Use the Laplace transform to solve the following initial value problem:

$y{}^{\u2033}+y=2t$

with$y\left(\frac{\pi}{4}\right)=\frac{\pi}{2}$ and $y}^{\prime}\left(\frac{\pi}{4}\right)=2-\sqrt{2$

with

Erika Bernard
2022-04-08
Answered

Use the Laplace transform to solve the following initial value problem:

$y{}^{\u2033}+y=2t$

with$y\left(\frac{\pi}{4}\right)=\frac{\pi}{2}$ and $y}^{\prime}\left(\frac{\pi}{4}\right)=2-\sqrt{2$

with

You can still ask an expert for help

llevochalecoiozq

Answered 2022-04-09
Author has **15** answers

Laplace-transforming both sides of $y{}^{\u2033}+y=2t$

$s}^{2}Y\left(s\right)-{y}_{0}s-{v}_{0}+Y\left(s\right)=\frac{2}{{s}^{2}$

Hence

$Y\left(s\right)={y}_{0}\left(\frac{s}{{s}^{2}+1}\right)+({v}_{0}-2)\left(\frac{1}{{s}^{2}+1}\right)+\frac{2}{{s}^{2}}$

Taking the inverse Laplace transform,

$y\left(t\right)={y}_{0}\mathrm{cos}\left(t\right)+({v}_{0}-2)\mathrm{sin}\left(t\right)+2t$

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

${y}_{0}+{v}_{0}=2{\textstyle \phantom{\rule{2em}{0ex}}}{\textstyle \phantom{\rule{2em}{0ex}}}{\textstyle \phantom{\rule{2em}{0ex}}}{y}_{0}-{v}_{0}=2$

Thus${y}_{0}=2$ and ${v}_{0}=0$ , and

$y\left(t\right)=2\mathrm{cos}\left(t\right)-2\mathrm{sin}\left(t\right)+2t$

Hence

Taking the inverse Laplace transform,

From the conditions

Thus

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