Consider the initial value problem for $a{y}^{\text{'}\text{'}}+b{y}^{\text{'}}+cy=f(t),y(0)=0,y^{\text{'}}(0)=0a,b,cf(t)f(t)y(t)Y(s)=(s)F(s)(s)=\frac{1}{a{s}^2+bs+c}f(t)=9ty(t)=2(e^{-3t}-1)+t(e^{-3t}+5),t0Y(s)=L\{y(t)\}F(s)=L\{f(t)\}Y(s)=F(s)=(s)(s)=\frac{2(\frac{1}{s+3}+\frac{1}{s})+(\frac{1}{(s+3{)}^2}+\frac{5}{s^2})}{\frac{9}{s^2}}0$ :

$a{y}^{\text{'}\text{'}}+b{y}^{\text{'}}+cy=f(t),y(0)=0,y^{\text{'}}(0)=0$

where $a,b,c$ are constants and $f(t)is$ a known function. $We$ can view this problem $as$ defining a linear system, where $f(t)is$ a known input and the corresponding solution $y(t)is$ the output. Laplace transforms $of$ the input and output functions satisfy the multiplicative relation

$Y(s)=(s)F(s)$

where $(s)=\frac{1}{a{s}^2+bs+c}is$ the system transfer function.

Suppose $an$ input $f(t)=9t,$ when applied $to$ the linear system above, produces the output $y(t)=2(e^{-3t}-1)+t(e^{-3t}+5),t0.$

$a.$ Find $Y(s)=L\{y(t)\}$ and $F(s)=L\{f(t)\}.$

$Y(s)=$

$$

$$

$F(s)=$

$b.$ Use your answer $to$ part $(a)to$ find the system transfer function, $(s).$

$(s)=\frac{2(\frac{1}{s+3}+\frac{1}{s})+(\frac{1}{(s+3{)}^2}+\frac{5}{s^2})}{\frac{9}{s^2}}$

$$

$(formulas).$