clovnerie0q

2022-09-03

Using Laplace transform, solve the system:
${w}^{\prime }+y=\mathrm{sin}\left(x\right)$
${y}^{\prime }-z={e}^{x}$
${z}^{\prime }+w+y=1$
where w(0)=0 and z(0)=y(0)=1.

Colin Dougherty

Are w, y and z functions of x?
If so, hint:
${w}^{″}+{y}^{\prime }=\mathrm{cos}x$ from eq 1
$\to {w}^{″}+{e}^{x}+z=\mathrm{cos}x$
$\to {w}^{‴}+{e}^{x}+{z}^{\prime }=-\mathrm{sin}x$
$\to {w}^{‴}+{e}^{x}+\left(1-w-y\right)=-\mathrm{sin}x$
$\to {w}^{‴}-w=-\mathrm{sin}x-{e}^{x}-1+y$
$\to {w}^{‴}-w=-\mathrm{sin}x-{e}^{x}-1+\left(\mathrm{sin}\left(x\right)-{w}^{\prime }\right)$
$\to {w}^{‴}+{w}^{\prime }-w=-\mathrm{sin}x-{e}^{x}-1+\mathrm{sin}\left(x\right)$
$\to {w}^{‴}+{w}^{\prime }-w=-{e}^{x}-1$
Can you apply Laplace transform now?
Apply Laplace to get:
${s}^{3}F\left(s\right)-{s}^{2}w\left(0\right)-{s}^{\prime }{w}^{\prime }\left(0\right)-{s}^{0}{w}^{″}\left(0\right)+{s}^{\prime }F\left(s\right)-{s}^{0}w\left(0\right)-F\left(s\right)=-L\left({e}^{x}\right)-L\left(1\right)$

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