In this introductory course on Ordinary Differential Equations, we first provide basic terminologies on the theory of differential equations and then proceed to methods of solving various types of ordinary differential equations. We handle first order differential equations and then second order linear differential equations. We also discuss some related concrete mathematical modeling problems, which can be handled by the methods introduced in this course. The lecture is self contained. However, if necessary, you may consult any introductory level text on ordinary differential equations. For example, "Elementary Differential Equations and Boundary Value Problems by W. E. Boyce and R. C. DiPrima from John Wiley & Sons" is a good source for further study on the subject. The course is mainly delivered through video lectures. At the end of each module, there will be a quiz consisting of several problems related to the lecture of the week.
3-10 Ricatti's Equation - Ex. 6
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來自 Korea Advanced Institute of Science and Technology(KAIST) 的課程
Introduction to Ordinary Differential Equations
193 個評分
Korea Advanced Institute of Science and Technology(KAIST)
193 個評分
從本節課中
FIRST ORDER DEFERENTIAL EQUATION 2
與講師見面
Kwon, Kil HyunProfessor
Department of Mathematical Sciences
Now, let's introduce one more interesting equation,
So-called Ricatti differential equation.
It's a nonlinear first order differential equation
of the form y prime = p(x) + q(x)y + r(x)y squared, okay?
We are assuming r(x) is not identical to 0, okay?
Otherwise, In other words,
if r(x) is equal to identical to 0, then the remaining equation is
a simple linear first order differential equation, right?
Now, assume that somehow we find one particular
solution, say y1, okay, y sub 1.
Assume that y sub 1 is the solution of this Ricatti differential equation.
Then, y = y1 + u is also
a solution of the same equation if and
only if u prime- (q + 2 ry sub
1) times u is = ru squared.
Can you recognize this due equation for you?
Okay?
This is a Bernoulli equation,
where n is equal 2, right?
And we know how to solve the Bernoulli equation.
So starting from this first order, non-linear differential equation,
so called the Ricatti differential equation.
If we happen to know any one particular solution,
and by setting y = y1 + u, we can solve the given
Ricatti equation completely, okay?
Let's look at the example, okay.
Here's the one, the nice example of the Ricatti differential equation,
say y prime = 2x squared + 1 over x y- 2y squared,
And I claimed that this Ricatti differential
equation has a solution given by constant k times x.
We don't know what the constant k is.
Find the k first, then using this fact,
solve this Ricatti equation completely.
So, said y1 is = k of x, plugging that expression into this one.
What is the y prime?
That is ky is equal to 2x squared plus 1 over x times k of x,
that is k,- 2y1 squared, that is- 2k squared.
k and k cancelled out, you are going to get 2-
2k squared times x squared = 0, okay?
Are you following me?
Right?
So k and k canceled out.
You are going to get 0 = 2x squared- 2k squared
x squared, so 2x squared is a common, and
1- k squared, and that is equal to 0, okay?
That is a possible when k is equal to + or- 1, right, okay?
We do not need the both of them, okay?
We do not need the both of them.
So I'm taking k is equal +1, okay, that means what?
y is equal to x, is a solution.
If you wish then, you can confirm it.
y is equal to x, is a solution to this Ricatti differential equation.
Then what I said before, okay, if we know one particular solution
of this equation, then there is another solution
of the type y is equal to the known solution plus unknown u, right?
The known solution is now x, so I said, y = x + u, okay?
Plugging that expression into the differential equation.
And to simplify it, then you are going to get a first order differential equation
for the new unknown u, which is u prime + (4x
minus 1 over x) u, that is equal to negative 2 times u squared.
This is a Bernoulli equation, okay, with n is equal 2.
So set w = u to the -1, okay?
And by the simple computation, The differential equation for
w becomes w prime + (1 over x- 4x) times w = 2, okay?
Which is then linear first order differential equation for
w, and which has an integrating factor, okay?
Exponential of integral 1 over x- 4x dx, right?
That is, in fact, x times exponential negative 2x squared.
So multiply this integrating factor on both sides of this equation,
you will get derivative of, right,
x times e to negative 2x squared times w,
that is equal to 2 times of x times exponential minus 2x squared, right?
So through the integration, okay,
x times exponential -2x squared times w,
which is the integral of 2x exponential
minus 2x squared dx, right, okay?
Can you see it right then?
Okay, here, I'm writing it.
x times e to the negative 2x squared times w,
this is equal to interval of 2xe to the -2x squared and
dx, and + some arbitrary constant c, right.
okay?
What is the anti-derivative of this one?
Right, okay?
That is equal to, right, e to the negative 2x squared,
And negative one-half, right?
Okay, let me check it, derivative of it,
Okay, + c, so divided through by x times exponential,
x times e to the negative 2x squared, okay?
You are going to get, okay?
w = 1 over x times (c times e to the 2x squared- one-half), right?
So that means, what is w?
w = u to the negative 1, right?
So u =, okay, okay, x over c times
e to the 2x squared- one-half.
And what is u?
U is equal to, what is y?
y = x + u, right?
So that y = x plus u, x plus, This,
that is the same as 2x over, okay?
2c times e to the 2x squared- 1.
But we can replace 2c by just c, because when c is an arbitrary constant,
2c is another arbitrary constant, right?
So I can use the c down there, okay?
So that's the general solution.
