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Math  /  Calculus

QuestionDetermine the inverse Laplace transform of the function below. 9s2+4\frac{9}{s^{2}+4}

Studdy Solution

STEP 1

1. The function 9s2+4 \frac{9}{s^2 + 4} is a rational function in the Laplace domain.
2. We will use known inverse Laplace transforms to find the inverse of the given function.
3. The inverse Laplace transform of ωs2+ω2 \frac{\omega}{s^2 + \omega^2} is sin⁡(ωt) \sin(\omega t) .

STEP 2

1. Identify the standard form of the inverse Laplace transform.
2. Rewrite the given function to match the standard form.
3. Apply the inverse Laplace transform.

STEP 3

Identify the standard form of the inverse Laplace transform. We know that the inverse Laplace transform of ωs2+ω2 \frac{\omega}{s^2 + \omega^2} is sin⁡(ωt) \sin(\omega t) .

STEP 4

Rewrite the given function 9s2+4 \frac{9}{s^2 + 4} to match the standard form. Notice that s2+4 s^2 + 4 can be written as s2+(2)2 s^2 + (2)^2 . The standard form requires the numerator to be ω \omega , which is 2 2 in this case. We can factor out a constant to match the form:
9s2+4=94⋅4s2+4=94⋅2⋅2s2+22 \frac{9}{s^2 + 4} = \frac{9}{4} \cdot \frac{4}{s^2 + 4} = \frac{9}{4} \cdot \frac{2 \cdot 2}{s^2 + 2^2}

STEP 5

Apply the inverse Laplace transform using the standard form. The inverse Laplace transform of 2s2+22 \frac{2}{s^2 + 2^2} is sin⁡(2t) \sin(2t) . Therefore, the inverse Laplace transform of the original function is:
L−1{9s2+4}=94⋅sin⁡(2t) \mathcal{L}^{-1} \left\{ \frac{9}{s^2 + 4} \right\} = \frac{9}{4} \cdot \sin(2t)
The inverse Laplace transform of the given function is:
94sin⁡(2t) \boxed{\frac{9}{4} \sin(2t)}

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