Use the Laplace transform co solve the following initial value problem:

$y^{\text{'}\text{'}}+5y^{\text{'}}-36y=0,y(0)=-3,y^{\text{'}}(0)=1$

a$.$ First, using $Y$ for the Laplace transform of $y(t),$ i$.$e$.,Y=L\{y(t)\},$ find the equation you get by taking the Laplace transform of the differential equation

$=0$

b$.$ Now solve for

$Y(s)=-\frac{(3s+14)}{(s+9)(s-4)}$

c$.$ Write the above answer in ies partial fraction decomposition, $Y(s)=\frac{A}{s+a}+\frac{b}{s1b}$ where

Consider the initial value problem

$y^{\text{'}\text{'}}+9y-cos(3y),y(0)=2,y^{\text{'}}(0)=8$

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$).$

$(s^{2}Y(s)-8s-8)+9Y(s)-\frac{s}{s^2+9},$ help $(formulas)$

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

$Y(s)=L\{y(t)\}=\frac{2s+8}{s^2+9}+\frac{s}{(s^2+9{)}^2}$

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

$y(t)=$

$$