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

Step 1
Given initial value problem is $y"+4y=4t$
$y(0)=1$
$y^{\prime }(0)=5$
We find solution by using the Laplace transform
Step 2
Solution:
Given initial value problem is $y"+4y=4ty(0)=1y^{\prime }(0)=5$
Taking Laplace transform on both side we get
$L\{y^{″}+4y\}=L\left\{4t\right\}$
$\Rightarrow L\left\{y^{″}\right\}+4L\left\{y\right\}=4L\left\{t\right\}$
$\Rightarrow (s^{2}y(s)-sy(0)-y^{\prime }(0))+4y(s)=\frac{4}{s}$
$\Rightarrow s^{2}y(s)-s(1)-5+4y(s)=\frac{4}{s^2}$
$\Rightarrow s^{2}y(s)-s-5+4y(s)=\frac{4}{s^2}$
$=(s^2+4)y(s)=\frac{4}{s^2}+s+5$
$\Rightarrow y(s)=\frac{4}{s^2(s^2+4)}+\frac{s}{s^2+4}+\frac{5}{s^2+4}$
Now e apply inverse Laplace transform to get solution of the IVP
$y(t)y(t)=L^{-1}[\frac{4}{s^2(s^2+4)}+\frac{s}{(s^2+4)}+\frac{5}{(s^2+4)}]$
$L^{-1}\left[\frac{4}{s^2(s^2+4)}\right]+L^{-1}\left[\frac{s}{s^2+4}\right]+L^{-1}\left[\frac{5}{s^2+4}\right]$
$L^{-1}[\frac{1}{s^2}-\frac{1}{s^2+4}]+L^{-1}\left[\frac{s}{s^2+4}\right]+L^{-1}\left[\frac{5}{s^2+4}\right]\dots \{\text{partial fraction }\frac{4}{s^2(s^2+4)}=\frac{1}{s^2}-\frac{1}{s^2+4}\}$
$L^{-1}\left[\frac{1}{s^2}\right]-L^{-1}\left[\frac{1}{s^2+4}\right]+L^{-1}\left[\frac{s}{s^2+4}\right]+5L^{-1}\left[\frac{1}{s^2+4}\right]$
$=t-\frac{1}{2}\sin (2t)+\cos (2t)+\frac{5}{2}\sin (2t)$
$=t+\frac{4}{2}\sin (2t)+\cos (2t)$
$y(t)=t+2\sin (2t)+\cos (2t)$
Therefore solution of the IVP $y^{″}+4y=4t,y(0)=1,y^{\prime }(0)=5\text{ is }y(t)=t+2\sin (2t)+\cos (2t)$
Step 3
Answer:
The solution of the IVP

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