Question Use the Laplace transform to solve the given equation. (Enter your answers as a comma-separated list. Hint: There are two solutions to a square root.) f(t)f(t - 1)dt = 5040t? FC) = 0

\begin{array}{l}\int_{0}^{t} f(\tau) f(t-\tau) d \tau=5040 t^{7} \\\Rightarrow \quad(f * f)(t)=5040 t^{7} \text {. } \\\text { [- con volution of } \\f \& g \text { is } f_{t} * g \\\left.=\int_{0}^{*} f(\tau) g(t-r) d \tau\right) \\\end{array}We shall solve this problem using Laplace transform.So, taking Laplace transform we get \rightarrow\begin{array}{l}L\{(f * f)(t)\}=5040 L\left\{t^{7}\right\} . \\\Rightarrow L\{f\} L\{f\}=5040 \cdot \frac{7 !}{s^{7+1}} \\\Rightarrow F(s) \cdot f(s)=\frac{5040 \times 7 !}{s^{8}} \quad[\because L\{f(t)\} \\\Rightarrow[F(s)]^{2}=\frac{5040 \times 7 !}{5^{8}} \\=F(s)] \\\Rightarrow \quad f(s)= \pm \frac{5040}{54} \\\Rightarrow L\{f(t)\}= \pm \frac{5040}{5^{4}} \\\Rightarrow \quad f(t)= \pm 5040 L^{-1}\left\{\frac{1}{54}\right\} \\\Rightarrow f(t)= \pm 5040 \cdot \frac{t^{3}}{31} \\= \pm 5040 \cdot \frac{t^{3}}{6} \\\therefore f(t)=840 t^{3},-840 t^{3} \\\end{array} ...