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Fundamentals Of Differential Equations And Boundary Value Problems
Found in: Page 375
Fundamentals Of Differential Equations And Boundary Value Problems

Fundamentals Of Differential Equations And Boundary Value Problems

Book edition 9th
Author(s) R. Kent Nagle, Edward B. Saff, Arthur David Snider
Pages 616 pages
ISBN 9780321977069

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Short Answer

In Problems 1-10, determine the inverse Laplace transform of the given function.

s12s2+s+6

The inverse Laplace transform for the given function is

L1s12s2+s+6=12e14tcos474t54794e14tsin474t

See the step by step solution

Step by Step Solution

Step 1: Determining the inverse laplace transform

  • For a given transfer function H, the Inverse Laplace Transform takes the output Y(s) and determines what X(s) it is in terms of (s).
  • Consider a function F(s), if there is a function f(t) that is continuous on [0,)and satisfies L{f}=Fthen we say that f(t) is the inverse Laplace transform of F(s) and employ the notation
  • f=L1{F}
  • L1n!(sa)n+1=eattn,n=1,2,

Step 2: Find inverse laplace transform for the given function

The given function is s12s2+s+6

Simplify s12s2+s+6 as:

s12s2+s+6=s12s2+12s+3=12s1s2+12s+3=12s1s2+12s+116+3116=12s1s+142+47162

Further simplify the equation as follows:

s12s2+s+6=12s+1454s+142+4742=12s+1454s+142+4742=12s+14s+142+474254794474s+142+4742

Find the inverse Laplace transform of

s12s2+s+6=12s+14s+142+474254794474s+142+4742using L1b(sa)2+(b)2=eatsinbt and L1sa(sa)2+(b)2=eatcosbt as:

L1s12s2+s+6=L112s+14s+142+474254794474s+142+4742=L112s+14s+142+4742L154794474s+142+4742=12L1s+14s+142+474254794L1474s+142+4742=12e14tcos474t54794e14tsin474t

Therefore, the inverse Laplace transform for the given function is

L1s12s2+s+6=12e14tcos474t54794e14tsin474t

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