Y^''-4 y^'+5 y=e^5 t+t sin3 t-cos3 t

Name: Problem 35, Chapter 4: Fundamentals of Differential Equations
Uploaded: 2019-10-25T02:51:25-08:00
Duration: 7 min 13 s
Channel: Brian Ketelobeter
Description: y^''-4 y^'+5 y=e^5 t+t sin3 t-cos3 t

Key Concepts

Superposition Principle

This principle states that the general solution of a linear differential equation is the sum of the general solution of the homogeneous equation and any particular solution of the non-homogeneous equation. It is fundamental in solving linear differential equations as it allows one to deal with complicated forcing terms by addressing simpler components separately.

Method of Undetermined Coefficients

This method is used to find a particular solution for non-homogeneous linear differential equations when the non-homogeneous term is of a specific type (such as exponentials, polynomials, sines, and cosines, or their products). The form of the trial solution is chosen based on the nature of the forcing term, and coefficients are determined by substituting the trial solution into the differential equation.

Non-Homogeneous Differential Equations

These involve a non-zero term on the right-hand side of the equation (often called the forcing term), which makes the equation non-homogeneous. The overall solution is typically found by adding a particular solution that accounts for the non-homogeneous part to the general solution of the corresponding homogeneous equation.

Linear Differential Equations

These are differential equations in which the dependent variable and all its derivatives appear linearly. The coefficients of the variable terms may be constants, and the superposition principle applies, allowing the solution to be expressed as the sum of a homogeneous (complementary) solution and a particular solution.

Characteristic Equation

For linear differential equations with constant coefficients, the characteristic equation is obtained by assuming a solution of the form e^(rt) and substituting it into the homogeneous part of the equation. The roots of this polynomial (which can be real or complex) determine the form of the homogeneous solution.

Transcript

00:02 Good day, ladies and gentlemen.

00:05 Today we're considering problem number of 35 in section 4 .5, which asked us to find the form, i guess the form of the particular solution to the given equation, and of course it was not supposed to solve it.

00:23 The first step that we want to do then is consider the homogeneous equation.

00:32 And we can find the homogeneous solution as this form.

00:39 And if you don't remember how to solve that, please just go back to earlier sections.

00:48 And we've covered this before.

00:51 But the next step we want to do is break.

00:56 So this is the right hand side here.

00:59 Break the right -hand side into what the book calls non -homogeneities, which in particular are functions, i guess, that you can solve using the method of undetermined coefficients.

01:25 And it turns out in this case that we have three of them and with these three the key observation here is that none of these have a relationship to the relationship to the roots of the auxiliary equation in particular then see the root to the auxiliary equation here was who plus i and in each of these cases the auxiliary equation that would give this one this as a root would be five this one would be three i and this one would be three i and this one would be three i so so you can see that the so that since none of these have have any relationship to 2 plus i, we can just apply the method of 6 .4 to find particular solutions to each of these.

02:50 And this is just straightforward that we use the these forms, because this is actually what the method of undetermined coefficients basically tells us what to apply.

03:15 And you'll notice that we can, in fact, just combine these, this y2 and y.

03:21 One because the constant d1 and b one, really that's just repeating the same thing.

03:30 You're just adding constants.

03:32 So there's, there is, you don't really need to have this second one there, but for completeness, i decided to conclude that.

03:44 And then we apply the superposition principle, and we see that the yp, p, here is just the sum of the three of those, and that looks like this.

04:02 And i mean, it's a superposition principle because the function here, f is just the sum of the three, this is f1, of course, f2 and f3...

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