# Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below. y"+y=g(t) , y(0)=-4 , y'(0)=0 where g{{left({t}right)}}={leftlbracebegin{matrix}{t}&{t}<{4}{5}&{t}>{4}end{matrix}right.} Y(s)-?

Solve for Y(s), the Laplace transform of the solution y(t) to the initial value problem below.
$y"+y=g\left(t\right),y\left(0\right)=-4,{y}^{\prime }\left(0\right)=0$
where $g\left(t\right)=\left\{\begin{array}{cc}t& t<4\\ 5& t>4\end{array}$
Y(s)-?
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Step 1 Basics
Here we use Second Shifting Property and Laplace Transform of Derivative as follows
Step 1
Laplace transform of derivative
$L\left\{f\left(t\right)\right\}={s}^{2}L\left\{f\right\}-\mathsf{0}-{f}^{\prime }\left(0\right)$
second shifting property
$L\left\{f\left(t-{t}_{0}\right)\right\}={e}^{-s{t}_{0}}L\left\{f\left(t\right)\right\}$
Step 2 Explanation
$y"+y=g\left(t\right),y\left(0\right)=-4,{y}^{\prime }\left(0\right)=0$
where $g\left(t\right)=\left\{\begin{array}{cc}t& t<4\\ 5& t>4\end{array}$
Taking Laplace transform both side of above equation , we get
$L\left\{y\right\}+L\left\{y\right\}=L\left\{g\left(t\right)\right\}$
$\left[{s}^{2}L\left\{y\left(t\right)\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)\right]+L\left\{y\right\}=L\left\{tu\left(t\right)+5u\left(t-u\right)\right]$
where u(t) is the step function

shifting property as $L\left\{f\left(t-{t}_{0}\right)\right\}={e}^{-s{t}_{0}}L\left\{f\left(t\right)\right\}$
${s}^{2}L\left\{y\right\}+4s+L\left\{y\right\}=\frac{1}{{s}^{2}}+{e}^{-4s}\left(\frac{5}{s}\right)$
$L\left\{y\right\}\left[{s}^{2}+1\right]=\frac{1}{{s}^{2}}+\frac{5{e}^{-4s}}{s}-4s$
$L\left\{y\right\}=\frac{1}{{s}^{2}\left({s}^{2}+1\right)}+\frac{5{e}^{-4s}}{s\left({s}^{2}+1\right)}-4\frac{s}{{s}^{2}+1}$
$y\left(s\right)=\frac{1}{{s}^{2}\left({s}^{2}+1\right)}+\frac{5{e}^{-4s}}{s\left({s}^{2}+1\right)}-4\frac{s}{{s}^{2}+1}$