1 Answer

Quantaliinuxite · Answer 1 · 2023-03-31T16:30:51+0000

6a. By the convolution theorem,

$L\{t^3\star e^(5t)\} = L\{t^3\} * L\{e^(5t)\} = \frac6{s^4} * \frac1{s-5} = \boxed{\frac5{s^4(s-5)}}$

6b. Similarly,

$L\{e^(3t)\star \cos(t)\} = L\{e^(3t)\} * L\{\cos(t)\} = \frac1{s-3} * \frac s{1+s^2} = \boxed{\frac s{(s-3)(s^2+1)}}$

7. Take the Laplace transform of both sides, noting that the integral is the convolution of
e^t and
f(t) .

$\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau$

$\implies \displaystyle F(s) = \frac3s - 4 F(s) G(s)$

where
g(t) = e^t . Then
$G(s) = \frac1{s-1}$ , and

$F(s) = \frac3s - \frac4{s-1} F(s) \implies F(s) = (\frac3s)/((s+3)/(s-1)) = 3(s-1)/(s(s+3))$

We have the partial fraction decomposition,

$(s-1)/(s(s+3)) = \frac13 \left(-\frac1s + \frac4{s+3}\right)$

Then we can easily compute the inverse transform to solve for f(t) :

$F(s) = -\frac1s + \frac4{s+3}$

$\implies \boxed{f(t) = -1 + 4e^(-3t)}$

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