 # Solve the following differential equations using the Laplace transform and the unit step function y"+4y=g(t) y(0)=-1 y'(0)=0 , text{ where } g(t)=begin{cases}t &, tleq 25 & ,t > 2end{cases} y"-y=g(t) y(0)=1 y'(0)=2 , text{ where } g(t)=begin{cases}1 &, tleq 3t & ,t > 3end{cases} Carol Gates 2021-02-02 Answered
Solve the following differential equations using the Laplace transform and the unit step function
$y"+4y=g\left(t\right)$
$y\left(0\right)=-1$

$y"-y=g\left(t\right)$
$y\left(0\right)=1$
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Step 1 Given :The differential equations using the Laplace Transform and the unit step function
$y"+4y=g\left(t\right)$
$y\left(0\right)=-1$

Step 2
The given initial value problem is
$y"+4y=g\left(t\right)$
$y\left(0\right)=-1$

Write the step using the Heaviside step functions.
If f(t) is a function of t for $t\ge 0$ whose Laplace Transform F(s) exists , then for any constant $a\ge 0$ , the function

has the Laplace Transform ${e}^{-as}F\left(s\right)$
$g\left(t\right)=\left(2-u\left(t-2\right)\right)+5u\left(t-2\right)=2-u\left(t-2\right)+5u\left(t-2\right)$
$g\left(t\right)=2+4u\left(t-2\right)$ So, the initial value problem can be written as
$y"+4y=2+4u\left(t-2\right),y\left(0\right)=-1,{y}^{\prime }\left(0\right)=0$
Taking the Laplace Transform of both sides , we get
${s}^{2}\overline{y}-sy\left(0\right)-{y}^{\prime }\left(0\right)+4\overline{y}=\frac{2}{s}+4{e}^{-2s}$
${s}^{2}\overline{y}-s\left(-1\right)+4\overline{y}=\frac{2}{s}+4{e}^{-2s}$
${s}^{2}\overline{y}+s+4\overline{y}=\frac{2}{s}+4{e}^{-2s}$
$\left({s}^{2}+4\right)\overline{y}=\frac{2}{s}+4{e}^{-2s}-s$
$\overline{y}=\frac{2}{s\left({s}^{2}+4\right)}+4\frac{{e}^{-2s}}{{s}^{2}+4}-\frac{s}{{s}^{2}+4}$
Taking Inverse Laplace Transform , we get

$L\left(1\right)=\frac{1}{s},{L}^{-1}\left(\frac{s}{{s}^{2}+4}\right)=\mathrm{cos}at,{L}^{-1}\left(\frac{1}{{s}^{2}+4}\right)=\frac{1}{a}\mathrm{sin}at$
${L}^{-}1\left[\overline{y}\right]={L}^{-1}\left[\frac{2}{s\left({s}^{2}+4\right)}\right]+4{L}^{-1}\left[\frac{{e}^{-2s}}{{s}^{2}+4}\right]-{L}^{-1}\left[\frac{s}{{s}^{2}+4}\right]$
$y=\frac{2}{2}{\int }_{0}^{t}\mathrm{sin}2tdt+\frac{4{e}^{-2s}}{2}\mathrm{sin}2t-\mathrm{cos}2t$

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