Math  /  Calculus

QuestionStandard 18: Problem 3 (1 point) Find the inverse Laplace transform f(t)=L1{F(s)}f(t)=\mathcal{L}^{-1}\{F(s)\} of the function F(s)=(5s2+64+9ss2+36)F(s)=-\left(\frac{5}{s^{2}+64}+\frac{9 s}{s^{2}+36}\right). f(t)=L1{(5s2+64+9ss2+36)}=f(t)=\mathcal{L}^{-1}\left\{-\left(\frac{5}{s^{2}+64}+\frac{9 s}{s^{2}+36}\right)\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 a function that has two parts: one with s2+64s^2 + 64 and another with s2+36s^2 + 36. Watch out! Don't forget the negative sign in front of the whole expression, and make sure to use the correct inverse Laplace transform formulas for each part!

STEP 2

1. Separate the terms
2. Transform the first term
3. Transform the second term
4. Combine the results

STEP 3

We're given F(s)=(5s2+64+9ss2+36)F(s) = -\left(\frac{5}{s^{2}+64}+\frac{9 s}{s^{2}+36}\right).
Let's **distribute** that negative sign to both terms inside the parentheses.
This makes it easier to work with each term individually.

STEP 4

So, we have F(s)=5s2+649ss2+36F(s) = -\frac{5}{s^{2}+64} - \frac{9s}{s^{2}+36}.
This helps us see the two parts more clearly!

STEP 5

The first term is 5s2+64-\frac{5}{s^{2}+64}.
Notice that 6464 is 828^2, so we can rewrite this as 5s2+82-\frac{5}{s^{2}+8^{2}}.

STEP 6

We know the inverse Laplace transform of as2+a2\frac{a}{s^{2}+a^{2}} is sin(at)\sin(at).
In our case, a=8a = \mathbf{8}, so the inverse Laplace transform of 1s2+82\frac{1}{s^{2}+8^{2}} is sin(8t)\sin(8t).

STEP 7

Since we have 5s2+82-\frac{5}{s^{2}+8^{2}}, we **multiply** the inverse Laplace transform of 1s2+82\frac{1}{s^{2}+8^{2}} by 5\mathbf{-5}.
This gives us 5sin(8t)=5sin(8t)-5 \cdot \sin(8t) = -5\sin(8t).

STEP 8

Our second term is 9ss2+36-\frac{9s}{s^{2}+36}.
We can rewrite 3636 as 626^2, giving us 9ss2+62-\frac{9s}{s^{2}+6^{2}}.

STEP 9

The inverse Laplace transform of ss2+a2\frac{s}{s^{2}+a^{2}} is cos(at)\cos(at).
Here, a=6a = \mathbf{6}, so the inverse Laplace transform of ss2+62\frac{s}{s^{2}+6^{2}} is cos(6t)\cos(6t).

STEP 10

We have 9ss2+62-\frac{9s}{s^{2}+6^{2}}, so we **multiply** the inverse Laplace transform of ss2+62\frac{s}{s^{2}+6^{2}} by 9\mathbf{-9}.
This gives us 9cos(6t)=9cos(6t)-9 \cdot \cos(6t) = -9\cos(6t).

STEP 11

Now, let's **add** the inverse Laplace transforms of both terms together.

STEP 12

We found that the inverse Laplace transform of the first term is 5sin(8t)-5\sin(8t) and the inverse Laplace transform of the second term is 9cos(6t)-9\cos(6t).
Adding these together gives us f(t)=5sin(8t)9cos(6t)f(t) = -5\sin(8t) - 9\cos(6t).

STEP 13

The inverse Laplace transform of F(s)F(s) is f(t)=5sin(8t)9cos(6t)f(t) = -5\sin(8t) - 9\cos(6t).

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