5-2 Superposition Principle

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Del curso dictado por Korea Advanced Institute of Science and Technology

Introduction to Ordinary Differential Equations

63 calificaciones

Korea Advanced Institute of Science and Technology

Introduction to Ordinary Differential Equations

63 calificaciones

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.

De la lección

LINEAR SECOND ORDER EQUATIONS 1

5-1 Linear differential equations7:29

5-2 Superposition Principle7:13

Conoce a los instructores

Kwon, Kil Hyun
Professor
Department of Mathematical Sciences

Okay, here is the fourth theorem, concerning to

the linear homogeneous differential equation, okay?

Which is so called the superposition principle for homogeneous equation.

Homogenous means what?

The right hand side b(x) = 0, like this one, 0.

Then we call this linear differential equation is homogeneous, okay?

So, consider the [INAUDIBLE] order homogeneous differential equation say (4),

and we assume that there are finitely many solution of this

linear homogeneous differential equation, okay?

From y sub 1 to y sub k, okay?

Then our claim is the arbitrarily linear combination, okay?

Sum of c sub i times y sub i, where c sub i is arbitrary constant.

Is also solution of the same differential equation, okay?

That claim I recorded as a superposition principle for

homogeneous differential equation, okay?

Its proof is very simple, by the linearity 3 way, okay.

What is the linearity?

If you apply such a finite linear combination of functions,

and apply the linear operator L, then we have this

is the same as sum of c sub i times L[y sub i].

We were assuming at the beginning that y sub i,

they're all solutions of l over y is equal to 0, right?

So this quantity is a 0, right?

L of yi = 0 times any number is a 0?

Some of them will be equal to 0, right?

That's the end of the proof, right?

Okay, so very simple observation, okay?

For example, consider this,

Initial value problem for variable coefficient homogenous

second order ordinary differential equation.

x squared time y double prime- x times y prime + y = 0.

And satisfying two initial conditions,

say y(1) = 1 and y prime of 1 = -1, okay?

After confirming that t sub 1 = x and

y sub 2 = x times the log of x.

Both are solutions of this differential equation

on the interval from zero to infinity, okay?

Showing that both with y sub 1 and y sub 2 down there,

we'll solve these solutions of these equation.

That's an easy computation, I'll leave it to you, okay?

Then what we can say by the superposition principle that I've just introduced,

any of the linear combination, say,

y = c sub 1 times x + c sub 2 times x times log of x,

is also a solution of the same differential equation, okay?

Any of this functions for arbitrary two constants c sub 1 and

the c sub, they satisfy this a second order homogeneous

differential equation part of a super position principle.

Now for this family of solution,

apply the two given boundary condition

y(1) = 1 and y prime of 1 = -1, right?

So first 1 = y(1) when x = 1, plug in 1 into

this equation then, you have a c sub 1 only, right?

Because log of 1 = 0, okay?

That already means the c sub 1 must be = 1, okay?

Now look at the initial condition, y prime of 1 = -1 so

you should have -1 = y prime(1), okay?

From this y, what is the derivative?

So let's compute it here okay?

So now here we have y = c sub 1

times x + c sub 2 times x logx.

We need to revert this y'= c1 + c2

logx + c2 times x times derivative of

log of x, that is 1(x), right?

Then plug in x = 1, plugging from here,

compute it when x = 1, okay?

How much is this one?

Log 1 = 0 so, you get = c1 + 0 + when x is equal to 1, you get c2, right?

So we get this equation, -1 = y prime of 1 and

that should be = c1 + c2 and

we already know that c1 = 1.

So that means that c2 = -2, right?

So when c1 = 1 and the c2 = -2,

you get function x- 2x times log of x, okay?

It must satisfy both differential and

these two initial conditions, right?

So this is a solution, right?

This is a solution to the given initial variable problem.

In fact this function is the only one possible solution to

the given initial very problem as the next theorem shows, okay?

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