# Solve the following differential equation by the laplace transform method. 6y^{(4)}+7y'''+21y"+28y'-12y=t(\cos(2t)+te^{\frac{-3t}{2}}).

Solve the following differential equation by the laplace transform method
$6{y}^{\left(4\right)}+7y{}^{‴}+21y28{y}^{\prime }-12y=t\left(\mathrm{cos}\left(2t\right)+t{e}^{\frac{-3t}{2}}\right)$
$y\left(0\right)=0,{y}^{\prime }\left(0\right)=1,y"\left(0\right)=0,{y}^{‴}\left(0\right)=-1$

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Ezra Herbert
$6{y}^{\left(4\right)}+7y{}^{‴}+21y28{y}^{\prime }-12y=t\left(\mathrm{cos}\left(2t\right)+t{e}^{\frac{-3t}{2}}\right)$
$L\left\{6{y}^{\left(4\right)}+7{y}^{‴}+21y"+28{y}^{\prime }-12y\right\}=L\left\{t\left(\mathrm{cos}\left(2t\right)+t{e}^{\frac{-3t}{2}}\right)\right\}$
$6L\left\{{y}^{‴}\left(t\right)\right\}+7L\left\{{y}^{‴}\left(t\right)\right\}+21L\left\{y"\left(t\right)\right\}+28L\left\{{y}^{\prime }\left(t\right)\right\}+12L\left\{y\right\}=L\left\{t\left(\mathrm{cos}\left(2t\right)+t{e}^{\frac{-3t}{2}}\right)\right\}$
$6\left({s}^{4}L\left\{y\right\}-{s}^{3}y\left(0\right)-{s}^{2}{y}^{\prime }\left(0\right)-sy"\left(0\right)-{y}^{‴}\left(0\right)\right)+7\left({s}^{3}L\left\{y\right\}-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-y"\left(0\right)\right)+21\left({s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)+28L\left(sL\left\{y\right\}-y\left(0\right)\right)-12L\left\{y\right\}=-\left(-\frac{\left(s+2\right)\left(s-2\right)}{\left({s}^{2}+4{\right)}^{2}}-\frac{16}{\left(2s+3{\right)}^{3}}\right)$
Plug the initial conditions , $y\left(0\right)=0,{y}^{\prime }\left(0\right)=1,y0\right)=0,y{}^{‴}\left(0\right)=-1$
$6\left({s}^{4}L\left\{y\right\}-{s}^{3}y\left(0\right)-{s}^{2}{y}^{\prime }\left(0\right)-sy"\left(0\right)-{y}^{‴}\left(0\right)\right)+7\left({s}^{3}L\left\{y\right\}-{s}^{2}y\left(0\right)-s{y}^{\prime }\left(0\right)-y"\left(0\right)\right)+21\left({s}^{2}L\left\{y\right\}-sy\left(0\right)-{y}^{\prime }\left(0\right)\right)+28\left(sL\left\{y\right\}-y\left(0\right)\right)-12L\left\{y\right\}=-\left(-\frac{\left(s+2\right)\left(s-2\right)}{\left({s}^{2}+4{\right)}^{2}}-\frac{16}{\left(2s+3{\right)}^{3}}\right)$