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 systematic method used to convert a non-homogeneous differential equation into a higher order homogeneous one by applying a differential operator that 'annihilates' the forcing function. This process not only simplifies the solution process but also guides the choice of the trial solution for the particular part by clarifying the structure needed to account for the non-homogeneous terms.

Method of Undetermined Coefficients

This technique involves assuming a form for the particular solution based on the structure of the forcing term. The unknown coefficients are then determined by substituting this assumed solution into the differential equation, ensuring that it satisfies the equation. This method is particularly effective when the non-homogeneous term is a combination of polynomials, exponentials, and trigonometric functions.

Homogeneous (Complementary) Solution and Characteristic Equation

The complementary solution is derived from the associated homogeneous equation by finding the roots of its characteristic equation. When these roots are repeated, the solution basis is modified (typically involving multiplication by x) to ensure linear independence among the solution components, forming the complete set needed for the homogeneous solution.

Non-homogeneous Differential Equations

These are differential equations that include a forcing term (a non-zero function on the right-hand side), which means the solution consists of a homogeneous part (solving the associated homogeneous equation) and a particular solution that accounts for the external input. Understanding this structure is crucial when tackling such problems using methods tailored to each part.

Transcript

00:01 Okay, to solve this problem here, first we're going to rewrite this in differential operator form to simplify things.

00:09 We get 4d plus 4 of y is going to be equal to 5xe to the negative 2x.

00:19 So the corresponding polynomial or auxiliary equation p of r is going to be equal to r squared plus 4r plus 4.

00:32 Set that equal to zero.

00:34 This can actually be factored into a perfect square like so.

00:39 So square root of four is two, and then twice of that is four.

00:45 So we get this.

00:47 So we get roots of negative two with multiplicity two, use of the square term here.

00:55 So we get our complementary solution, sorry, our corresponding homogeneous equation, by the way, is this here, we set equal to 0.

01:10 So y of c of x is going to be c1, e to the negative 2x, and then plus c2x, e to the negative 2x, like so.

01:22 The negative or the x comes from the multiplicity 2.

01:25 Now for the right -hand side, f of x is equal to 5x, e to negative 2x, we need to find the annihilator for this function here.

01:36 So the annihilator for this here, we have an r is equal to negative 2, and then this x tells us that we have a multiplicity of 2.

01:47 So our annihilator, a of d, it's going to be, so we start with d and then minus negative 2, so d plus 2, and then our multiplicity is 2.

01:57 Applying the annihilator to this here, remember this can also be factored into d plus 2 squared.

02:05 So d plus 2 squared and then times d squared plus 4d plus 4 of y is equal to, and then this annihilates this here.

02:20 Now, this can also be factored into d plus 2 squared like so.

02:26 So we'll end up just getting d plus 2 to the 4th, y is equal to 0.

02:33 Our general solution then is going to take the form y of x is equal to so c1e to negative 2x plus c2 x e to the negative 2x.

02:43 We start with our complementary solution.

02:46 And then we need to add our two multiplicities, our two extra multiplicities of negative 2.

02:52 So remember this has the equation r plus 2 to the 4th equals 0.

02:58 So we're going to have r equals negative 2.

03:01 And then we need multiplicity of 4.

03:04 So we're going to add an a0 term.

03:07 So an x squared, e to the negative 2x.

03:10 This is for, this is because of our multiplicity here.

03:15 And then now plus an a1 and then x cubed, e to the negative 2x.

03:21 That's going to be our last multiplicity.

03:22 The rest of this here, this is our triad solution.

03:28 So now we're going to, i'll plug in our trial solution into our original differential equation here.

03:36 So we're going to have, let's do that on the next page here.

03:40 We're going to have d squared plus 4d plus 4 of yp is going to be equal to 5x, e to the negative 2x.

03:58 Okay.

04:00 Now, remember our trial solution, so let's plug that in, 4d, plus 4, this is equal to a not x, e or x squared, e to the negative 2x, plus a1x cubed, e to the negative 2x, and this is equal to 5x, e to the negative 2x.

04:36 So first, we need to, or next, we need to find the second, the second, and first derivatives of this.

04:43 So let's first find, let's just find yp prime.

04:48 Let's see that separately.

04:50 So yp prime, we're going to take the derivative of this first, and we're going to need to use product rule.

04:58 So we're going to take first times derivative second, or a second times derivative of first is going to be 2a0x, e to the negative 2x, and then we're going to have minus 2a0 x squared e to the negative 2x.

05:14 And then with this here, we're going to have a plus 3a1x squared, e to the negative 2x, and then minus 2a1x cubed e to the negative 2x...

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