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Using Laplace Transforms to Solve IVPs with Discontinuous Forcing Functions Using Laplace Transforms to Solve IVPs with Discontinuous Forcing Functions

Using Laplace Transforms to Solve IVPs with Discontinuous Forcing Functions - PowerPoint Presentation

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Using Laplace Transforms to Solve IVPs with Discontinuous Forcing Functions - PPT Presentation

MAT 275 Example Find the solution of the IVP Solution Rewrite the forcing function using the notation Now apply the Laplace Transform Operator to both sides and simplify   c ASUSoMSS Scott Surgent Report errors to surgentasuedu ID: 674705

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Slide1

Using Laplace Transforms to Solve IVPs with Discontinuous Forcing Functions

MAT 275Slide2

Example:

Find the solution of the IVP

Solution: Rewrite the forcing function using the notation:Now apply the Laplace Transform Operator to both sides and simplify:

 

(c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu

2Slide3

Now isolate

:

The solution is the inversion of the above expressions:We’ll work on the term without the first. Note that the denominator is an irreducible quadratic over the reals, so we complete the square:This form exactly fits . Thus, . (c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu3Slide4

From the last slide, we have

Now we’ll work on finding

. The will result in appearing in the final result. So we mentally note this fact, then “ignore” it for the next few steps, as we rewrite into smaller fractions using partial fraction decomposition:Equating the numerators, we have .Collecting terms according to powers of , we have Thus, , and . (c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu4Slide5

So now we have

Completing the square on the second term, we have

. Thus, we need to have in the numerator:Finally, we have that .The whole solution is pieced together on the next slide.

 

(c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu

5Slide6

We have

.

From slide 3, we had .From the last slide, we had .Note that and

.

This gives

, where we must state the shift of 4 units to the right. The entire solution is

.

 

(c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu

6Slide7

The solution of

is

.When , then and we have .When , then and we have

At

, the function is continuous.

The graph is on the next slide.

 

(c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu

7Slide8

Graph of:

 (c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu8Slide9

Example:

Solve

Solution: We need to write the forcing function using notation.When , we have , which does not need a leading “u” for now.When , we need to “turn off” and “turn on” . Thus we have:Note that when , then , so that the last line is , which simplifies to , just like in the original statement.The IVP is now written

 

(c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu

9Slide10

We have

Apply the Laplace Transform operator to both sides:

 

(c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu

10Slide11

We now invert

. For

, we use partial fractions to simplify the expression:The numerator at upper right is written in terms of powers of :Equating numerators, we have

.

Thus,

, and since

, then

. Since

, then

, forcing

.

We have,

,

.

 

(c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu

11Slide12

Now we find

. The expression

decomposes asThus, .This gives , where .The solution of

is

 

(c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu

12Slide13

The graph of

is

 (c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu13