Question

use $f*g=int_{0}^{t}f(\tau)g(t - \tau)d\tau$ and $mathcal{l}^{-1}{f(s)g(s)}=f*g$ to solve the given initial - value problem. let $f(s)=mathcal{l}{f(t)}$.
$y-8y+16y = f(t)$, $y(0)=0$, $y(0)=6$
$y(t)=(quad)+int_{0}^{t}f(\tau)cdot(quad)d\tau$

Explanation:

Step1: Take Laplace transform

Take Laplace transform of \(y'' - 8y' + 16y=f(t)\). Using \(L\{y''\}=s^{2}Y(s)-sy(0)-y'(0)\), \(L\{y'\}=sY(s)-y(0)\) and \(L\{y\}=Y(s)\), \(L\{f(t)\}=F(s)\).
Since \(y(0) = 0\) and \(y'(0)=6\), we have \((s^{2}Y(s)- 6)-8(sY(s)) + 16Y(s)=F(s)\).

Step2: Solve for \(Y(s)\)

Rearrange the equation: \(Y(s)(s^{2}-8s + 16)-6=F(s)\).
Factor \(s^{2}-8s + 16=(s - 4)^{2}\), so \(Y(s)=\frac{6}{(s - 4)^{2}}+\frac{F(s)}{(s - 4)^{2}}\).

Step3: Take inverse Laplace transform

Use \(L^{-1}\{\frac{n!}{(s - a)^{n+1}}\}=t^{n}e^{at}\) and \(L^{-1}\{F(s)G(s)\}=f*g\).
For \(\frac{6}{(s - 4)^{2}}\), \(n = 1,a = 4\), \(L^{-1}\{\frac{6}{(s - 4)^{2}}\}=6te^{4t}\).
For \(\frac{F(s)}{(s - 4)^{2}}\), \(G(s)=\frac{1}{(s - 4)^{2}}\), \(L^{-1}\{\frac{1}{(s - 4)^{2}}\}=te^{4t}\), so \(L^{-1}\{\frac{F(s)}{(s - 4)^{2}}\}=\int_{0}^{t}f(\tau)e^{4(t-\tau)}(t - \tau)d\tau\).

Answer:

\(y(t)=6te^{4t}+\int_{0}^{t}f(\tau)e^{4(t-\tau)}(t - \tau)d\tau\)