Determine the general solution to the given differential equation. Derive your trial solution us

Determine the general solution to the given differential...

Key Concepts

Linear Differential Equations with Constant Coefficients

This concept involves differential equations where the coefficients of the derivatives are constants. Such equations are typically solved by assuming solutions of an exponential form, which leads to the formulation of a characteristic equation. The structure of the homogeneous solution is based on the roots of this characteristic equation, and the solution to the entire problem is expressed as the sum of the homogeneous (complementary) solution and a particular solution corresponding to the non-homogeneous part.

Characteristic Equation

The characteristic equation is formed by replacing each derivative in the linear differential equation with a power of a variable (commonly r), transforming the differential equation into a polynomial equation. The roots of this polynomial determine the form of the homogeneous solution. In cases with repeated roots, the solution involves additional multiplicative factors (usually polynomials in x) to account for the multiplicity, ensuring linear independence among the solution components.

Homogeneous and Particular Solutions

For a linear non-homogeneous differential equation, the general solution is the sum of the homogeneous solution, which solves the equation with zero on the right-hand side, and a particular solution, which is a specific solution that satisfies the entire equation. Identifying both components is crucial for capturing the full behavior of the system described by the differential equation.

Method of Annihilators

The annihilator method is a systematic approach for finding a particular solution to non-homogeneous linear differential equations. It involves applying a differential operator (annihilator) to both sides of the equation that 'annihilates' or nullifies the non-homogeneous terms. This converts the equation into a higher-order homogeneous equation, which can be solved using the standard method of finding the roots of the characteristic equation.

Method of Undetermined Coefficients and Trial Solution

The method of undetermined coefficients is used to guess the form of the particular solution based on the non-homogeneous term of the differential equation. When the non-homogeneous term is similar to a term in the homogeneous solution (a case of resonance), the trial solution must be adjusted by multiplying by an appropriate power of x to ensure the terms are linearly independent. This is essential when using the annihilator technique, as the operator used must address the presence of repeated or resonant factors.

Transcript

00:01 To start this problem, i'm going to first rewrite this in differential operator form, so that's d -cube plus 3d squared, plus 3d plus 1.

00:08 Y is equal to 2e to the negative x plus 3e to the 2x.

00:14 Since the coefficients here are 1 -331, i can immediately tell that this is going to be a polynomial to cube.

00:24 These are the coefficients of the binomial expansion for n equals 3.

00:31 So that's going to be equal to d plus 1 cubed of y is equal to 2e to the negative x plus 3e to the 2x.

00:43 So now i'm going to solve the corresponding complementary or find the corresponding complementary function.

00:51 So that's going to be d plus 1 cube, y, c is equal to 0.

00:56 That's just setting the right hand side here equal to 0.

00:58 Now the corresponding auxiliary equation, p of r, is going to be equal to r plus 1 cubed.

01:06 We set that equal to 0.

01:07 So r is going to be negative 1 with a multiplicity of 3.

01:14 So our general solution, or not general, but our complementary solution, y of c, is going to be equal to c1, e to the negative x, and then plus c2x, e to the negative x, and then plus c3x squared e to the negative x.

01:34 Okay.

01:34 Now, we need to get rid of this right -hand side here by finding its annihilator.

01:39 So first we have f of x is equal to 2e to the negative x plus 3e to the 2x.

01:47 For this part of the solution, we have that r is equal to negative 1.

01:51 For here, we have r is equal to 2.

01:52 So our annihilator 1, a of d, is going to be d minus r.

01:59 So that's going to be plus 1.

02:01 Then we have a second annihilator, a2 of d, is going to be d minus 2.

02:07 So our annihilator in total is going to be d plus 1 and then d minus 2.

02:14 So then we can multiply or apply this operator to our original.

02:19 So we'll actually get, and i'm going to rewrite this, i'm going to switch this, so that we get d minus 2, d plus 1, and then d plus 1 cubed, y is going to be 0, since this annihilates this solution here.

02:37 Okay, now we can rewrite this as d minus 2, and then d plus 1 to the 4th, y is equal to 0.

02:45 So we get an extra r is equal to 2, and then a negative 1 with multiplicity 4.

02:50 So the complementary solution already has the negative 1 multiplicity 3.

02:56 So now our general solution will have the form.

03:02 So we write down our complementary solution first, x, e to the negative x, plus c3, x squared, e to the negative x.

03:11 Then i'm going to write down our a plus a0 and then x cubed e to the negative x.

03:21 So this is for the multiplicity 4 of our solution.

03:25 And then we have also a plus, then a1, e to the 2x, like so.

03:36 So now we need to plug in our trial solution.

03:41 This is our trial solution, y of p here, of x, y p of x.

03:50 So then now we need to plug this into our original equation here.

03:57 Sorry, this one here.

04:00 Okay.

04:02 So that is going to be, so a d plus 1 cubed times, then we have a0x cubed, e to the negative x plus a1, and then we have an, sorry, e to the 2x, like so.

04:28 Let me go back and make sure that's correct.

04:31 Okay.

04:31 And then now we do is equal to 2e to the negative x plus 3e to the 2x.

04:44 Okay.

04:46 So we need to apply d plus 1.

04:48 D plus 1 first to this.

04:50 And because we can write this out as d plus 1 times d plus 1.

04:55 So we're going to take the derivatives one at a time.

05:00 So we'll have, okay, we'll apply 1 first.

05:02 So that means we're taking the derivative of this.

05:06 So to do so, we can do this.

05:08 The derivative of this times this...