Question

If the Laplace transform of $ \int_0^t \frac{((1+2t)^2-1)e^{3t}}{t} dt = \frac{A}{S-3} + \frac{B}{(S-3)^2} + \frac{C}{(S-3)} $ then $3(A+B+C)=$

• A. 8
• B. 4
• C. 12
• D. 15

Hint:

• Simplify the time-domain function first before attempting Laplace Transform.
• \(\mathcal{L}\{t^n e^{at}\} = \frac{n!}{(s-a)^{n+1}}\).
• \(\mathcal{L}\{\int_0^t f(\tau)d\tau\} = \frac{F(s)}{s}\).
• \(\mathcal{L}\{f(t)/t\} = \int_s^\infty F(\sigma)d\sigma\) (if limit exists).
• Ensure the structure of the given partial fraction expansion matches the expected form of the Laplace Transform.

Answer 1

The Correct Option is B

$$ \mathcal{L}\left\{\int_0^t \frac{((1+2\tau)^2 - 1)e^{3\tau}}{\tau} d\tau\right\} = \frac{A}{s} + \frac{B}{(s-3)^2} + \frac{C}{s-3} $$

Let \( f(\tau) = \frac{((1+2\tau)^2 - 1)e^{3\tau}}{\tau} \).
Expand the numerator: \( ((1+2\tau)^2 - 1) = (1 + 4\tau + 4\tau^2 - 1) = 4\tau + 4\tau^2 \). 
So, \( f(\tau) = \frac{(4\tau + 4\tau^2)e^{3\tau}}{\tau} = \frac{4\tau(1+\tau)e^{3\tau}}{\tau} = 4(1+\tau)e^{3\tau} = 4e^{3\tau} + 4\tau e^{3\tau} \), for \( \tau \neq 0 \).

Let \( h(t) = 4e^{3t} + 4te^{3t} \). 
The Laplace transform of an integral is given by \( \mathcal{L}\left\{\int_0^t h(\tau) d\tau\right\} = \frac{1}{s} \mathcal{L}\{h(t)\} \).

First, we find the Laplace transform of \( h(t) \): 
\( \mathcal{L}\{h(t)\} = \mathcal{L}\{4e^{3t} + 4te^{3t}\} = 4\mathcal{L}\{e^{3t}\} + 4\mathcal{L}\{te^{3t}\} \) 
Using standard Laplace transforms: 
\( \mathcal{L}\{e^{at}\} = \frac{1}{s-a} \), and \( \mathcal{L}\{t^n e^{at}\} = \frac{n!}{(s-a)^{n+1}} \) 
So, \( \mathcal{L}\{e^{3t}\} = \frac{1}{s-3} \), \( \mathcal{L}\{te^{3t}\} = \frac{1}{(s-3)^2} \) 
Hence, \( \mathcal{L}\{h(t)\} = \frac{4}{s-3} + \frac{4}{(s-3)^2} \)

Therefore, \( \mathcal{L}\left\{\int_0^t h(\tau) d\tau\right\} = \frac{1}{s} \left(\frac{4}{s-3} + \frac{4}{(s-3)^2}\right) = \frac{4}{s(s-3)} + \frac{4}{s(s-3)^2} \)

We are given: \( \frac{A}{s} + \frac{B}{(s-3)^2} + \frac{C}{s-3} \) 
So, \( \frac{4}{s(s-3)} + \frac{4}{s(s-3)^2} = \frac{A}{s} + \frac{B}{(s-3)^2} + \frac{C}{s-3} \)

Combine the terms on the left-hand side: 
\( \frac{4(s-3) + 4}{s(s-3)^2} = \frac{4s - 12 + 4}{s(s-3)^2} = \frac{4s - 8}{s(s-3)^2} \)

Now do partial fraction decomposition: 
\( \frac{4s - 8}{s(s-3)^2} = \frac{A}{s} + \frac{B}{(s-3)^2} + \frac{C}{s-3} \)

Multiply both sides by \( s(s-3)^2 \): 
\( 4s - 8 = A(s-3)^2 + Bs + Cs(s-3) \)

Set \( s = 0 \):
\( -8 = 9A \Rightarrow A = -\frac{8}{9} \)

Set \( s = 3 \):
\( 4 = 3B \Rightarrow B = \frac{4}{3} \)

Expand: 
\( 4s - 8 = A(s^2 - 6s + 9) + Bs + C(s^2 - 3s) = (A+C)s^2 + (-6A + B - 3C)s + 9A \)

Comparing coefficients: \( A + C = 0 \Rightarrow C = \frac{8}{9} \)

Final coefficients:
\( A = -\frac{8}{9}, \quad B = \frac{4}{3}, \quad C = \frac{8}{9} \)

Find \( 3(A + B + C) \):
\( A + B + C = -\frac{8}{9} + \frac{4}{3} + \frac{8}{9} = \frac{4}{3} \)
So, \( 3(A + B + C) = 3 \times \frac{4}{3} = 4 \)

The final answer is: \( \boxed{4} \)