6-8 Homogeneous linear equations – Example

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來自 Korea Advanced Institute of Science and Technology 的課程

Introduction to Ordinary Differential Equations

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Korea Advanced Institute of Science and Technology

Introduction to Ordinary Differential Equations

126 個評分

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.

從本節課中

Linear Second order equations 2

6-8 Homogeneous linear equations – Example4:57

6-9 Higher order equations6:51

與講師見面

Kwon, Kil Hyun
Professor
Department of Mathematical Sciences

Let us confirm what we have introduced so far through examples.

Let us solve the following.

The constant coefficient second order homogeneous equation.

First, Y squared minus 4Y is equal to zero.

You can read immediately,

it's characteristic equation is R squared minus four,

factorize it, then you are going to get two distinctive real roots,

negative two and plus two.

So that E to the negative 2X and E to the 2X,

they are two linearly independent solution of this differential equation.

So their general solution is Y is equal to C1 times E to

the negative 2X plus C2 times E to the 2X.

Sometimes,

rather than expressing the general solution in this form,

it might, the Y use for some times to use the following two functions.

Add this two linear independent functions and divide it by two,

you get so-called the hyperbolic cosine of two of X.

On the other hand,

subtract these two functions and divide it by two,

you will get so-called the hyperbolic sine of two of X.

By the superposition principle,

hyperbolic cosine 2X, hyperbolic sine 2X,

they are all also two linearly independent solutions.

So, we have another choice of the general solution of the given problem.

Say Y is equal to C1,

hyperbolic cosine of two of X and the Cs of two hyperbolic sine of two of X.

Here, we have some typo,

I need not cosine hyperbolic of X,

but I need the hyperbolic cosine two of X and the hyperbolic sine of two of X.

Second problem,

Y double prime plus 4Y plus four is equal to zero.

It's corresponding characteristic equation will be R squared plus four plus four,

factorize it, you will get R plus two square that is equal to zero.

That means we have only one double root,

say R1 is equal to negative two. That means what?

E to the negative 2X is one solution.

Another solution which is linearly independent from E to the negative 2X,

we can obtain through the reduction of order.

And we will get in fact,

the second linearity independent solution will be X times E to the negative 2X.

So, general solution will be the linear combination.

In other words, Y is equal to E to the negative 2X times C1 plus C2 of X.

Finally, lets consider one more example.

Say Y double prime minus 2Y prime plus two is equal to zero.

Again, we can read out this characteristic equation very simply,

R squared minus 2R plus two is equal to zero.

Solve this quadratic equation.

You will get two complex conjugate to root this.

Say, one plus I and one minus I.

So here you have your two solutions,

R is equal to one plus or minus I.

And that means E to the X and from this part here,

we have cosine X or sine of X.

From this we are in the imaginary parts,

so that we have a general solution will be Y is equal to the E to

the X times C1 cosine X plus C2 sine of X.

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