# Solve {(y''-3y'+2y=g),(y(0)=y'(0)=0):} whereas g(t)={(0,t>pi),(sin t,0<=t<=pi):}

Solve
$\left\{\begin{array}{rl}& {y}^{″}-3{y}^{\prime }+2y=g\\ & y\left(0\right)={y}^{\prime }\left(0\right)=0\end{array}$
whereas
$g\left(t\right)=\left\{\begin{array}{ll}0,& t>\pi \\ \mathrm{sin}t,& 0\le t\le \pi .\end{array}$
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Mateo Tate
Should be:
$\begin{array}{c}{y}^{″}-3{y}^{\prime }+2y=g\\ {s}^{2}Ly-3sLy+2Ly={\int }_{0}^{\pi }\mathrm{sin}t{e}^{-ts}dt\\ {s}^{2}Ly-3sLy+2Ly=\frac{1+{e}^{-\pi s}}{1+{s}^{2}}\\ Ly\left({s}^{2}-3s+2\right)=\frac{1+{e}^{-\pi s}}{1+{s}^{2}}\\ Ly=\frac{1+{e}^{-\pi s}}{\left(1+{s}^{2}\right)\left(s-1\right)\left(s-2\right)}=\left(1+{e}^{-\pi s}\right)\left(\frac{3}{10}\frac{s}{1+{s}^{2}}+\frac{1}{10}\frac{1}{1+{s}^{2}}+\frac{1}{5}\frac{1}{s-1}-\frac{1}{2}\frac{1}{s-2}\right)\end{array}$
Let $Lz=\frac{3}{10}\frac{s}{1+{s}^{2}}+\frac{1}{10}\frac{1}{1+{s}^{2}}+\frac{1}{5}\frac{1}{s-1}-\frac{1}{2}\frac{1}{s-2}$
Then $z=\left(\frac{3}{10}\mathrm{cos}t+\frac{1}{10}\mathrm{sin}t+\frac{1}{5}{e}^{t}-\frac{1}{2}{e}^{2t}\right)u\left(t\right)$
So we have $Ly=\left(1+{e}^{-\pi s}\right)Lz=Lz+{e}^{-\pi s}Lz$
Therefore $y\left(t\right)=z\left(t\right)+z\left(t-\pi \right)$
Substitute to find: