T^2 y^''(t)+7 t y^'(t)-7 y(t)=0

Name: Problem 9, Chapter 4: Fundamentals of Differential Equations
Uploaded: 2019-11-13T16:57:16-08:00
Duration: 7 min 34 s
Channel: Brian Ketelobeter
Description: t^2 y^''(t)+7 t y^'(t)-7 y(t)=0

Key Concepts

Second-Order Linear Differential Equations

Second-order linear differential equations involve the second derivative of the unknown function and can have variable or constant coefficients. Understanding their structure and solution methods, especially in cases with variable coefficients like in Euler-Cauchy equations, is fundamental in the study of differential equations.

Trial Solution Technique

The method of trial solutions involves substituting an assumed form of the solution, such as y = t^r, into the differential equation. This substitution reduces the differential equation to an algebraic equation in terms of r, simplifying the process of finding the general solution.

Characteristic Equation

For Euler-Cauchy equations, a trial solution of the form y = t^r is assumed, which leads to a polynomial equation known as the characteristic or auxiliary equation. The roots of this polynomial determine the form of the general solution to the differential equation.

Euler-Cauchy Equation

An Euler-Cauchy equation, also known as an equidimensional equation, is a type of linear differential equation in which the coefficients are powers of the independent variable. These equations have the form t^n y^(n) + ... = 0 and are typically solved using methods that exploit the structure of the variable coefficients.

Transcript

00:03 Good day, folks.

00:06 Today we're considering problem numbers nine on section 4 .7, and it is asking us to find the general solution to the given differential equation.

00:22 Now, you'll notice that this is the form of a what we call a koshi euler equation.

00:32 And in this form, we know that we take a guess, if you will, of solutions of the form, y of t equals t to the r.

00:47 Now, when you, when you guess this kind of solution and you put it into the ordinary difference equation here, then you get, equation of this form.

01:03 So right here, so this is really just substituting t to the r and here, and, you know, through multiplication, everything, we're going to go hit this.

01:18 And in particular, then, by factoring out the t to the r, we have this.

01:25 Now, of course, since we, you know, since t, is not going to be identically zero, you know.

01:36 We know that this guy here must be zero.

01:43 So in particular, then, the way i have this written is more in line with the way the book writes.

01:54 That's why i wrote it like this.

01:57 And you get this kind of equation.

02:01 And this here, is as we had before that character what's called the characteristic equation and again i wrote it like this because this in the book the a you know you have the a the form of the equation in the book is b minus a and c here so i just wrote it like this so that to make it clear where the book formulation of that equation comes from.

02:40 And again, this characteristic equation is going to turn up numerous times throughout this book or throughout this section, but it will always end up looking like this.

02:54 And then, you know, like we did earlier, the roots like we saw earlier in earlier sections, we saw that the roots of an equation of the characteristic equation tells us about solutions.

03:13 And in particular, then, when you try to find the roots here, you're going to find that negative 7 and 1 are roots.

03:26 And since we have two real roots here, then the solution we get, is just like this form.

03:35 It's just c1, t to the negative 7 plus c2t, i guess t to the first or just t.

03:42 That is our homogeneous solution to this equation.

03:48 And it just comes down to the fact that it's a koshi -oiler equation.

03:57 Now, i just want to point out, i guess, that this is a form of equation...

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