- Let $\operatorname{y}\left(t\right)$ be a continuous function on $\left[0,\infty\right)$ whose Laplace transforms exists. If $\operatorname{y}\left(t\right)$ satisfies \begin{equation} \int_{0}^{t}\left[1 - \cos\left(t - \tau\right)\right] \operatorname{y}\left(\tau\right) \,\mathrm{d}\tau = t^{4}\label{1}\tag{1} \end{equation} then $\operatorname{y}\left(1\right) = $
- I was able to find the answer by differentiating ( Leibniz integration $\mbox{rule )}\left(\ref{1}\right)$ twice to get $$ \int_{0}^{t}\cos\left(t - \tau\right) \operatorname{y}\left(\tau\right) \,\mathrm{d}\tau = 12t^{2}\,, $$ and subtracting with original equation and differentiating again to get $\operatorname{y}\left(t\right) = 4t^{3} + 24t^{2}$ and hence $\operatorname{y}\left(1\right) = 28$.
One's allowed only less than a minute per question in the exam, and that makes me doubt if there's a simpler way to do it $!$.
Is there a simpler way to do it $?$. Thanks in advance.
Laplace transform of the convolution of two functions is the product of the Laplace transforms of the functions. Thus, take the LT of both sides:
Let
$$\int_0^{\infty} dt \, y(t) e^{-s t} = Y(s)$$ $$\int_0^{\infty} dt \, (1-\cos{t}) e^{-s t} = \frac1{s} - \frac{s}{1+s^2} = \frac1{s(1+s^2)}$$
$$\int_0^{\infty} dt \, t^4 e^{-s t} = \frac{4!}{s^5} $$
Thus,
$$\frac{Y(s)}{s(1+s^2)} = \frac{24}{s^5} \implies Y(s) = 24 \left (\frac1{s^2} + \frac1{s^4} \right )$$
We can invert this LT very easily to recover $y(t)$. It is
$$y(t) = 24 t + 4 t^3$$
and $y(1)=\cdots$