Laplace transforms

$f(t)$	$F(s)$
$e^{at}$	$\frac{1}{s-a}$
$t^n$	$\frac{n!}{s^{n+1}}$
$\cos (at)$	$\frac{s}{s^2+a^2}$
$\sin(at)$	$\frac{a}{s^2+a^2}$
$u_c(t)$	$\frac{e^{-cs}}{s}$
$\delta(t-c)$	$e^{-cs}$

$$\mathcal{L}(c_1f_1+c_2f_2)=c_1F_1(s)+c_2F_2(s)$$ $$ \mathcal{L}(f')=s F(s)-f(0)$$ $$ \mathcal{L}(f'')=s^2 F(s)-s f(0)-f'(0)$$ $$ \mathcal{L}(f(t-c)u_c(t))=e^{-cs} F(s)$$ $$ \mathcal{L}(e^{at} f(t))=F(s-a)$$

Notation

• $F=\mathcal{L}(f)$ means $F(s)=\int_0^\infty e^{-st}f(t)dt$
• Heaviside function $u_c(t)=\begin{cases}0 & t\in[0,c) \\ 1& t\geq c\end{cases}$
• Dirac delta $\delta(t-c)$

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Solve $$y''+y= \delta(t-\tfrac{\pi}{2}), \quad y(0)=1,y'(0)=0$$

1. $y(t)=\cos (t) + u_{\pi/2}(t)(1-\cos (t))$
2. $y(t)=(1-u_{\pi/2}(t))\sin t$
3. $y(t)=(1-u_{\pi/2}(t))\cos t$
4. $y(t)=\sin t + u_{\pi/2}(t)\sin t $
5. None of the above

Show Solution

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Answer: C.
Solution $s^2 Y(s)-s+Y(s)=e^{-\pi s}$ so $Y(s)=\frac{s}{s^2+1}+ \frac{e^{-\pi/2 s}}{s^2+1} $ We get $\begin{multline*}y(t)=\cos t +\sin(t-\pi/2) )u_{\pi/2}(t) \\=\cos t -\cos(t) u_{\pi/2}(t) =\begin{cases} \cos t & t\in(0,\pi/2]\\ 0 & t>\pi/2 \end{cases}\end{multline*}$
Another one?

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Solve the differential equation $y''+y=\delta(t-\pi), y(0)=0, y'(0)=1$.

1. $y(t)=\cos (t) + u_\pi(t)(1-\cos (t))$
2. $y(t)=(1-u_\pi(t))\sin t$
3. $y(t)=\sin (t) + u_\pi(t)(1-\sin (t))$
4. $y(t)=\sin t + u_\pi(t)\sin t $
5. None of the above

Show Solution

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Answer: B.
Solution The Laplace transform is $Y(s)=\frac{1}{s^2+1}+\frac{e^{-\pi s}}{s^2+1}$. Therefore the solution is $y(t)=\sin t + u_\pi(t)\sin (t-\pi)=(1-u_\pi(t))\sin t$. Thus $$y(t)=\left\{\begin{array}{cl} \sin t &\hbox{ if } t\leq \pi \\ 0 &\hbox{ if } t>\pi \\ \end{array}\right.$$

Another one?

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Solve $$y''+y= \delta(t-\pi), \quad y(0)=1, y'(0)=1$$

1. $y(t)=\cos (t) + u_{\pi}(t)(1-\cos (t))$
2. $y(t)=(1-u_{\pi}(t))\sin t$
3. $y(t)=\cos t +(1-u_{\pi}(t))\sin t$
4. $y(t)=\sin t + u_{\pi}(t)\sin t $
5. None of the above

Show Solution

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Answer: C.
Solution $s^2 Y(s)-s-1+Y(s)=e^{-\pi s}$ so $Y(s)=\frac{1}{s^2+1}+\frac{s}{s^2+1}+\frac{e^{-\pi s}}{s^2+1} $ We get $\begin{multline*}y(t)=\sin t+\cos t +\sin(t-\pi) u_{\pi}(t) \\=\cos t +(1-u_{\pi}(t))\sin t =\begin{cases} \sin t+ \cos t & t\in(0,\pi]\\ \cos t& t\geq \pi \end{cases}\end{multline*}$
Another one?