x) = C·e−1/x = y(x)
All is correct.

Example 2

tan(x)·y'(x) = y²(x)
Let's represent y'(x) as a ratio of differentials dy/dx, our equation will look like
tan(x)·dy/dx = y²(x)
Now we can separate argument x and function y into different sides of an equation
dy/y² = dx/tan(x)
Integrate both sides
dy/y² = dx/tan(x)
which results in
−1/y + C = cos(x)dx/sin(x)
or, equivalently, since
cos(x)·dx = dsin(x),
it can be transformed into
−1/y + C = dsin(x)/sin(x)
The integral on the right can be calculated and the result is
−1/y + C = ln(sin(x))
(where C is any constant) or, since we have to find an expression for y in terms of x, we can transform it into
y = −1/[ln(sin(x))+C]
It would look better if we bring the constant under a logarithm, getting
y = −1/ln(C·sin(x))

Let's check this result.
y'(x) = [1/ln²(C·sin(x))] · [1/(C·sin(x))] · C·cos(x) = cot(x)/ln²(C·sin(x))
tan(x)·y'(x) = 1/ln²(C·sin(x)) = y²(x)
All is correct.

As you see, in all examples above there is a constant that can take any value, as in the case of indefinite integrals. That's because we explicitly use integration as a tool to solve our differential equation. That's why the term "solving" as related to differential equations sometimes is replaced with term "integrate". So, to integrate a differential equation means to solve it.

Without any additional information, as we see, a differential equation can have infinite number of solutions. But we need only one, that corresponds to some practical problem, from which this equation was obtained. Therefore, we need some condition imposed on our solution to determine this constant that is present in the general solution.

Consider Example 1 above
x²·y'(x) = y(x)
and its solution
y = C·e−1/x
This solution represents a whole family of functions, each satisfying our differential equation.
To determine a particular solution we are interested in, we have to define what we are interested in using some additional information about function y(x). For example, we know that our function y(x) equals to 1 if x=1.
Let's substitute this into a general solution to our differential equation to find the value of constant C needed to satisfy our condition.
y(1) = C·e−1/1 = 1
from which we can find constant C:
C·e-1 = 1
C/e = 1
C=e
Therefore, particular solution we are looking for is
y = e·e−1/x = e1-(1/x)

In the Example 2 let's determine constant C by a condition y(π/2)=1
That results in the following
y(π/2) = −1/ln(C·sin(π/2)) = 1
ln(C) = −1
C = 1/e
So, our particular solution is
y = −1/ln(sin(x)/e)