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Math  /  Calculus

QuestionFind the inverse Laplace transform of F(s)=ln⁡(1+1s2)F(s)=\ln \left(1+\frac{1}{s^{2}}\right) a. 2−2cos⁡(t)t\frac{2-2 \cos (t)}{t} b. sin⁡(t)t\frac{\sin (t)}{t} C. cos⁡(t)t\frac{\cos (t)}{t} d. 2−2sin⁡(t)t\frac{2-2 \sin (t)}{t}

Studdy Solution

STEP 1

1. We are given the Laplace transform F(s)=ln⁡(1+1s2) F(s) = \ln \left(1+\frac{1}{s^{2}}\right) .
2. We need to find the inverse Laplace transform of F(s) F(s) .
3. We are given four possible options for the inverse Laplace transform.

STEP 2

1. Recognize the form of the given Laplace transform.
2. Use known Laplace transform properties or theorems to simplify.
3. Identify the inverse Laplace transform from the given options.

STEP 3

Recognize that F(s)=ln⁡(1+1s2) F(s) = \ln \left(1+\frac{1}{s^{2}}\right) can be related to the Laplace transform of a known function. The function inside the logarithm suggests a series expansion approach.

STEP 4

Recall the series expansion for ln⁡(1+x) \ln(1+x) :
ln⁡(1+x)=x−x22+x33−x44+⋯ \ln(1+x) = x - \frac{x^2}{2} + \frac{x^3}{3} - \frac{x^4}{4} + \cdots
Apply this to ln⁡(1+1s2) \ln \left(1+\frac{1}{s^{2}}\right) :
ln⁡(1+1s2)=1s2−12s4+13s6−⋯ \ln \left(1+\frac{1}{s^{2}}\right) = \frac{1}{s^2} - \frac{1}{2s^4} + \frac{1}{3s^6} - \cdots

STEP 5

Recognize that each term in the series corresponds to a known Laplace transform:
- The inverse Laplace transform of 1s2 \frac{1}{s^2} is t t . - The inverse Laplace transform of 1s4 \frac{1}{s^4} is t36 \frac{t^3}{6} . - The inverse Laplace transform of 1s6 \frac{1}{s^6} is t5120 \frac{t^5}{120} .
Thus, the inverse Laplace transform of the series is:
t−t32⋅6+t53⋅120−⋯ t - \frac{t^3}{2 \cdot 6} + \frac{t^5}{3 \cdot 120} - \cdots

STEP 6

Recognize that the series represents the function:
sin⁡(t)t \frac{\sin(t)}{t}
This is because the series expansion of sin⁡(t)t \frac{\sin(t)}{t} is:
sin⁡(t)t=1−t23!+t45!−⋯ \frac{\sin(t)}{t} = 1 - \frac{t^2}{3!} + \frac{t^4}{5!} - \cdots

STEP 7

Identify the correct option from the given choices. The inverse Laplace transform is:
b. sin⁡(t)t \frac{\sin(t)}{t}
The inverse Laplace transform of F(s)=ln⁡(1+1s2) F(s) = \ln \left(1+\frac{1}{s^{2}}\right) is:
sin⁡(t)t \boxed{\frac{\sin(t)}{t}}

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