$\backslash$table$[[$Entered$,$Answer Preview,Result$],[Y(s{)}^{**}[(s^2)+36],Y(s)(s^2+36),$correct$],[[\frac{1}{s^2}]-(\frac{e^{-5^{**}s}}{s^2})-[e^{-5^{**}s}{]}^{**}(\frac{5}{s}),\frac{1}{s^2}-\frac{e^{-5s}}{s^2}-e^{-5s}\frac{5}{s},$correct$],[\frac{1-[e^{-5^{**}s}]-5^{**}{s}^{**}[e^{-5^{**}s}]}{(s^2{)}^{**}[(s^2)+36]},\frac{1-e^{-5s}-5s{e}^{-5s}}{s^2(s^2+36)},$correct$],[\frac{6^{**}t-6^{**}{t}^{**}(t-5)+{36}^{**}(t-5{)}^{**}cos(6^{**}(t-5))-sin(6^{**}t)+(t-5{)}^{**}sin(6^{**}(t-5))}{216},\frac{6t-6t(t-5)+36(t-5)cos(6(t-5))-sin(6t)+(t-5)sin(6(t-5))}{216},$incorrect$]]$

At least one of the answers above is NOT correct.

$(1$ point$)$ Consider the initial value problem

$y^{\text{'}\text{'}}+36y=g(t),y(0)=0,y^{\text{'}}(0)=0$

where $g(t)=\{\begin{array}{ccc}t& if& 0t<mn>5</mn>\\ 0& if& 5t<mi></mi>\end{array}$

a$.$ Take the Laplace transform of both sides of the given differential equation to create the corresponding algebraic equation. Denote the Laplace transform of $y(t)$ by $Y(s).$ Do not move any terms from one side of the equation to the other $($until you get to part $($b$)$ below$).$

$$

$$ help $($formulas$)$

b$.$ Solve your equation for $Y(s).$

$Y(s)=L\{y(t)\}=\frac{1-e^{-5(s)}-5s{e}^{-5(s)}}{s^2(s^2+36)}$

c$.$ Take the inverse Laplace transform of both sides of the previous equation to solve for $y(t).$

$y(t)=\frac{6t-6t(t-5)+36(t-5)cos(6(t-5))-sin(6t)+(t-5)sin(6(t-5))}{216}$