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

Determine the general solution to the given differential...

Key Concepts

Characteristic Equation

For linear differential equations with constant coefficients, the characteristic equation is a polynomial obtained by substituting an exponential trial solution into the homogeneous equation. The roots of this polynomial dictate the nature of the homogeneous solution, indicating whether the solution components are real, repeated, or complex, and thereby guiding the appropriate construction of the general solution.

Annihilator Method

The annihilator method is an approach to systematically determine a trial solution for the particular part of a non-homogeneous differential equation. It does so by applying a differential operator (annihilator) that transforms the non-homogeneous term to zero. This allows one to form an extended homogeneous equation whose solution space contains the particular solution, thereby facilitating its determination.

Homogeneous and Particular Solutions

A key strategy for solving linear differential equations is to decompose the solution into two parts: one that satisfies the homogeneous equation and one that satisfies the non-homogeneous equation. The homogeneous solution is typically found by solving the characteristic equation, whereas the particular solution is derived by a technique that fits the form of the non-homogeneous term without overlapping with the homogeneous solution.

Method of Undetermined Coefficients

This method involves assuming a trial solution form for the particular solution based on the type of non-homogeneous term present (such as polynomials, exponentials, or trigonometric functions). The undetermined coefficients in this assumed solution are then calculated by substituting the trial solution into the differential equation and equating coefficients.

General Solution of Differential Equations

In the context of linear differential equations, the general solution is the sum of the complementary (or homogeneous) solution and a particular solution. The complementary solution solves the homogeneous version of the differential equation (where the right-hand side is zero), while the particular solution addresses the non-homogeneous part. This principle underscores the linearity and superposition properties of such equations.

Transcript

00:01 To start this problem, let's rewrite this in differential operator form.

00:05 So it's going to look like this to d squared minus 5d minus 6.

00:10 Y is equal to 4x squared.

00:13 We're going to find the corresponding solution to the homogeneous equation, which means we're going to first find the complementary solution, which means setting the right -hand side equal to 0.

00:25 Now this has the corresponding auxiliary equation, r -q plus 2r squared minus 5r minus 6.

00:36 We set that equal to 0.

00:39 So to factor that, or to find the roots, i'm going to attempt to factor this by using synthetic division.

00:45 According to the rational root theorem, since this is 1 here, we can test the factors of 6.

00:50 So it's going to be plus or minus 6 or plus or minus 2 or plus or minus 3.

00:53 So i'm going to try one first.

00:56 So if i do one, then multiply this one by one, then add these two.

01:04 So then i get three.

01:05 Then now multiply this here.

01:07 So we get three.

01:09 Then now add these two, and that becomes negative two.

01:13 So multiply this, and we get negative two.

01:17 That becomes negative eight.

01:18 Since this last one is not zero, then that means one is not a factor.

01:24 So let's try two next.

01:28 So 2, 2 here, that becomes 4, this becomes 8.

01:33 So we get 3, and then 6, 0.

01:36 Great, so that means 2, 2 is a factor of this polynomial here.

01:42 So we can factor r minus 2, and then the remainder, or these are going to be the coefficients of the remainder.

01:49 So r squared plus 4r plus 3 is equal to 0.

01:53 We can try to factor this some more.

01:55 Even.

01:57 So for example, we'll take a look at the factors of three and see if they can add up to four.

02:03 So for example, three gives three plus one, or the factors of three are just one and three and one plus three is equal to four.

02:11 So we can factor this even more to r minus two and then r plus one and r plus three.

02:18 This is equal to zero.

02:20 Now, this here, we can get the roots directly, is 2, negative 1, negative 3.

02:30 All right.

02:31 So then our complementary solution, y of c of x, is equal to c1e to 2x, plus c2e to the negative x, plus c3e to the negative 3x.

02:46 Okay.

02:47 Now we need to find the annihilator so we can take care of this right hand side here.

02:53 So f of x is equal to 4x squared.

02:59 We can say that there is going to be an e to the zero out front.

03:03 So that means we have an a is equal to zero.

03:07 So our solution, when we have a real root like that, it's just going to be d minus a.

03:15 So d minus zero.

03:17 Then also, this x squared here tells us that we have a multiplicity of three.

03:24 So we're going to take this cube is going to be our annihilator.

03:30 Now we apply this annihilator to our original function.

03:34 So let's apply it to this factored form here.

03:39 So we're going to have d cubed of d minus 2, d plus 1, and d plus 3, y is equal to 0.

03:49 This is going to have roots r is equal to 0, 2, 1, and negative 3.

03:56 And then this 0 is going to have multiplicity 3.

04:01 So our general solution is going to take the form.

04:05 Now we start with our complementary solution, c1e to the 2x, 2x, 2x plus c2e to the negative x, plus c3e to the negative 3x.

04:19 And then now we're going to do our 0 multiplicity 3.

04:22 So that's going to be plus an a0 plus a1x.

04:28 And then plus a2x squared.

04:33 So this part up here, or this last part here, is our trial solution, yp of x.

04:39 So now we're going to plug this in, we'll plug this into our original differential equation, which is this one here, and set it equal to this.

04:49 So remember our original differential equation took the form, so d minus 2, d plus 1, d minus 3, and then we're plugging in a0 plus a1x plus a2x squared.

05:07 And we're setting that equal to 4x squared like so.

05:12 Okay, sorry, d plus 3.

05:15 Okay.

05:16 Let me make sure that's correct.

05:19 Right.

05:20 Okay.

05:22 So now, all right, let's apply this first operator here.

05:27 So just bring down to d minus 2, d plus 1.

05:30 Then we take the derivative of that and then add 3 plus that.

05:34 So it becomes a not.

05:37 Let's see here.

05:39 Okay, a1 of x and then plus 2, a2x squared.