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

Determine the general solution to the given differential...

Key Concepts

Annihilator Technique

The annihilator technique is a method used to determine a trial solution for non-homogeneous linear differential equations by applying an operator that 'annihilates' or reduces the non-homogeneous term to zero. By finding an operator that nullifies the forcing function, the non-homogeneous differential equation can be transformed into a higher-order homogeneous one, simplifying the process of finding a particular solution.

Differential Operator Notation

Differential operators, such as D representing differentiation, provide a compact way to express differential equations. This operator notation enables algebraic manipulations of the equations, including factorization and application of techniques like the annihilator method.

Homogeneous and Particular Solutions

Any non-homogeneous linear differential equation can be solved by finding two parts: the complementary (or homogeneous) solution that solves the corresponding homogeneous equation, and the particular solution that accounts for the non-homogeneous part. The general solution to the differential equation is the sum of these two parts.

Linear Differential Equations

Linear differential equations are equations involving an unknown function and its derivatives, where the function and its derivatives appear to the power of one and are not multiplied together. The linearity property allows for the superposition of solutions, making it possible to express the overall solution as the sum of the homogeneous solution and a particular solution.

Transcript

00:01 Okay, so to find the general solution, first let's find the complementary solution by solving the corresponding homogeneous equation, which means we set this first right -hand side equal to zero.

00:13 So then the corresponding auxiliary equation, p of r, to do r times r plus three, r plus three here, this is equal to.

00:24 Now we have r plus three is going to be equal to zero.

00:32 So we have r is equal to 0, and then r is equal to negative 3.

00:38 So the corresponding complementary function or solution yc is going to be the c1, and then this is e to the 0.

00:48 It's just going to be 1, so it's just going to be c1.

00:51 Then we have plus c2, e to the negative 3x, like so.

00:58 Now we look at the right hand side here, f of x, and we need to find the annihilators of this function.

01:04 So that's going to be multiplied out, or is equal to 5x, plus x e to the x.

01:13 So here we have an, this is an r is equal to 1, and here we have an r is equal to 0.

01:23 R is equal to 0.

01:25 We can think of it as being multiplied by e to the 0x.

01:31 So then our first annihilator, a1 for this part here, it's just going to be equal to, it's going to be equal to d and then minus r, which is zero.

01:44 And then this x tells us we have multiplicity two.

01:50 Multiplicity two.

01:52 So we're going to take d squared.

01:54 Now a2, we're going to start with d minus r, so d minus 1.

02:00 And again, this x tells us we have multiplicity two.

02:04 So we're going to take that and we're going to square it.

02:09 So then now we multiply our original equation by a of d.

02:15 So our annihilator first, a of d is equal to d squared times d minus 1 squared.

02:25 So we're going to have that when we apply this here, we get d squared, d minus 1 squared, d minus 1 squared, d, d plus 3, 1.

02:40 And then is equal to this gets annihilated by this here.

02:47 So that's just equal to zero on this side.

02:50 This has a corresponding polynomial equation of, well, this d squared and d cube combined together to become r cubed, and then we have d minus 1 squared, and then d plus 3.

03:04 So our general solution, sorry, this has roots 0 multiplicity 3, and then 1, multiplicity 2, and then negative 3.

03:19 So our general solution, y of x, is going to be first our complementary solution, c1, plus c1 plus c2e to the negative 3x.

03:35 So that takes care of this part, and then one of these multiplicities are equal 0.

03:40 So we still need to do two more multiplicities of 0.

03:43 So we're going to have plus, plus, and then instead of c's, now we're going to start with a.

03:51 So we'll do a, not x first, plus a1, x squared.

03:57 So these three parts here correspond to our zero multiplicity three.

04:02 Now we just need our one multiplicity two.

04:04 So that's going to be plus a2, e to the 2x, and then our plus a3x, x, e to the 2x, and then our plus a3, x, to the 2 or sorry, e to the x.

04:17 Sorry, it should be 1 with multiplicity 2.

04:21 E to the x, e to the x, e to the x like so.

04:26 So now this part here, this is our trial solution.

04:33 So now we're going to plug this into our differential equation here.

04:39 So if we do that, we have d times d plus 3 of, and then we have a, not x plus a 1 x squared and then plus a 2 e to the x and then plus a 2 e to the x and then plus a 3 x e to the x and this is equal to 5x plus x e to the x x e to the x.

05:29 Now let's rearrange this.

05:33 Let's do the d first...

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