Math / Calculus

QuestionStandard 17: Problem 6 (1 point)
Find the inverse Laplace transform $f(t)=\mathcal{L}^{-1}\{F(s)\}$ of the function $F(s)=\frac{6 s-23}{s^{2}-8 s+20}$ . $f(t)=\mathcal{L}^{-1}\left\{\frac{6 s-23}{s^{2}-8 s+20}\right\}=$ $\square$ 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 given function, which means figuring out what original function would produce this given function when the Laplace transform is applied to it. Watch out! Remember completing the square is our friend!
Also, don't forget those handy Laplace transform formulas.

STEP 2

1. Prepare the denominator
2. Rewrite the numerator
3. Split the fraction
4. Apply inverse Laplace transform

STEP 3

We'll complete the square for the denominator $s^2 - 8s + 20$ .
Take half of the coefficient of the $s$ term (which is $-8$ , so half is $-4$ ), square it (giving us $(-4)^2 = 16$ ), and add and subtract this value within the expression.
This gives us $s^2 - 8s + 16 - 16 + 20$ .

STEP 4

Now, we can rewrite $s^2 - 8s + 16$ as $(s-4)^2$ .
Our denominator becomes $(s-4)^2 + 4$ , which is $(s-4)^2 + 2^2$ .
This form is much more friendly for inverse Laplace transforms!

STEP 5

We want the numerator to play nicely with our new denominator, which has a $(s-4)$ .
Let's rewrite the numerator $6s - 23$ as $6(s-4) + 1$ .
We chose this because we know terms involving $(s-4)$ will work well with our denominator when we apply the inverse Laplace transform.

STEP 6

Now, we have $\frac{6(s-4) + 1}{(s-4)^2 + 2^2}$ .
Let's split this into two fractions: $\frac{6(s-4)}{(s-4)^2 + 2^2} + \frac{1}{(s-4)^2 + 2^2}$ .
This makes it easier to apply the inverse Laplace transform formulas.

STEP 7

We can further rewrite the second fraction to perfectly match the formula for the inverse Laplace transform of sine.
We multiply and divide by 2, the value in the denominator squared, to get $\frac{6(s-4)}{(s-4)^2 + 2^2} + \frac{1}{2} \cdot \frac{2}{(s-4)^2 + 2^2}$ .

STEP 8

Now, we can use the inverse Laplace transform formulas!
Recall that $\mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^2 + b^2}\right\} = e^{at}\cos(bt)$ and $\mathcal{L}^{-1}\left\{\frac{b}{(s-a)^2 + b^2}\right\} = e^{at}\sin(bt)$ .

STEP 9

Applying these formulas to our expression $\frac{6(s-4)}{(s-4)^2 + 2^2} + \frac{1}{2} \cdot \frac{2}{(s-4)^2 + 2^2}$ with $a=4$ and $b=2$ , we get $6e^{4t}\cos(2t) + \frac{1}{2}e^{4t}\sin(2t)$ .

STEP 10

The inverse Laplace transform is $f(t) = 6e^{4t}\cos(2t) + \frac{1}{2}e^{4t}\sin(2t)$ .

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Studdy Solution

STEP 1

STEP 2

1. Prepare the denominator
2. Rewrite the numerator
3. Split the fraction
4. Apply inverse Laplace transform

STEP 3

We'll complete the square for the denominator $s^2 - 8s + 20$ .
Take half of the coefficient of the $s$ term (which is $-8$ , so half is $-4$ ), square it (giving us $(-4)^2 = 16$ ), and add and subtract this value within the expression.
This gives us $s^2 - 8s + 16 - 16 + 20$ .

STEP 4

Now, we can rewrite $s^2 - 8s + 16$ as $(s-4)^2$ .
Our denominator becomes $(s-4)^2 + 4$ , which is $(s-4)^2 + 2^2$ .
This form is much more friendly for inverse Laplace transforms!

STEP 5

We want the numerator to play nicely with our new denominator, which has a $(s-4)$ .
Let's rewrite the numerator $6s - 23$ as $6(s-4) + 1$ .
We chose this because we know terms involving $(s-4)$ will work well with our denominator when we apply the inverse Laplace transform.

STEP 6

Now, we have $\frac{6(s-4) + 1}{(s-4)^2 + 2^2}$ .
Let's split this into two fractions: $\frac{6(s-4)}{(s-4)^2 + 2^2} + \frac{1}{(s-4)^2 + 2^2}$ .
This makes it easier to apply the inverse Laplace transform formulas.

STEP 7

We can further rewrite the second fraction to perfectly match the formula for the inverse Laplace transform of sine.
We multiply and divide by 2, the value in the denominator squared, to get $\frac{6(s-4)}{(s-4)^2 + 2^2} + \frac{1}{2} \cdot \frac{2}{(s-4)^2 + 2^2}$ .

STEP 8

Now, we can use the inverse Laplace transform formulas!
Recall that $\mathcal{L}^{-1}\left\{\frac{s-a}{(s-a)^2 + b^2}\right\} = e^{at}\cos(bt)$ and $\mathcal{L}^{-1}\left\{\frac{b}{(s-a)^2 + b^2}\right\} = e^{at}\sin(bt)$ .

STEP 9

Applying these formulas to our expression $\frac{6(s-4)}{(s-4)^2 + 2^2} + \frac{1}{2} \cdot \frac{2}{(s-4)^2 + 2^2}$ with $a=4$ and $b=2$ , we get $6e^{4t}\cos(2t) + \frac{1}{2}e^{4t}\sin(2t)$ .

STEP 10

The inverse Laplace transform is $f(t) = 6e^{4t}\cos(2t) + \frac{1}{2}e^{4t}\sin(2t)$ .

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Studdy Solution

STEP 1

STEP 2

1. Prepare the denominator2. Rewrite the numerator3. Split the fraction4. Apply inverse Laplace transform

STEP 3

STEP 4

Now, we can rewrite s2−8s+16s^2 - 8s + 16s2−8s+16 as (s−4)2(s-4)^2(s−4)2.Our denominator becomes (s−4)2+4(s-4)^2 + 4(s−4)2+4, which is (s−4)2+22(s-4)^2 + 2^2(s−4)2+22.This form is much more friendly for inverse Laplace transforms!

STEP 5

STEP 6

STEP 7

STEP 8

STEP 9

STEP 10

The inverse Laplace transform is f(t)=6e4tcos⁡(2t)+12e4tsin⁡(2t)f(t) = 6e^{4t}\cos(2t) + \frac{1}{2}e^{4t}\sin(2t)f(t)=6e4tcos(2t)+21​e4tsin(2t).

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1. Prepare the denominator
2. Rewrite the numerator
3. Split the fraction
4. Apply inverse Laplace transform

Now, we can rewrite $s^2 - 8s + 16$ as $(s-4)^2$ .
Our denominator becomes $(s-4)^2 + 4$ , which is $(s-4)^2 + 2^2$ .
This form is much more friendly for inverse Laplace transforms!

The inverse Laplace transform is $f(t) = 6e^{4t}\cos(2t) + \frac{1}{2}e^{4t}\sin(2t)$ .