Studdy AI Logo
Math  /  Calculus

QuestionDetermine L−1{ F}\mathscr{L}^{-1}\{\mathrm{~F}\}. F(s)=3s3−4s2−7s+28s3(s−4)F(s)=\frac{3 s^{3}-4 s^{2}-7 s+28}{s^{3}(s-4)}

Studdy Solution

STEP 1

1. We are given the function F(s)=3s3−4s2−7s+28s3(s−4) F(s) = \frac{3s^3 - 4s^2 - 7s + 28}{s^3(s-4)} .
2. We need to find the inverse Laplace transform, denoted as L−1{F(s)} \mathscr{L}^{-1}\{F(s)\} .

STEP 2

1. Perform partial fraction decomposition on F(s) F(s) .
2. Identify the inverse Laplace transforms for each term in the decomposition.
3. Sum the inverse Laplace transforms to find L−1{F(s)} \mathscr{L}^{-1}\{F(s)\} .

STEP 3

Perform partial fraction decomposition on F(s) F(s) :
F(s)=3s3−4s2−7s+28s3(s−4) F(s) = \frac{3s^3 - 4s^2 - 7s + 28}{s^3(s-4)}
Assume:
F(s)=As+Bs2+Cs3+Ds−4 F(s) = \frac{A}{s} + \frac{B}{s^2} + \frac{C}{s^3} + \frac{D}{s-4}
Combine the fractions:
F(s)=A(s2)(s−4)+B(s)(s−4)+C(s−4)+D(s3)s3(s−4) F(s) = \frac{A(s^2)(s-4) + B(s)(s-4) + C(s-4) + D(s^3)}{s^3(s-4)}

STEP 4

Equate numerators:
3s3−4s2−7s+28=A(s2)(s−4)+B(s)(s−4)+C(s−4)+D(s3) 3s^3 - 4s^2 - 7s + 28 = A(s^2)(s-4) + B(s)(s-4) + C(s-4) + D(s^3)
Expand and collect like terms:
3s3−4s2−7s+28=As3−4As2+Bs2−4Bs+Cs−4C+Ds3 3s^3 - 4s^2 - 7s + 28 = As^3 - 4As^2 + Bs^2 - 4Bs + Cs - 4C + Ds^3
3s3−4s2−7s+28=(A+D)s3+(−4A+B)s2+(−4B+C)s−4C 3s^3 - 4s^2 - 7s + 28 = (A + D)s^3 + (-4A + B)s^2 + (-4B + C)s - 4C

STEP 5

Set up equations for coefficients:
1. A+D=3 A + D = 3
2. −4A+B=−4-4A + B = -4
3. −4B+C=−7-4B + C = -7
4. −4C=28-4C = 28
Solve these equations to find A,B,C, A, B, C, and D D .

STEP 6

Solve the equations:
From equation 4:
C=−7 C = -7
Substitute C=−7 C = -7 into equation 3:
−4B−7=−7 -4B - 7 = -7 −4B=0 -4B = 0 B=0 B = 0
Substitute B=0 B = 0 into equation 2:
−4A=−4 -4A = -4 A=1 A = 1
Substitute A=1 A = 1 into equation 1:
1+D=3 1 + D = 3 D=2 D = 2
So, A=1 A = 1 , B=0 B = 0 , C=−7 C = -7 , D=2 D = 2 .

STEP 7

Write the partial fraction decomposition:
F(s)=1s+0s2+−7s3+2s−4 F(s) = \frac{1}{s} + \frac{0}{s^2} + \frac{-7}{s^3} + \frac{2}{s-4}
Simplify:
F(s)=1s−7s3+2s−4 F(s) = \frac{1}{s} - \frac{7}{s^3} + \frac{2}{s-4}

STEP 8

Identify the inverse Laplace transforms for each term:
1. L−1{1s}=1 \mathscr{L}^{-1}\left\{\frac{1}{s}\right\} = 1
2. L−1{−7s3}=−7⋅t22=−72t2 \mathscr{L}^{-1}\left\{\frac{-7}{s^3}\right\} = -7 \cdot \frac{t^2}{2} = -\frac{7}{2}t^2
3. L−1{2s−4}=2e4t \mathscr{L}^{-1}\left\{\frac{2}{s-4}\right\} = 2e^{4t}

STEP 9

Sum the inverse Laplace transforms:
L−1{F(s)}=1−72t2+2e4t \mathscr{L}^{-1}\{F(s)\} = 1 - \frac{7}{2}t^2 + 2e^{4t}
The inverse Laplace transform is:
1−72t2+2e4t \boxed{1 - \frac{7}{2}t^2 + 2e^{4t}}

Was this helpful?

Studdy solves anything!

banner

Start learning now

Download Studdy AI Tutor now. Learn with ease and get all help you need to be successful at school.

ParentsInfluencer programContactPolicyTerms
TwitterInstagramFacebookTikTokDiscord