6a. By the convolution theorem,
[tex]L\{t^3\star e^{5t}\} = L\{t^3\} \times L\{e^{5t}\} = \dfrac6{s^4} \times \dfrac1{s-5} = \boxed{\dfrac5{s^4(s-5)}}[/tex]
6b. Similarly,
[tex]L\{e^{3t}\star \cos(t)\} = L\{e^{3t}\} \times L\{\cos(t)\} = \dfrac1{s-3} \times \dfrac s{1+s^2} = \boxed{\dfrac s{(s-3)(s^2+1)}}[/tex]
7. Take the Laplace transform of both sides, noting that the integral is the convolution of [tex]e^t[/tex] and [tex]f(t)[/tex].
[tex]\displaystyle f(t) = 3 - 4 \int_0^t e^\tau f(t - \tau) \, d\tau[/tex]
[tex]\implies \displaystyle F(s) = \dfrac3s - 4 F(s) G(s)[/tex]
where [tex]g(t) = e^t[/tex]. Then [tex]G(s) = \frac1{s-1}[/tex], and
[tex]F(s) = \dfrac3s - \dfrac4{s-1} F(s) \implies F(s) = \dfrac{\frac3s}{\frac{s+3}{s-1}} = 3\dfrac{s-1}{s(s+3)}[/tex]
We have the partial fraction decomposition,
[tex]\dfrac{s-1}{s(s+3)} = \dfrac13 \left(-\dfrac1s + \dfrac4{s+3}\right)[/tex]
Then we can easily compute the inverse transform to solve for f(t) :
[tex]F(s) = -\dfrac1s + \dfrac4{s+3}[/tex]
[tex]\implies \boxed{f(t) = -1 + 4e^{-3t}}[/tex]
