\(\text{Step 1 }\)

\(\text{ (a) }\)

\(\text{Formula used: }\)

\(\text{Let f and g be two functions. The convolution of f with g is defined as } (f*g)(t)=\int_0^t f(w)g(t-w)dw\)

\(\text{Calculation: }\)

\(\text{Here given that } f(t)=e^t \text{ and } g(t)=e^{-2t}\)

\(\text{Then the convolution of f with g is given below:}\)

\((f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau\)

\(=\int_0^t (e^{w})(e^{-2(t-w)})dw\)

\(\int_0^t (e^{3w})(e^{-2w})dw\)

\(=\frac{1}{3}e^{-2t}\left[e^{3w}\right]_0^t\)

\(=\frac{1}{3}e^{-2t}\left[e^{3t}-1\right]\)

\(\text{Step 2}\)

\(\text{ (b) }\)

\(\text{Here, the Laplace transform of f is }L(f(t))=L(e^t)=\frac{1}{s-1}\)

\(\text{The Laplace transform of g is } L(g(t))=L(e^{-2t})=\frac{1}{s+2}\)

\(\text{Now, find the inverse of the convolution of f with g as follows: }\)

\((f*g)(t)=L^{-1}\left\{F(s)G(s)\right\}\)

\(=L^{-1}\left\{\frac{1}{s-1} \times \frac{1}{s+2}\right\}\)

\(=L^{-1}\left\{\frac{1}{s-1}\right\} * L^{-1}\left\{\frac{1}{s+2}\right\}\)

\(=e^t * e^{-2t}\)

\(\text{ (a) }\)

\(\text{Formula used: }\)

\(\text{Let f and g be two functions. The convolution of f with g is defined as } (f*g)(t)=\int_0^t f(w)g(t-w)dw\)

\(\text{Calculation: }\)

\(\text{Here given that } f(t)=e^t \text{ and } g(t)=e^{-2t}\)

\(\text{Then the convolution of f with g is given below:}\)

\((f*g)(t)=\int_0^t f(\tau)g(t-\tau)d\tau\)

\(=\int_0^t (e^{w})(e^{-2(t-w)})dw\)

\(\int_0^t (e^{3w})(e^{-2w})dw\)

\(=\frac{1}{3}e^{-2t}\left[e^{3w}\right]_0^t\)

\(=\frac{1}{3}e^{-2t}\left[e^{3t}-1\right]\)

\(\text{Step 2}\)

\(\text{ (b) }\)

\(\text{Here, the Laplace transform of f is }L(f(t))=L(e^t)=\frac{1}{s-1}\)

\(\text{The Laplace transform of g is } L(g(t))=L(e^{-2t})=\frac{1}{s+2}\)

\(\text{Now, find the inverse of the convolution of f with g as follows: }\)

\((f*g)(t)=L^{-1}\left\{F(s)G(s)\right\}\)

\(=L^{-1}\left\{\frac{1}{s-1} \times \frac{1}{s+2}\right\}\)

\(=L^{-1}\left\{\frac{1}{s-1}\right\} * L^{-1}\left\{\frac{1}{s+2}\right\}\)

\(=e^t * e^{-2t}\)