Question

Consider the initial value problem

y′′+y=g(t),y(0)= 0,y′(0)=0,
{0​ if 0≤t<4
where g(t)={t 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) by 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

The Laplace transform of the differential equation
`\( y'' + y = g(t) \)` with initial conditions
`\( y(0) = 0 \)` and
`\( y'(0) = 0 \)` is expressed as
`\( s^2Y(s) - sy(0) - y'(0) + Y(s) = G(s) \),` where
`\( Y(s) \)` denotes the Laplace transform of
`\( y(t) \)` and
`\( G(s) \)` represents the Laplace transform of
`\( g(t) \).`This forms the corresponding algebraic equation in the Laplace domain without rearranging terms until the subsequent part of the problem.

Step-by-step explanation:

The given differential equation
`\( y'' + y = g(t) \)` is subjected to a Laplace transform, resulting in
`\( s^2Y(s) - sy(0) - y'(0) + Y(s) = G(s) \)`. Here,
`\( Y(s) \)` signifies the Laplace transform of
`\( y(t) \)` while
`\( G(s) \)` represents the Laplace transform of
`\( g(t) \).` The Laplace transform of the differential equation yields an algebraic equation in the Laplace domain, preserving the derivatives and initial conditions in terms of
`\( Y(s) \)`,
`\( y(0) \)`, and
`\( y'(0) \).`

The transformation of the differential equation to the Laplace domain facilitates solving differential equations using algebraic methods, streamlining the process by converting differential equations into simpler algebraic equations. Retaining the terms in the form
`\( s^2Y(s) - sy(0) - y'(0) + Y(s) = G(s) \)` before further manipulation aids in maintaining the integrity of the initial conditions within the Laplace domain, enabling subsequent steps to solve for
`\( Y(s) \)` by rearranging terms and solving algebraically.