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.
6-7 Homogeneous linear equations with constant coefficients
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來自 Korea Advanced Institute of Science and Technology 的課程
Introduction to Ordinary Differential Equations
58 個評分
Korea Advanced Institute of Science and Technology
58 個評分
從本節課中
Linear Second order equations 2
與講師見面
Kwon, Kil HyunProfessor
Department of Mathematical Sciences
Yeah, we now handle the problem of finding a general solution for
homogeneous linear differential equation,
okay, with constant coefficients, okay?
With constant coefficients.
To make things a lot simple,
we restrict our service to the case of the order two.
So consider second order homogeneous linear
equation with constant coefficients which I write
it as Ay double prime + by prime + c is equal to 0.
Where the a is a non-zero constant and b and c they are all real constants.
So the problem we are concerned for the time being is the constant
coefficients second order homogeneous differential equation.
A y double prime + b y prime + c is equal to 0.
By inspection, we may guess that that differential
equation may have a solution of the form of e to the r of x.
What's the special property of this exponential function?
Its derivative is r times e to the rx.
Its second derivative will be r squared times e to the r(x), right?
And so on, okay?
So let us say y is equal to e to the rx in the differential equation 1.
Here is the differential equation one.
ay double prime + b prime + cy = 0,
a is a non-zero constant, and b,
c are arbitrary real constants, right?
In this equation, set y is equal to e to the rx, right?
Set y is equal to e to the rx.
Then what you're going to get,
Is derivative is r times e to the rx,
second derivative will be r squared times e to the r of x.
So, in all those y or y prime or yw prime they always
contain the term e to the rx as a common factor.
So take this common factor out, e to the rx,
then the other parts will be ar squared + br + c.
And that must be equal to zero if this is the solution to that
differential equation.
That means what?
For each of the rx to be a solution of this second order,
constant of coefficient, differential equation,
r must be here, root of that quadratic equation, okay?
ar squared + br + c = 0, okay?
This quadratic equation we call it as the characteristic equation or
the auxiliary equation corresponding to the differential equation 1, okay?
So, in a sense solving this differential equation is reduced
to two algebraic problems, solving the root of this quadratic equation,
say the characteristic equation of the differential equation, okay?
As you know, depending on the choice of the constant a, b,
and c the type of roots of this quadratic equation will be different.
And we know that the roots of that quadratic equation,
let me write it here, characteristic equation.
Say ar squared + br + c = 0.
Try to solve it then.
We know that the root of this equation will be r
is equal to 1 over 2a times negative b plus or
minus square root of b squared minus 4ac.
And their nature depends on the sign of the so-called discriminant.
The quantity of this is square root the same, so for
discriminant of the quadratic equation, b squared minus 4ac and
we know where the characteristic equation 2.
This characteristic equation has two distinct real roots,
r1 and r2 when b squared minus 4ac is strictly positive.
It is one double real root,
say r1 is equal to r2 when the b squared minus 4ac = 0,
and it has a two complex conjugate roots say,
alpha plus minus beta, where alpha and beta are real constants.
In particular we can choose beta to be a positive constant, okay?
When b squared minus 4ac is a negative, right?
So we have three distinct cases for
the roots of the characteristic equation.
So let's consider those three cases separately.
First, the e distinct case.
When we have two distinct real roots.
When this characteristic equation has two distinct real roots r1 not equal to r2.
Then e to the r1x is the solution, right?
And er2x is also a solution and
because r1 is not the same as r2,
you can compute distance functions,
and you can come from the Ruskin is never zero.
So then there is e to the r1 x and the e to the r2 of x,
they form also called fundamental set of solutions of differential equation.
And that means the general solution of
differential equation will be y = c1 times
e to the r1 of x and plus c2e times r2x.
That's the easiest case.
The second case, we have only one real double
root because b squared minus 4ac = 0.
So when this characteristic equation two has only one double root,
r1 = r2, then number must be through the roots formula, okay?
That number must be minus b/2a, right?
And we know then y1 = e to the r1 of x is a solution, right?
And we need to know another solution which is linearly
independent from this y1, okay?
The second solution we can obtain by the reduction of order,
say y2, which is of the form y1 times some another
function, g(x), which is not a constant.
Through the necessary computation in this case then,
we know that y sub 2 is equal to e to the r1x times g(x),
and the g(x) is given by the formula.
And because r1 is equal negative b over 2a,
plugging that onto this form and computing this intergration,
then we are going to get, this will be equal to x, right?
With the choice of suitable constant, so then it is what?
y2, which is x times e to the r1 of x,
which is linearly independent of y1 and
also satisfies the differential equation.
Because y1 and the y2 are linearly independent,
our general solution of differential equation will
be e to the arrow and x times c + 1 + c2 where c1 and
c2 are two of the equal constants, right?
Finally the case three, so
case three when the discriminate b case minus 4ac is negative.
We have two complex conjugates root sub categoristic equation.
So the two roots are r1 = alpha + i beta and
r2 = alpha minus i beta, where the alpha is real and
the beta is some positive constant, right?
Then y sub 1, which is e to the alpha + i beta x, and
the y sub 2, e to the alpha minus i beta times x.
They're two linearly independent but
complex-valued solutions of the differential equation.
But we prefer to have real-valued solutions,
because our original differential equation is a real coefficient,
real constant coefficient, second order homogenous differential equation.
Let me remind you here, the differential equation
one is ay double prime + by prime + cy = 0.
That's a differential equation.
Where the a is any non-zero real number, b and c are the real numbers, right.
So starting from this real constant coefficients,
second differential equation,
we prefer to have a real-valued solution.
But we get two linearly independent solution which is a complex valued.
So let's take a real and imaginary part of the solution y sub 1.
Say because this is a complex value of something it's a real part.
Real part of y sub 1 is equal to y1 + complex conjugate of y1 over 2.
That gives us the real part of it.
But in fact complex conjugator y1, that is equal to exactly y2.
You can confirm it by looking at this y sub 1 and the y sub 2.
Y sub 2 is a complex conjugate of y sub 1.
And what is it?
Using the Euler's formula, if you expanded this expression out,
then you can extract just the real part and the imaginary part.
Just the real part will be e to the alpha x times cosine beta of x.
By the similar token, this imaginary part of y sub 1,
which is y sub 1 minus complex conjugate y sub 1 over 2i,
which is the same as y sub 1 minus y sub 2 over 2i.
That is, in fact, e to the r of x times the sin beta of x, right?
Look at this expression in the middle, okay?
Linear combinations.
Real part of y1 and imaginary part of y1
are linear combination of two solutions y1 and
y2 linear combinations.
Then by the superposition principle for the homogeneous differential equation,
because both the y1 and the y2 are solutions of this differential equation.
Their linear combination, in fact which is a real part of y sub 1,
is also a solution of the same differential equation.
And the same is true for the imaginary parts.
In other words, e to the alpha x cosine beta of x.
And e to the alpha x times sine beta of x.
They're all solutions of this given second order
homogeneous differential equation, okay?
By the superposition principle.
Moreover, they're real valued, both functions
are real valued now and they are linearly independent.
You can check it simply by computing first kin of these two functions.
So we know have a what, two real valued linearly
independent solutions of the given differential equation.
And then this is general solution,
will be just a linear combination of this and that,
which we can write it as, because e to the alpha x is always common.
So y is equal to e to the alpha x times c1 cosine
beta of x + c2 sine beta of x, right?
This is the general solution of the given equation on n,
the b squared minus 4ac is strictly negative, right?
