# Use the Laplace transform to solve the following

Use the Laplace transform to solve the following initial value problem:
$y{}^{″}+y=2t$
with $y\left(\frac{\pi }{4}\right)=\frac{\pi }{2}$ and ${y}^{\prime }\left(\frac{\pi }{4}\right)=2-\sqrt{2}$
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Laplace-transforming both sides of $y{}^{″}+y=2t$
${s}^{2}Y\left(s\right)-{y}_{0}s-{v}_{0}+Y\left(s\right)=\frac{2}{{s}^{2}}$
Hence
$Y\left(s\right)={y}_{0}\left(\frac{s}{{s}^{2}+1}\right)+\left({v}_{0}-2\right)\left(\frac{1}{{s}^{2}+1}\right)+\frac{2}{{s}^{2}}$
Taking the inverse Laplace transform,
$y\left(t\right)={y}_{0}\mathrm{cos}\left(t\right)+\left({v}_{0}-2\right)\mathrm{sin}\left(t\right)+2t$
From the conditions $y\left(\frac{\pi }{4}\right)=\frac{\pi }{2}$ and ${y}^{\prime }\left(\frac{\pi }{4}\right)=2-\sqrt{2}$ ,we get
${y}_{0}+{v}_{0}=2\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}{y}_{0}-{v}_{0}=2$
Thus ${y}_{0}=2$ and ${v}_{0}=0$, and
$y\left(t\right)=2\mathrm{cos}\left(t\right)-2\mathrm{sin}\left(t\right)+2t$