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Math  /  Calculus

Questionfind L−1{s2−40s−393(s−11)(s2+4s+13)}L^{-1}\left\{\frac{s^{2}-40 s-393}{(s-11)\left(s^{2}+4 s+13\right)}\right\}

Studdy Solution

STEP 1

What is this asking? We need to find the *inverse Laplace transform* of a given expression, which means we're going from the *s-domain* back to the *time-domain*!
Think of it like translating from one language to another, but with math! Watch out! Don't forget to **carefully factor** and **decompose** the expression.
A little slip-up can lead to a completely different answer!

STEP 2

1. Partial Fraction Decomposition
2. Inverse Laplace Transform

STEP 3

Alright, let's **break down** this fraction!
We're going to use *partial fraction decomposition*.
This is like taking apart a complicated machine to see all its simpler parts.
We want to rewrite our expression as a sum of simpler fractions.

STEP 4

Our given expression is s2−40s−393(s−11)(s2+4s+13)\frac{s^{2}-40 s-393}{(s-11)\left(s^{2}+4 s+13\right)}.
We can decompose it into the form: s2−40s−393(s−11)(s2+4s+13)=As−11+Bs+Cs2+4s+13\frac{s^{2}-40 s-393}{(s-11)\left(s^{2}+4 s+13\right)}=\frac{A}{s-11}+\frac{B s+C}{s^{2}+4 s+13} where AA, BB, and CC are constants we need to find.

STEP 5

To find these constants, we'll **multiply both sides** by the denominator (s−11)(s2+4s+13)(s-11)(s^2 + 4s + 13).
This gives us: s2−40s−393=A(s2+4s+13)+(Bs+C)(s−11)s^{2}-40 s-393=A\left(s^{2}+4 s+13\right)+(B s+C)(s-11)

STEP 6

Now, let's be clever!
If we let s=11s = \mathbf{11}, the second term on the right-hand side disappears!
This gives us: 112−40⋅11−393=A(112+4⋅11+13)11^2 - 40 \cdot 11 - 393 = A(11^2 + 4 \cdot 11 + 13) 121−440−393=A(121+44+13)121 - 440 - 393 = A(121 + 44 + 13)−712=178A-712 = 178ASo, A=−4A = \mathbf{-4}!

STEP 7

Now, let's **expand** the equation from 2.1.3 and **group** the terms: s2−40s−393=As2+4As+13A+Bs2−11Bs+Cs−11Cs^2 - 40s - 393 = As^2 + 4As + 13A + Bs^2 - 11Bs + Cs - 11C s2−40s−393=(A+B)s2+(4A−11B+C)s+(13A−11C)s^2 - 40s - 393 = (A+B)s^2 + (4A-11B+C)s + (13A-11C)

STEP 8

By **comparing coefficients**, we get the following equations: A+B=1A + B = 1 4A−11B+C=−404A - 11B + C = -4013A−11C=−39313A - 11C = -393

STEP 9

We already know A=−4A = \mathbf{-4}.
Substituting this into the first equation gives us −4+B=1-4 + B = 1, so B=5B = \mathbf{5}.

STEP 10

Substituting A=−4A = \mathbf{-4} into the third equation gives us −52−11C=−393-52 - 11C = -393, so −11C=−341-11C = -341, and C=31C = \mathbf{31}.

STEP 11

So, our decomposed fraction is: −4s−11+5s+31s2+4s+13\frac{-4}{s-11}+\frac{5 s+31}{s^{2}+4 s+13}

STEP 12

Now, let's do the *inverse Laplace transform*!
This is like translating back from the other language.
We'll take the inverse Laplace transform of each part of our decomposed fraction.

STEP 13

We know that L−1{1s−a}=eat\mathcal{L}^{-1}\left\{\frac{1}{s-a}\right\}=e^{a t}.
So, L−1{−4s−11}=−4e11t\mathcal{L}^{-1}\left\{\frac{-4}{s-11}\right\}=-4 e^{11 t}.

STEP 14

For the second term, we need to **complete the square** in the denominator: s2+4s+13=(s+2)2+32s^2 + 4s + 13 = (s+2)^2 + 3^2.

STEP 15

We can rewrite the second term as: 5(s+2)+21(s+2)2+32=5s+2(s+2)2+32+73(s+2)2+32\frac{5(s+2)+21}{(s+2)^{2}+3^{2}}=5 \frac{s+2}{(s+2)^{2}+3^{2}}+7 \frac{3}{(s+2)^{2}+3^{2}}

STEP 16

Using the inverse Laplace transform properties, we get: L−1{5s+2(s+2)2+32+73(s+2)2+32}=5e−2tcos⁡(3t)+7e−2tsin⁡(3t)\mathcal{L}^{-1}\left\{5 \frac{s+2}{(s+2)^{2}+3^{2}}+7 \frac{3}{(s+2)^{2}+3^{2}}\right\}=5 e^{-2 t} \cos (3 t)+7 e^{-2 t} \sin (3 t)

STEP 17

Combining the results, we get the **final inverse Laplace transform**: −4e11t+5e−2tcos⁡(3t)+7e−2tsin⁡(3t)-4 e^{11 t}+5 e^{-2 t} \cos (3 t)+7 e^{-2 t} \sin (3 t)

STEP 18

The inverse Laplace transform is −4e11t+5e−2tcos⁡(3t)+7e−2tsin⁡(3t)-4 e^{11 t}+5 e^{-2 t} \cos (3 t)+7 e^{-2 t} \sin (3 t).

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