iarc6io

2022-07-17

Using Laplace transforms calculate the solution of the ordinary differential equation $\frac{{d}^{2}y}{d{t}^{2}}-2\frac{dy}{dt}+y={e}^{-t}$, subject to the initial conditions y(0)=0 and $\frac{dy}{dt}\left(0\right)=1$. Select the correct answer from the following list.
$a\right)y\left(t\right)=\frac{3t{e}^{t}}{2}+\frac{{e}^{-t}}{4}-\frac{{e}^{t}}{4}\phantom{\rule{0ex}{0ex}}b\right)y\left(t\right)=t{e}^{t}+\frac{{t}^{2}{e}^{t}}{2}\phantom{\rule{0ex}{0ex}}c\right)y\left(t\right)=\frac{t{e}^{-t}}{2}-\frac{{e}^{-t}}{4}+\frac{{e}^{t}}{4}\phantom{\rule{0ex}{0ex}}d\right)y\left(t\right)=t{e}^{-t}+\frac{{t}^{2}{e}^{-t}}{2}\phantom{\rule{0ex}{0ex}}e\right)y\left(t\right)=\frac{t{e}^{-t}}{2}-\frac{{e}^{-t}}{2}+\frac{{e}^{t}}{2}$

sweetwisdomgw

Auxillary equation and given by
${m}^{2}=2m+1=0⇒\left(m-1{\right)}^{2}=0⇒m=1,1$
complementary solution is
${y}_{2}\left(x\right)=\left({c}_{1}+{c}_{2}t\right){e}^{t}$ where ${c}_{1}$ and ${c}_{2}$ are some constant
Particular integral
$\frac{1}{{D}^{2}-2D+1}{e}^{-t}=\frac{1}{\left(-1{\right)}^{2}-2\left(-1\right)+1}{e}^{-t}=\frac{1}{4}{e}^{-t}$
complete solution (y)
$y\left(t\right)=\left({c}_{1}+{c}_{2}t\right){e}^{t}+\frac{1}{4}{e}^{-t}+\frac{1}{4}{e}^{-t}$
using integral conclusion
$y\left(0\right)={c}_{1}+\frac{1}{4}⇒{c}_{1}=-\frac{1}{4}\phantom{\rule{0ex}{0ex}}{y}^{\prime }\left(t\right)=\left({c}_{1}+{c}_{2}t\right){e}^{t}+{e}^{t}×{c}_{2}-\frac{1}{4}{e}^{-t}\phantom{\rule{0ex}{0ex}}{y}^{\prime }\left(0\right)={c}_{1}{e}^{0}+{c}_{2}-\frac{1}{4}{e}^{-0}⇒{c}_{1}+{c}_{2}-\frac{1}{4}\phantom{\rule{0ex}{0ex}}⇒y\left(t\right)=-\frac{1}{4}{e}^{t}+\frac{3}{2}t{e}^{t}+\frac{1}{4}{e}^{-t}$

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