Question

Consider the initial value problem
y′′+36y=g(t),y(0)=0,y′(0)=0,y″+36y=g(t),y(0)=0,y′(0)=0,
where g(t)={t0if 0≤t<4if 4≤t<[infinity]. g(t)={t if 0≤t<40 if 4≤t<[infinity].
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)y(t) by Y(s)Y(s). Do not move any terms from one side of the equation to the other (until you get to part (b) below).

Answer 1

Answer:

$s^2Y(s)+38Y(s)=g(s)$

Step-by-step explanation:

Given the second order differential equation with initial condition.

$y''+36y=g(t); \ y(0)=0, \ y'(0)=0, \ and \ y''(0)=1$

Take the Laplace transform of each side of the equation.

~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~

$\boxed{\left\begin{array}{ccc}\text{\underline{Laplace Transforms of DE's:}}\\\\L\{y''\}=s^2Y(s)-sy(0)-y'(0)\\\\\ L\{y'\}=sY(s)-y(0) \\\\ L\{y\}=Y(s)\end{array}\right}$

Taking the Laplace transform of the DE.

$y''+36y=g(t); \ y(0)=0, \ y'(0)=0, \ and \ y''(0)=1\\\\\Longrightarrow L\{y''\}+38L\{y\}=L\{g(t)\}\\\\\Longrightarrow s^2Y(s)-s(0)-0+38Y(s)=g(s)\\\\\Longrightarrow \boxed{\boxed{s^2Y(s)+38Y(s)=g(s)}}$

Thus, the Laplace transform has been applied.