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.
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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
.
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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.
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7Slide8
Graph of:
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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
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9Slide10
We have
Apply the Laplace Transform operator to both sides:
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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,
,
.
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11Slide12
Now we find
. The expression
decomposes asThus, .This gives , where .The solution of
is
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12Slide13
The graph of
is
(c) ASU-SoMSS - Scott Surgent. Report errors to surgent@asu.edu13