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Math  /  Algebra

QuestionFind L−1{1s2+12s+36}L^{-1}\left\{\frac{1}{s^{2}+12 s+36}\right\}

Studdy Solution

STEP 1

What is this asking? We need to find the inverse Laplace transform of the given function. Watch out! Don't forget to complete the square in the denominator to simplify the expression!

STEP 2

1. Simplify the expression
2. Identify the inverse Laplace form
3. Find the inverse Laplace transform

STEP 3

Alright, let's **simplify the denominator** by completing the square!
We have s2+12s+36 s^2 + 12s + 36 .
To complete the square, we take the coefficient of s s , which is **12**, divide it by 2, and then square it.
So, (122)2=36 \left(\frac{12}{2}\right)^2 = 36 .
Now, let's rewrite the quadratic expression: s2+12s+36=(s+6)2 s^2 + 12s + 36 = (s + 6)^2 So, our original function becomes: 1s2+12s+36=1(s+6)2 \frac{1}{s^2 + 12s + 36} = \frac{1}{(s + 6)^2}

STEP 4

Next, we need to **identify the form** of the inverse Laplace transform.
The expression 1(s+6)2\frac{1}{(s + 6)^2} looks like a standard inverse Laplace transform form.
Remember, the inverse Laplace transform of 1(s−a)2\frac{1}{(s-a)^2} is t⋅eat t \cdot e^{at} .

STEP 5

Now, let's **find the inverse Laplace transform** using the identified form.
Here, a=−6 a = -6 .
So, the inverse Laplace transform of 1(s+6)2\frac{1}{(s + 6)^2} is: L−1{1(s+6)2}=t⋅e−6t L^{-1}\left\{\frac{1}{(s + 6)^2}\right\} = t \cdot e^{-6t}

STEP 6

The inverse Laplace transform of 1s2+12s+36\frac{1}{s^2 + 12s + 36} is t⋅e−6t t \cdot e^{-6t} .

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