Solve the given initial-value problem: y^''-y=9 x e^2 x, y(0)=0, y^'(0)=7.

Name: Problem 34, Chapter 8: Differential Equations and Linear Algebra
Uploaded: 2020-05-21T14:44:02-08:00
Duration: 9 min 24 s
Channel: Mike Gaerlan
Description: Solve the given initial-value problem: y^''-y=9 x e^2 x, y(0)=0, y^'(0)=7.

Solve the given initial-value problem: y^''-y=9 x e^2 x,...

Key Concepts

Method of Undetermined Coefficients

This is a technique used to find a particular solution to non-homogeneous linear differential equations when the non-homogeneous term is of a type (such as exponentials, polynomials, sines, and cosines or their products) that allows an appropriate ansatz. The coefficients in the guessed particular solution are determined by substituting the guess into the differential equation and equating coefficients.

Initial Value Problem

An initial value problem involves not only solving the differential equation but also finding specific values for the arbitrary constants by applying given initial conditions. This step ensures that the solution uniquely satisfies both the differential equation and the designated initial values, tailoring the general solution to a specific scenario.

Second Order Linear Differential Equations with Constant Coefficients

This concept deals with differential equations of the form y'' + ay' + by = f(x) where the coefficients a and b are constants. These equations are fundamental in understanding the behavior of many physical systems, and solving them involves finding the general solution to the corresponding homogeneous equation and then obtaining a particular solution to account for the non-homogeneous term.

Characteristic Equation

For homogeneous second order linear differential equations, the solution method involves forming the characteristic (auxiliary) equation by replacing the derivatives with powers of a variable (typically r). The roots of this algebraic equation determine the structure of the homogeneous solution, which can involve real distinct roots, repeated roots, or complex conjugate roots.

Transcript

00:01 Okay, so to solve this initial value problem, first we're going to need to find the general solution of this.

00:07 So writing that in differential operator form, that's going to be like this, so 9x, e to the 2x.

00:15 Now first or next, we need to find the complementary solution.

00:20 That's going to be d squared minus 1, yc is equal to 0.

00:25 So we set that right -hand side equal to 0 first.

00:29 Then our corresponding auxiliary equation, p of r, is going to be equal to r squared or sorry, r squared minus 1 is equal to 0.

00:39 So we get r is going to be equal to plus or minus 1.

00:43 So our complementary solution, y, c of x, is going to take the form c1, e to the x, plus c2e to the negative x, c2e to the negative x, like so.

00:56 Next, we need to annihilate this part.

01:00 Here.

01:01 So we have f of x is going to be equal to 9x, e to the 2x.

01:08 So here we notice that we have a root of 2 here.

01:15 And then we also have a multiplicity of 2 here.

01:20 This tells us that our annihilator is going to take the form of d minus 2 and then squared.

01:29 So applying this to both sides here, we'll actually just get that our trial solution y p of x is going to be equal to.

01:41 And then we have a1, e to the 2x.

01:47 And then we're also going to have a, sorry, a0, e to the 2x.

01:51 And then plus an a1x, e to the 2x, like so.

01:58 Which we can also rewrite as, so a0 plus a1x, e to the 2x, like so.

02:07 So our general solution is going to look like y of x is equal to c1 e to the x plus c2 e to the negative x plus a not plus a1 x e to the 2x.

02:27 So to solve for a not an a1, we're going to plug it into the original equation here.

02:35 So we need to find yp prime of x.

02:41 This is going to be equal to.

02:44 Now we do first times derivative of a second.

02:48 So that's going to be 2a not plus a1x, e to the 2x, and then plus the derivative of the first, which is going to be just a1, and then times e to the 2x.

03:07 So this can simplify to become, if we combine this with this here, we get 2a0 plus a1, and then plus 2a1 x, and then e to the 2x.

03:28 Then we also need to find y double prime, my p double prime, x.

03:35 So, again, the derivative will take this times the derivative of this.

03:41 So it's going to be two times that.

03:43 So we'll have 4a0 plus 2a1 plus 4a1x times e to the 2x, then plus the derivative of this times this.

03:56 So it's going to be plus 2a1e to the 2x.

04:02 Okay.

04:03 So we can further combine this, well, 2a1, e to the 2x to be here.

04:12 So again, that's going to be equal to 4a0 plus 4a1 plus 4a1x is e to the 2x like so.

04:25 So, okay, plug these in now here and we get.

04:32 So we had 4a0 plus 4a1.

04:36 Plus 4a1 x e to the 2x and then minus the original the original being what was it a not a not minus or plus a 1 x e to the 2x and then this is going to be equal to our right hand side which was 9x e to the 2x like so so let's combine these two so we have a minus a not here.

05:13 So that's going to become 3a0.

05:17 And then we have a minus a 1x.

05:19 So it's going to come from here.

05:21 So we have that plus 4a1 and then plus 3a1x.

05:27 And this is times e to the 2x.

05:29 It's equal to 9x, e to the 2x.

05:32 We can cancel out the e to the 2x from both sides...

Please give Ace some feedback