Math  /  Calculus

QuestionStandard 18: Problem 4 (1 point)
Find the inverse Laplace transform f(t)=L1{F(s)}f(t)=\mathcal{L}^{-1}\{F(s)\} of the function F(s)=10ss29F(s)=\frac{10 s}{s^{2}-9} f(t)=L1{10ss29}=L1{5s+3+5s3}=f(t)=\mathcal{L}^{-1}\left\{\frac{10 s}{s^{2}-9}\right\}=\mathcal{L}^{-1}\left\{\frac{5}{s+3}+\frac{5}{s-3}\right\}= \square help (formulas) Preview My Answers Submit Answers
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Studdy Solution

STEP 1

What is this asking? We need to find the *inverse Laplace transform* of F(s)=10ss29F(s) = \frac{10s}{s^2 - 9}, which means finding a function f(t)f(t) whose Laplace transform is F(s)F(s). Watch out! Remember those Laplace transform formulas!
We'll need them to connect F(s)F(s) back to f(t)f(t).

STEP 2

1. Rewrite the function
2. Inverse Laplace Transform

STEP 3

Let's **factor the denominator** of our function F(s)F(s).
We have s29s^2 - 9, which is a *difference of squares*.
So, we can rewrite it as (s3)(s+3)(s-3)(s+3).
This gives us F(s)=10s(s3)(s+3)F(s) = \frac{10s}{(s-3)(s+3)}.
This makes it look like something we can work with using *partial fractions*!

STEP 4

We want to **rewrite** 10s(s3)(s+3)\frac{10s}{(s-3)(s+3)} in the form As3+Bs+3\frac{A}{s-3} + \frac{B}{s+3}.
To find AA and BB, we multiply both sides by (s3)(s+3)(s-3)(s+3) to get 10s=A(s+3)+B(s3)10s = A(s+3) + B(s-3).
If we let s=3s=3, we get 103=A(3+3)+B(33)10 \cdot 3 = A(3+3) + B(3-3), which simplifies to 30=6A30 = 6A, so A=5A=5.
If we let s=3s=-3, we get 10(3)=A(3+3)+B(33)10 \cdot (-3) = A(-3+3) + B(-3-3), which simplifies to 30=6B-30 = -6B, so B=5B=5.
Awesome! Now we have F(s)=5s3+5s+3F(s) = \frac{5}{s-3} + \frac{5}{s+3}.

STEP 5

Now, we can **apply the inverse Laplace transform** to each term.
Remember the formula L1{1sa}=eat\mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\} = e^{at}.
Using this, we have: L1{5s3}=5L1{1s3}=5e3t \mathcal{L}^{-1}\left\{\frac{5}{s-3}\right\} = 5\mathcal{L}^{-1}\left\{\frac{1}{s-3}\right\} = 5e^{3t} and L1{5s+3}=5L1{1s(3)}=5e3t \mathcal{L}^{-1}\left\{\frac{5}{s+3}\right\} = 5\mathcal{L}^{-1}\left\{\frac{1}{s-(-3)}\right\} = 5e^{-3t}

STEP 6

Finally, we **add these results together** to get our final answer: f(t)=5e3t+5e3t f(t) = 5e^{3t} + 5e^{-3t} We can also rewrite this using the definition of hyperbolic cosine: f(t)=10e3t+e3t2=10cosh(3t) f(t) = 10 \cdot \frac{e^{3t} + e^{-3t}}{2} = 10 \cosh(3t)

STEP 7

The inverse Laplace transform of F(s)F(s) is f(t)=10cosh(3t)f(t) = 10\cosh(3t).

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