# Solve the following initial value problem by using Laplace transform. y"+4y=4t+12 , y(0)=4 , y'(0)=0

Solve the following initial value problem by using Laplace transform
${y}^{″}+4y=4t+12,y\left(0\right)=4,{y}^{\prime }\left(0\right)=0$

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odgovoreh

${y}^{″}+4y=4t+12,y\left(0\right)=4,{y}^{\prime }\left(0\right)=0$
Taking the Laplace transform on both side
$L\left\{y"\right\}+4L\left\{y\right\}=4L\left\{t\right\}+12L\left\{1\right\}$
${s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)+4L\left\{y\right\}=4\frac{1}{{s}^{2}}+12\frac{1}{s}$
$\left({s}^{2}+4\right)L\left\{y\right\}-4s-0=4\left(\frac{3s+1}{s}\right)$
$\left({s}^{2}+4\right)L\left\{y\right\}=4\left(\frac{3s+1}{s}+s\right)$
$\left({s}^{2}+4\right)L\left\{y\right\}=4\left(\frac{3s+1+{s}^{2}}{s}\right)$
$L\left\{y\right\}=4\left(\frac{{s}^{2}+3s+1}{s\left({s}^{2}+4\right)}\right)$
$=4\left(\frac{{s}^{2}+4+3s-3}{s\left({s}^{2}+4\right)}\right)$
$=4\left[\frac{1}{s}+\frac{3s-3}{s\left({s}^{2}+4\right)}\right]$
$=4\left[\frac{1}{s}+\frac{1}{4}\left(\frac{12s-12}{s\left({s}^{2}+4\right)}\right)\right]$
$=4\left[\frac{1}{s}+\frac{1}{4}\left(\frac{3{s}^{2}-3{s}^{2}+12s-12}{s\left({s}^{2}+4\right)}\right)\right]$
$=4\frac{1}{s}+\frac{s\left(3s+12\right)-3\left({s}^{2}+4\right)}{s\left({s}^{2}+4\right)}$
$L\left\{y\right\}=\frac{4}{s}+\frac{3s+12}{{s}^{2}+4}-\frac{3}{s}$
$L\left\{y\right\}=\frac{1}{s}+3\frac{s}{{s}^{2}+4}+\frac{12}{{s}^{2}+4}$
$y\left(t\right)={L}^{-1}\left\{\frac{1}{s}\right\}+3{L}^{-1}\left\{\frac{s}{{s}^{2}+4}\right\}+6{L}^{-1}\left\{\frac{2}{{s}^{2}+4}\right\}$
$y\left(t\right)=1+3\mathrm{cos}\left(2t\right)+6\mathrm{sin}\left(2t\right)$