# Find the equation by applying the Laplace transform. displaystyle{y}^{{{left({4}right)}}}-{y}= sin{{h}}{t} y(0)=y'(0)=y"(0)=0 y'''(0)=1

Find the equation by applying the Laplace transform.
${y}^{\left(4\right)}-y=\mathrm{sin}ht$
$y\left(0\right)={y}^{\prime }\left(0\right)=y"\left(0\right)=0$
${y}^{‴}\left(0\right)=1$
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Step 1
The given equation is,
${y}^{4}-y=\mathrm{sinh}t\dots \left(1\right)$
The given initial condition is,
$y\left(0\right)=1\dots \left(2\right)$
$y\prime \left(0\right)=1\dots \left(3\right)$
$y{}^{″}\left(0\right)=1\dots \left(4\right)$
$y{}^{‴}\left(0\right)=1\dots \left(5\right)$
The Laplace transform is given as,
$L\left\{{F}^{n}\left(t\right)\right\}={s}^{n}L\left\{F\left(t\right)\right\}-{s}^{n-1}F\left(0\right)-{s}^{n-2}F\prime \left(0\right)-\dots -{F}^{n-1}\left(0\right)$
$L\left\{\mathrm{sin}hat\right\}=\frac{a}{{s}^{2}-{a}^{2}}$
Step 2
The given equation (1) is,
${y}^{4}-y=\mathrm{sinh}t$
Taking Laplace transform on both sides,
$L\left\{y{}^{⁗}-y\right\}=L\left\{\mathrm{sin}ht\right\}$
${s}^{4}L\left\{y\right\}-{s}^{3}y\left(0\right)-{s}^{2}y\prime \left(0\right)-sy{}^{″}\left(0\right)-y{}^{‴}\left(0\right)-L\left\{y\right\}=\frac{1}{{s}^{2}-1}$
On putting the values of equation (2), (3), (4) & (5) in the above equation,
${s}^{4}L\left\{y\right\}-{s}^{3}\left(1\right)-{s}^{2}\left(1\right)-s\left(1\right)-\left(1\right)-L\left\{y\right\}=\frac{1}{{s}^{2}-1}$
${s}^{4}L\left\{y\right\}-{s}^{3}-{s}^{2}-s-1-L\left\{y\right\}=\frac{1}{{s}^{2}-1}$
${s}^{4}L\left\{y\right\}-L\left\{y\right\}=\frac{1}{{s}^{2}-1}+{s}^{3}+{s}^{2}+s+1$
$\left({s}^{4}-1\right)L\left\{y\right\}=\frac{1+{s}^{5}-{s}^{3}+{s}^{4}-{s}^{2}+{s}^{3}-s+{s}^{2}-1}{{s}^{2}-1}$
$\left({s}^{4}-1\right)L\left\{y\right\}=\frac{{s}^{5}+{s}^{4}-s}{{s}^{2}-1}$