Math  /  Calculus

Question(1 point)
Consider the function F(s)=4s10s25s+4F(s)=\frac{4 s-10}{s^{2}-5 s+4}. a. Find the partial fraction decomposition of F(s)F(s) : 4s10s25s+4=+\frac{4 s-10}{s^{2}-5 s+4}=\square+\square b. Find the inverse Laplace transform of F(s)F(s). f(t)=L1{F(s)}=f(t)=\mathcal{L}^{-1}\{F(s)\}= \square help (formulas)

Studdy Solution

STEP 1

What is this asking? We're given a function F(s)F(s) and need to break it down into simpler fractions (partial fractions) and then find the inverse Laplace transform to get back to a function of tt, f(t)f(t). Watch out! Don't forget to factor the denominator correctly and be careful with signs when solving for the constants in the partial fraction decomposition!

STEP 2

1. Factor the Denominator
2. Partial Fraction Decomposition
3. Inverse Laplace Transform

STEP 3

Alright, let's **factor** that denominator!
We've got s25s+4s^2 - 5s + 4.
We're looking for two numbers that multiply to **4** and add up to **-5**.
Those numbers are **-1** and **-4**!

STEP 4

So, we can rewrite the denominator as (s1)(s4)(s-1)(s-4).
Boom! Our function now looks like this: F(s)=4s10(s1)(s4)F(s) = \frac{4s - 10}{(s-1)(s-4)}

STEP 5

Now for the fun part!
We want to express F(s)F(s) as the sum of two simpler fractions: 4s10(s1)(s4)=As1+Bs4\frac{4s - 10}{(s-1)(s-4)} = \frac{A}{s-1} + \frac{B}{s-4} Where AA and BB are constants we need to find.

STEP 6

To solve for AA and BB, let's multiply both sides by (s1)(s4)(s-1)(s-4) to get rid of the denominators: 4s10=A(s4)+B(s1)4s - 10 = A(s-4) + B(s-1)

STEP 7

Let's be clever!
If we set s=1s = 1, the BB term disappears! 4(1)10=A(14)4(1) - 10 = A(1-4) 6=3A-6 = -3AA=2A = 2Woohoo! We found AA!

STEP 8

Now, let's set s=4s = 4 to make the AA term vanish: 4(4)10=B(41)4(4) - 10 = B(4-1) 6=3B6 = 3BB=2B = 2Awesome! We also found BB!

STEP 9

So, our partial fraction decomposition is: 4s10(s1)(s4)=2s1+2s4\frac{4s - 10}{(s-1)(s-4)} = \frac{2}{s-1} + \frac{2}{s-4}

STEP 10

Time to bring it home!
We need to find the inverse Laplace transform of our decomposed function.
Remember the formula L1{1sa}=eat\mathcal{L}^{-1}\{\frac{1}{s-a}\} = e^{at}.

STEP 11

Applying this formula to both terms, we get: f(t)=L1{F(s)}=L1{2s1}+L1{2s4}f(t) = \mathcal{L}^{-1}\{F(s)\} = \mathcal{L}^{-1}\{\frac{2}{s-1}\} + \mathcal{L}^{-1}\{\frac{2}{s-4}\} f(t)=2et+2e4tf(t) = 2e^{t} + 2e^{4t}

STEP 12

a. 2s1+2s4\frac{2}{s-1} + \frac{2}{s-4} b. f(t)=2et+2e4tf(t) = 2e^{t} + 2e^{4t}

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