# Solution of the following initial value problem using the Laplace transform y"+4y=4t y(0)=1 y'(0)=5

Solution of the following initial value problem using the Laplace transform
$y"+4y=4t$
$y\left(0\right)=1$
${y}^{\prime }\left(0\right)=5$
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Step 1
Given initial value problem is $y"+4y=4t$
$y\left(0\right)=1$
${y}^{\prime }\left(0\right)=5$
We find solution by using the Laplace transform
Step 2
Solution:
Given initial value problem is $y"+4y=4ty\left(0\right)=1{y}^{\prime }\left(0\right)=5$
Taking Laplace transform on both side we get
$L\left\{{y}^{″}+4y\right\}=L\left\{4t\right\}$
$⇒L\left\{{y}^{″}\right\}+4L\left\{y\right\}=4L\left\{t\right\}$
$⇒\left({s}^{2}y\left(s\right)-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)+4y\left(s\right)=\frac{4}{s}$
$⇒{s}^{2}y\left(s\right)-s\left(1\right)-5+4y\left(s\right)=\frac{4}{{s}^{2}}$
$⇒{s}^{2}y\left(s\right)-s-5+4y\left(s\right)=\frac{4}{{s}^{2}}$
$=\left({s}^{2}+4\right)y\left(s\right)=\frac{4}{{s}^{2}}+s+5$
$⇒y\left(s\right)=\frac{4}{{s}^{2}\left({s}^{2}+4\right)}+\frac{s}{{s}^{2}+4}+\frac{5}{{s}^{2}+4}$
Now e apply inverse Laplace transform to get solution of the IVP
$y\left(t\right)y\left(t\right)={L}^{-1}\left[\frac{4}{{s}^{2}\left({s}^{2}+4\right)}+\frac{s}{\left({s}^{2}+4\right)}+\frac{5}{\left({s}^{2}+4\right)}\right]$
${L}^{-1}\left[\frac{4}{{s}^{2}\left({s}^{2}+4\right)}\right]+{L}^{-1}\left[\frac{s}{{s}^{2}+4}\right]+{L}^{-1}\left[\frac{5}{{s}^{2}+4}\right]$

${L}^{-1}\left[\frac{1}{{s}^{2}}\right]-{L}^{-1}\left[\frac{1}{{s}^{2}+4}\right]+{L}^{-1}\left[\frac{s}{{s}^{2}+4}\right]+5{L}^{-1}\left[\frac{1}{{s}^{2}+4}\right]$
$=t-\frac{1}{2}\mathrm{sin}\left(2t\right)+\mathrm{cos}\left(2t\right)+\frac{5}{2}\mathrm{sin}\left(2t\right)$
$=t+\frac{4}{2}\mathrm{sin}\left(2t\right)+\mathrm{cos}\left(2t\right)$
$y\left(t\right)=t+2\mathrm{sin}\left(2t\right)+\mathrm{cos}\left(2t\right)$
Therefore solution of the IVP
Step 3