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.
8-1 Variation of parameters I
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來自 Korea Advanced Institute of Science and Technology 的課程
Introduction to Ordinary Differential Equations
63 個評分
Korea Advanced Institute of Science and Technology
63 個評分
從本節課中
Linear Second order equations 4
與講師見面
Kwon, Kil HyunProfessor
Department of Mathematical Sciences
As a last two topic in chapter four,
I'd like to introduce one more very strong way of
obtaining a particular solution for the linear non-homogeneous differential equation.
Though the method of undetermined coefficients is rather simple,
but it has two strong constraints.
Consider differential equation L|y| = g(x).
Non-homogeneous one, okay?
L is the linear differential operator.
L is the linear differential operator.
What is L?
Is equal to a_n(x) and D the to the n plus a_n - 1(x),
D to the n-1 and
plus a_1(x) and D and plus a_0(x), right?
That is L, differential operator of order n, okay?
So, consider such a differential equation, non-homogeneous differential equation.
For the method of undetermined coefficients to be applied,
first your differential operator L, okay,
must have only constant of coefficients.
In other words, all those coefficients, a_n, a_n-1, a_1,
and a_0, they must be a constant, okay?
None of them can be a true function of x.
Second constraint is coming
from the right-hand side of that differential equation L|y| = g(x),
g(x) some times we quote it as the input function, okay?
This input function must have a differential polynomial annihilator.
It must have differential
polynomial annihilator.
We know that not every function has differential polynomial annihilator, okay?
For example, can you remind it, okay?
For example, the 1(x) or the sec(x) something like that.
They have no annihilator at all, right?
So, if this right-hand side of this input function g(x) happen to be 1(x),
then the method of undetermined the coefficients cannot be used.
So to overcome such difficulties,
here we introduce the second method,
the method of variation of parameters to obtain
a particular solution of such a non-homogeneous differential equation.
In this case, the differential operator may have
a variable coefficient and this input function g(x) can be arbitrary.
It may or may not have a differential polynomial annihilator.
To make the thing simple,
we consider only second order non-homogeneous differential equation.
So consider non-homogeneous linear second order differential equation with
possibly variable coefficients in its standard form y" + p(x)y' + q(x)y = g(x), right?
Assume that we have
two linear independent solution of corresponding homogeneous problem.
Corresponding homogeneous problem,
trivially that is equal to y" + p(x)y' + q(x)y = 0, right?
This is the corresponding homogeneous problem.
We assume that y_1 and y_2 are
two linear independent solutions of this homogeneous problem.
In other words, we have a complementary solution y_c = c_1. y_1 + c_2.y_2, right?
And the general solution of this reason given
non-homogeneous question will be the general solution of
differential equation one is y
is equal to complementary solution plus particular solution.
We already know what is this complementary solution y_c, right?
So only thing unknown is y_p now, okay?
The main idea of the method of variation
of parameters is to assume that there is
a particular solution of
this non-homogeneous problem in the form of y_p = u_1.y_1 + u_2.y_2, okay?
Where the u_1(x) and
the u_2(x) are two unknown functions to be determined, okay?
So that y_p becomes a particular solution of this original non-homogeneous problem.
Here in our guess for y_p, we've two unknowns the u_1 and u_2, okay?
So in order to determine two unknown functions,
the u_1 and the u_2,
we need two conditions.
Our guess is, you are looking for a particular solution of y_sub_p of
the given differential equation in the form of
y_p is equal to u_1 times y_1 plus, u_2 times y_2.
Since we have two unknowns,
as I said, u_1 and u_2,
we need two conditions to determine u_1 and u_2.
One condition trivially comes from
the condition that y_p is a particular solution of the given differential equation.
In other words, y_p double prime plus py_p prime,
plus qy_p is equal to g. Using this expression,
this equation will give us one equation for two unknowns,
u_1 and u_2, trivially.
We need one more equation.
That another equation, we will take it so that our next computation is to be easy.
Differentiating two.
Two is this.
This is the equation two.
Differentiating this y_p, you will get derivative of y_p will
be u_1 y_1 prime plus u_2 y_2 prime,
plus u_1 prime y_1,
and plus u_2 prime y_2.
It's easy by the product rule of differentiation.
From this expression, we require the second part to be equal to zero.
Say, u_1 prime y_1,
plus u_2 prime y_2,
to be equal to 0.
We simply require it without a proper reason at this moment.
Then, y_p prime will be substituted to prime 1.
In other words, u_1 y_1 prime plus u_2 y_2 prime.
So now, we have right away, from y_p prime,
this is equal to u_1 y_1 prime plus u_2 y_2 prime,
this is of course to plot,
plus u_1 prime y_1,
plus u_2 prime y_2.
We require this is equal to zero.
We simply require this is equal to zero.
Now what does that mean then?
That means, y_p is equal to only this part.
So, is a second derivative will be,
u_1 y_1 double prime,
u_2 y _2 double prime and plus u_1 prime y_1 prime,
plus u_2 prime y_2 prime,
by the product rule again.
So using this y_p,
and y_p prime, and y_p double prime,
make the left hand side of the differential equations, say,
y_p double prime plus p of x times
y_p prime plus true x of y
over p. Here's the computation,
y_p double prime plus p_y p prime,
plus q_y p. If we reorganize this sum of these three times then,
you are going to get u_1 times y_1 double prime plus p_y1 prime, plus q_y1,
plus u_2 times y_2 double prime, plus p_y2 prime,
plus q_y2, plus u_1 prime y_1 prime,
plus u_2 prime and y_2 prime.
Let's pay attention to the quantities in this bracket,
and the quantities in this bracket.
What can you say about these two things?
What I mean is y_1 double prime plus p_y1 prime plus
q_y1 and y_2 double prime plus p_y2 prime plus q of y_2.
What can you say about these two?
Can you remind what are those y_1 and y_2?
I said,
y_1 and y_2,
this is two linearly independent solutions
to corresponding homogeneous problem
y double prime plus p_x y prime,
plus true x y equal to zero.
In other words, y_1 double prime plus p_y1 prime, plus q_y1,
or y_2 double prime plus p_y2 prime plus q_y2,
both of them must be equal to 0,
because y_1 and y_2 are solutions of this corresponding homogeneous problem.
What does that mean?
The first bracket is equal to zero.
The second bracket, that is equal to zero.
It then means that in fact,
this term, this part,
and that part, they disappear.
So, we will get only u_1 prime times y_1 prime,
plus u_2 prime times y_2 prime because y_p,
this is a particular solution of our original problem.
Some of these three terms must be equal to g. Can you see why?
We have one more condition right here,
an unknown, the u_1 and the u_2.
We have a one on other such a condition, say,
u_1 prime y_1 plus u_2 prime y_2 is equal to 0.
So let me summarize those two equations here.
