Y^''(t)-(1)/(t) y^'(t)+(5)/(t^2) y(t)=0

Name: Problem 15, Chapter 4: Fundamentals of Differential Equations
Uploaded: 2019-11-13T17:43:53-08:00
Duration: 5 min 30 s
Channel: Brian Ketelobeter
Description: y^''(t)-(1)/(t) y^'(t)+(5)/(t^2) y(t)=0

Key Concepts

Second Order Linear Differential Equations

This concept refers to differential equations involving an unknown function and its derivatives up to the second order, with the property that the equation is linear in the unknown function and its derivatives. Such equations often admit methods for finding complementary and particular solutions based on the superposition principle and are fundamental in modeling physical phenomena like mechanical vibrations or electrical circuits.

Euler-Cauchy (Equidimensional) Equation

Euler-Cauchy equations are a special class of linear differential equations with variable coefficients that are powers of the independent variable. They have the characteristic form where each derivative term is multiplied by a corresponding power of the variable. Their structure allows for solution techniques that involve assuming a solution in the form of a power function, leading to an algebraic characteristic equation.

Power Function Substitution

This technique involves assuming a solution of the form y(t) = t^r for Euler-Cauchy equations. By substituting this trial solution into the differential equation, one obtains an algebraic equation in terms of r—the characteristic equation. Solving this equation provides the exponents that are used to construct the general solution.

Characteristic Equation and Complex Roots

The characteristic equation is derived by substituting the power function trial solution into the original differential equation. When this algebraic equation yields complex roots, the general solution is expressed in terms of trigonometric functions multiplied by a power function of the independent variable. This reflects oscillatory behavior in systems described by the differential equation.

Transcript

00:01 Good day, ladies and gentlemen.

00:05 Today we're looking at problem number 15 from section 4 .7, and it tells us to find the general solution for t less than zero of the given difference equation.

00:20 Well, now there's two different ways you could go about solving this, which is one, you could approach it using the technique.

00:33 Of the book, or sorry, you could go about using the p of t, q of t, and g of t.

00:47 In this case, p of t would be t to the negative first, q of t to the negative second, and g of t would of course be zero.

00:58 And then from there, you could go about solving the differential equation using the material from that way, which, of course, we will be doing very nearly in the near future.

01:14 Or you could have recognized that just by multiplying through by t squared, we get this guy here.

01:25 And now you'll recognize this guy, or maybe you should recognize it as being something a t squared b t plus b t plus c y in other words it is a kosher euler equation.

01:51 And since we're looking for solutions, t less than zero, of course, you know, there's no problem with multiplying through by the t squared because we don't lose or gain solutions to it.

02:06 So it's actually perfectly okay to do it.

02:10 And it is just a koshi euler equation.

02:13 In this case, we get a equals 1, b equals negative 1, and c equals 5.

02:19 And from there, we can plug those numbers into the form of the characteristic equation.

02:26 And we get this guy here, which involves the roots, r equals plus or minus 2i.

02:38 In particular, then i just note that the book uses alpha plus or minus beta i.

02:47 I just use a plus or minus b .i because i don't have a t for an alpha.

02:54 Okay.

02:54 So now we have the roots of the characteristic equation.

03:00 And from there, we know then how to get, we know the general solution...