 # Let x(t) be the solution of the initial-value problem(a) Find the Laplace transform F(s) of the forcing f(t).(b) Find the Laplace transform X(s) of the solution x(t).x"+8x'+20x=f(t)x(0)=-3x'(0) necessaryh 2020-12-25 Answered

Let x(t) be the solution of the initial-value problem
(a) Find the Laplace transform F(s) of the forcing f(t).
(b) Find the Laplace transform X(s) of the solution x(t).
$x"+8{x}^{\prime }+20x=f\left(t\right)$
$x\left(0\right)=-3$
${x}^{\prime }\left(0\right)=5$

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given differential equation is x
$x"+8{x}^{\prime }+20x=f\left(t\right)$
$x\left(0\right)=-3$
${x}^{\prime }\left(0\right)=5$

$L\left[{x}^{\prime }\right]=sX\left(s\right)-x\left(0\right)$
$L\left[{t}^{n}\right]=\frac{n!}{{s}^{n+1}},L\left\{{e}^{-at}\right\}=\frac{1}{s+a}$
$\text{Step 2}$
$\text{Taking Laplace tranform of equation}$
$L\left[x"\right]+8L\left[{x}^{\prime }\right]+20L\left[x\right]=L\left[{t}^{2}\right]+4{e}^{2}L\left\{{e}^{-t}\right\}$
${s}^{2}\left(X\left(s\right)\right)-sX\left(0\right)-{x}^{\prime }\left(0\right)+8\left[s\left(X\left(s\right)\right)-x\left(0\right)\right]+20X\left(s\right)=\frac{2!}{{s}^{2+1}}+\frac{4{e}^{2}}{s+1}$
$\therefore {s}^{2}X\left(s\right)-s\left(-3\right)-5+8\left(sX\left(s\right)-5\right)+20X\left(s\right)=\frac{2}{{s}^{3}}+\frac{4{e}^{2}}{s+1}$
$\therefore \left({s}^{2}+8s+20\right)X\left(s\right)+3s-5-40=\frac{2}{{s}^{3}}+\frac{4{e}^{2}}{s+1}$
$\therefore \left({s}^{2}+8s+20\right)X\left(s\right)+3s-45=\frac{2}{{s}^{3}}+\frac{4{e}^{2}}{s+1}$
$\therefore X\left(s\right)\left[{s}^{2}+8s+20\right]=\frac{2}{{s}^{3}}+\frac{4{e}^{2}}{s+1}+45-3s$
$\therefore x\left(s\right)=\frac{1}{{s}^{2}+8s+20}\left\{\left(45-3s\right)+\frac{2}{{s}^{3}}+\frac{4{e}^{2}}{s+1}\right\}$