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1

Let $f(t)=te^{−t}\sin(2t)=tg(t)$ and $g(t)=e^{−t}\sin(2t)=e^{−t}h(t)$ with $h(t)=\sin(2t)$. So $F(s)=-G'(s)$ and $G(s)=H(s+1)$ with $H(s)=\frac{2}{s^2+4}$.


3

Here is an approach. $\displaystyle \mathcal{L}(\sin 2t) = \frac{2}{s^2 + 2^2} = \frac{2}{s^2 + 4}$, using the table. $\displaystyle \mathcal{L} (e^{-t} \sin 2t) = \frac{2}{(s + 1)^2 + 4}$, using frequency shifting. $\displaystyle \mathcal{L}( t e^{-t} \sin 2t) = -\frac{d}{ds}\!\left(\frac{2}{(s + 1)^2 + 4}\right) = \frac{4 ...


1

Yes, your result is correct. I would also recommend solving it using the Heaviside Unit Step Function. We can write your piecewise function (see my response Using laplace transforms to solve a piecewise defined function initial value problem) as: $$f(t) = 0 - 0 u(t-2) + t u(t-2) = t u(t-2)$$ The Laplace Transform (see item 27) is: $$\mathscr{L} (t~ ...


0

The ODE is not a linear one. The Laplace transform method is not convenient. It is simpler to change of function first : $${\operatorname{d}\!x(t)\over\operatorname{d}\!t}=Y(t)$$ $$4 -0.1\: Y(t) - 0.01 \big(Y(t)\big)^2 = 0.9{\operatorname{d}\!Y(t)\over\operatorname{d}\!t}$$ Then, continue with the usual method to solve the ODE of separable kind.


1

Let $N(t)=M(t+\tau)$ and $G(t)=F(t+\tau)$ then, for every nonnegative $t$, $$N(t)=G(t)+\int_0^tN(t-x)\,\mathrm dF(x)$$ hence $$\tilde{N}(s)=\frac{\tilde{G}(s)}{1-\tilde{f}(s)}.$$


1

We could, also, use the definition \begin{align} \mathcal{L}^{-1}\{F(s)\} &= \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{st}}{(s^2 + a^2)^n}ds\\ &= \sum\text{Res} \end{align} We have $n$ poles at $s = \pm ia$ so $$ \mathcal{L}^{-1}\{F(s)\} = \lim_{s\to ia}\frac{1}{(n - ...


1

You can use the convolution theorem inductively: $$\mathcal L^{-1}\left[ \frac{1}{(s^2 + a^2)^n} \right] = \mathcal L^{-1}\left[ \frac{1}{(s^2 + a^2)^{n-1}} \right] * \mathcal L^{-1}\left[ \frac{1}{(s^2 + a^2)} \right]$$ and we have the base case. Hence $$\mathcal L^{-1}\left[ \frac{1}{(s^2 + a^2)^n} \right] = \underbrace{\frac{\sin(ax)}{a} * \cdots * ...


0

Taking the Laplace transform and applying the initial conditions $$ s^2Y(s)-sy(0)-y'(0)+2sY(s)-2y(0)+5Y(s)= s^2Y(s)-s+2sY(s)-2+5Y(s)=0 $$ So $$ Y(s)=\frac{s+2}{s^2+2s+5}=\frac{s+2}{(s+1)^2+4}=\underbrace{\frac{s+1}{(s+1)^2+4}}_{\mathcal{L}\{e^{-t}\cos(2t)u(t)\}}+\frac{1}{2}\underbrace{\frac{2}{(s+1)^2+4}}_{\mathcal{L}\{e^{-t}\sin(2t)u(t)\}} $$ that is $$ ...


2

Hints: The Laplace Transform yields: $\mathscr{L} (y''(t)) = s^2y(s) -s y(0) -y'(0)$ $\mathscr{L} (2 y'(t)) = 2(y(s) - y(0))$ $\mathscr{L} (5 y(t)) = 5y(s)$ Now, substitute in the initial conditions, solve for $y(s)$ and find the inverse Laplace Transform. If you have gotten to this point, please share $y(s)$ and we can provide further details as to ...


0

Multiplying, we get $$(As+B)(s^2+9) + (Cs+D)(s^2-12s+40) = 7s^2 + 9s + 3.$$ Expanding and collecting terms, we obtain \begin{align} 7s^2 + 9s+3 &= A s^3+9 A s+B s^2+9 B+C s^3-12 C s^2+40 C s+D s^2-12 D s+40 D \\ &= (A+C)s^3 + (B-12C+D)s^2 + (9A + 40C -12D)s + (9B + 40D). \end{align} This yields the linear system of equations $$ \begin{bmatrix} 1 ...


1

$$\mathcal{L}\{e^{-2t} u(t)\} = \int_{-\infty}^\infty e^{-2t} u(t) e^{-st} dt = \int_{0}^\infty e^{-2t} e^{-st} dt = \frac{1}{s+2}$$ $\mathcal{L}\{e^{-2t}\}$ doesn't converge for bilateral Laplace transformation.


1

You have a system $\mathbf X'=\mathbf{AX}$. Call $\mathbf x=\mathscr L\{\mathbf X\}$. Then if you apply Laplace transform to the initial differential equations you will obtain $s\mathbf x-\mathbf x(0)=\mathbf{Ax}$. This implies that $$(s\mathbf I-\mathbf A)\mathbf x=\mathbf x(0)\Longrightarrow \mathbf x = (s\mathbf I-\mathbf A)^{-1}\mathbf ...


2

We are given: $$X(t) = \begin{bmatrix} -3 & 0 & 2 \\ 1 & -1 & 0\\ -2 & -1 & 0\end{bmatrix} \begin{bmatrix} x(t) \\ y(t)\\ z(t)\end{bmatrix}, ~~ X(0) = \begin{bmatrix} 4 \\ -1\\ 2\end{bmatrix}$$ We can write this as: $$\tag 1 \begin{align} x' &= -3x + 2z \\ y' &= x-y \\ z' &= -2x - y \end{align}$$ Taking the Laplace ...


0

$\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle} \newcommand{\braces}[1]{\left\lbrace\, #1 \,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\, #1 \,\right\rbrack} \newcommand{\ceil}[1]{\,\left\lceil\, #1 \,\right\rceil\,} \newcommand{\dd}{{\rm d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\dsc}[1]{\displaystyle{\color{red}{#1}}} ...


1

We have $$L(\cosh t-1)=\frac{s}{s^2-1}-\frac{1}{s}$$ so $$L(\frac{\cosh t-1}{t})=\int_{s}^{\infty}(\frac{u}{u^2-1}-\frac{1}{u})du$$ and $$Lf(t)=\frac{1}{s}\int_{s}^{\infty}(\frac{u}{u^2-1}-\frac{1}{u})du=\frac{1}{s}\left[Ln\frac{\sqrt{u^2-1}}{u}\right]_{s}^{\infty}=\frac{1}{s}Ln\frac{s}{\sqrt{s^2-1}}$$


0

Hint (@Amzoti's comment, rephrased): Start with $$\mathcal{L}^{-1}\left(\frac1s\right),\quad \mathcal{L}^{-1}\left(\frac1{s^2}\right),\quad \mathcal{L}^{-1}\left(\frac1{s+1+2\mathrm i}\right),\quad \mathcal{L}^{-1}\left(\frac1{s+1-2\mathrm i}\right).$$


0

With the appearance of $e^{-s}$, you should expect that a Dirac delta comes into play here. The reason being that $$\int_0^{\infty} e^{-st} \delta(t-t')\,dt = e^{-st'}$$ if $t\in (0,\infty)$. In your case, we expect $t'=1$. This is only part of the puzzle. Since you have an appearance of $s$, a derivative is to be expected. I'll leave it to you to piece it ...


1

Since I am bad at memorizing the tables of the inverse Laplace transform, I prefer to use the Bromwich integral (the inverse Laplace transform). \begin{align} \mathcal{L}^{-1}\{f(s)\} &= \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{16se^{st}}{(s^2 + 4)^2}ds\\ &= \sum\text{Res} \end{align} The residues occur at $s = \pm 2i$ both of ...


1

Using a table, note the form: $$f(t) = t\sin(at)$$ $$F(s) = \frac{2as}{(s^2+a^2)^2}$$ Using this: $$f(t) = \mathcal{L}^{-1}\left(\frac{16s}{\left(s^2+4\right)^2}\right) = \mathcal{L}^{-1}\left(\frac{2*2*s}{\left(s^2+2^2\right)^2}*4\right) =4t\sin(2t)$$


0

If $t$ is restricted to the domain $0 \leq t \leq |x|$ for some $x \in \Bbb{R}$ (in particular, for $x = 2\pi$, the equation will always have a bounded solution, and in fact, given any finite initial conditions, all solutions will be bounded. Are you sure you were not thinking of a related problem like $$ f(t) = t \mbox{ for } 0 \leq t < 2 \pi \\ f(t) = ...


4

Differentiate (left side) - (right side) twice and you'll get $$ y'' + y = 0$$ so $y(t) = a \cos(t) + b \sin(t) $. Plug that in to your integral equation and you'll find that you need $a=1$, $b=0$.


0

The separation of variables method leads to solutions of the form $\sin(\omega x)\sinh(\omega y)$ or $\sin(\omega x)\cosh(\omega y)$ or $\cos(\omega x)\sinh(\omega y)$ or $\cos(\omega x)\cosh(\omega y)$ or other equivalent forms with exp fonctions, where $\omega$ is any constant. The conditions $\frac{\partial u}{\partial x}(0,y)=0$ and $\frac{\partial ...


2

$F(s)={1\over s-1}\implies f(t)=e^t$ so using the theorem you want $u(t-a)\color{blue}{f(t-a)}=u(t-1)f(t-1)=u(t-1)e^{t-1}$


2

I prefer to use the definition then tables to solve the inverse Laplace transform. \begin{align} \mathcal{L}^{-1}\{e^{-s}/(s - 1)\} &= \frac{1}{2\pi i}\int_{\gamma - i\infty}^{\gamma + i\infty}\frac{e^{-s}e^{st}}{s- 1}ds\\ &= \sum\text{Res}\\ &= \lim_{s\to 1}(s-1)\frac{e^{s(t-1)}}{s-1}\\ &= e^{t - 1}\mathcal{U}(t - 1) \end{align} Then from ...


0

Chat defeated me! You are pretty much finished $$sin^2(2t) = \frac{1}{2} - \frac{1}{2}cos(4t)$$ Transforming $$L(sin^2(2t)) = \frac{1}{2}L(1) - \frac{1}{2}L(cos(4t))=\frac{1}{2}(\frac{1}{s}) - \frac{1}{2}(\frac{s}{s^2+16})$$ Your x seems to have become a 2t since yesterday?


2

Using the shift theorem: $$\mathscr L\{f(t-a)\,\mathcal U(t-a)\}(s)=e^{-as}F(s).$$ Here $\mathcal U(t-a)$ is the Heaviside step function that is defined as $1$ for $t\geq a$, $0$ otherwise. What you've got is $e^{-as}F(s):a=3,F(s)=\dfrac{1}{s-2}.$ Therefore, applying the theorem yields $$e^{2(t-3)}\,\mathcal U(t-3).$$ So the proof for the shift theorem ...


2

We should start with the definition. $$ \mathcal{L}\{f(t)\} = \int_0^{\infty}f(t)e^{-st}dt $$ Then we can compute the Laplace transform as \begin{align} \mathcal{L}(u'') + 4\mathcal{L}(u) &= \mathcal{L}\{H(x)\} + \mathcal{L}\{H(x - \pi)\}\\ s^2U(s) - su(0) - u'(0) + 4U(s) &= \int_0^{\infty}e^{-st}dt + \int_{\pi}^{\infty}e^{-st}dt \end{align}


1

You need the inverse Laplace transform of $$ \dfrac{2}{(s^2 + 2 s + 5)^2}$$ This may help: $$ {\mathcal L} \{t e^{at} \cos(bt)\} = \dfrac{(s-a)^2 - b^2}{((s-a)^2+b^2)^2} = \dfrac{1}{(s-a)^2 + b^2} - \dfrac{2b^2}{((s-a)^2+b^2)^2}$$


0

We can take inverse Laplace transform by using the Bromwich integral. That is, \begin{align} \mathcal{L}^{-1}\{Y(s)\} &= \frac{1}{2\pi i}\lim_{T\to\infty}\int_{\gamma -iT}^{\gamma +iT}Y(s)e^{st}ds\\ &= \sum\text{Res} \end{align} where $$ Y(s) = \frac{2}{(s^2 + 2s + 5)^2} + \frac{2s + 3}{s^2 + 2s + 5} $$ Then the poles of $s$ are at $s = -1\pm 2i$. ...


1

We could use the inverse Laplace transform integral/Bromwich Integral/Mellin-Fourier integral/Mellin's inverse formula (many names) and then use Residue theory. The pole is at $s = 1/2$ of order three. The Bromwich contour is a line from $\gamma - i\infty$ to $\gamma + i\infty$ and then a partial circle connecting the line. \begin{align} \frac{1}{2\pi ...


1

I'm busy, what's this all about? (Tl;dr) You have a small calculation error, but yes, you would be able to do it because $\mathcal{L}^{-1}$ is a linear operator. Let's take a step back and think about what we are doing, shall we? :) One of the Laplace transform properties is that $$\mathcal{L}^{-1} (F(s-a)) = e^{at} \mathcal{L}^{-1}(F(s)).$$ This is ...


0

You are correct. However, $\Phi(s)$ is just a constant, and does not depend on $s_0$, so $$\lim_{s_{0} \to \infty}\Phi(s)=\Phi(s).$$


1

Your work agrees exactly with Wolfram Alpha's. Try expanding the denominator of your expression and the denominator of Wolfram Alpha's expression.


1

I'm assuming, as you seem to have implied, that $A,B,C,D,E,F$ are arbitrary, and, in particular, that $E$ and $F$ can be defined as some combination of $A,B,C,D$. $$ \frac{A}{Bx^{\alpha+1}}=\frac{A}{Bx^{\alpha+1}}\cdot\frac{x^\beta}{x^\beta}= \frac{Ax^\beta}{Bx^{\alpha+1}x^\beta}=\frac{Ax^\beta}{Bx^{\alpha+\beta+1}} $$ And, similarly, $$ ...


1

Assuming that I properly understood (which is not sure, I must confess) : let $$P=\frac{A}{Bx^{\alpha+1}}$$ $$Q=\frac{C}{Dx^{\alpha+\beta}}$$ So $$P^{\frac{1}{\alpha+1}}=\Big(\frac{A}{B}\Big)^{\frac{1}{\alpha+1}}\frac 1x$$ $$P^{\frac{1}{\alpha+1}}Q=\Big(\frac{A}{B}\Big)^{\frac{1}{\alpha+1}}\Big(\frac CD ...


1

Just a simple mistake of the common but annoying "How could I have missed that?" type. $$\sin t = t + \frac{t^3}{6} + O(t^5),$$ so you should look at the integral $$\int_0^\delta te^{-xt}\,dt = -\frac{\delta e^{-\delta x}}{x} + \frac{1}{x}\int_0^\delta e^{-xt} = \frac{1-e^{-\delta x} - \delta xe^{-\delta x}}{x^2}\sim \frac{1}{x^2}.$$


0

Because $y(0)=y'(0)=0$, the LT of the ODE yields $$(s^2+2 s+2) Y(s) = e^{-\pi s} + a e^{-T s}$$ where $Y$ is the LT of $y$. Thus we have that $$y(t) = \frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \frac{e^{(t-\pi) s} + a e^{(t-T) s}}{s^2+2 s+2} $$ which, by the residue theorem, is $$e^{-(t-\pi)} \sin{(t-\pi)} \theta(t-\pi) + a e^{-(t-T)} ...


0

You have a $-s-3+6=-s+3$ on the left-hand side, so when it goes to the right-hand side, you get $$ Y(s)=\frac{\color{blue}{s-3}}{s^2-6s+9}+\frac{G(s)}{s^2-6s+9}. $$ Then, since $s^2-6s+9=(s-3)^2$, \begin{align} Y(s)&=\frac{s-3}{s^2-6s+9}+\frac{G(s)}{s^2-6s+9}\\ &=\frac{s-3}{(s-3)^2}+\frac{G(s)}{(s-3)^2}\\ &=\frac{1}{s-3}+\frac{G(s)}{(s-3)^2}.\\ ...


0

I edited Robert's answer. However it was not accepted. Here there is a more explicit explanation of his idea: Consider the change of variables: $$\boldsymbol X(t)=\boldsymbol Y(t-t_0).$$ Therefore the system $$\cases{\boldsymbol X'(t)=\boldsymbol A\boldsymbol X(t)\\\boldsymbol X(t_0)= \boldsymbol b}$$ is equivalent to the system $$\cases{\boldsymbol ...


1

To solve $X' = A X$ with $X(t_0) = b$ given, solve $Y' = A Y$ with $Y(0) = b$, and take $X(t) = Y(t - t_0)$.


0

I suspect you have a typo in what you posted since $u(-t)=0$ for $t>0$. Laplace transforms assume the underlying function is causal. Otherwise, computing the Laplace transform of the zero function is trivial. Now, if you mean $u(t)$ (as I suspect) which is the unit step function, just compute its Laplace transform straightaway from the definition: $$ ...


0

This looks wrong to me. Assuming by $u$ you mean the unit step function $u(t) = 1$ if $t \geq 0$ and $u(t) = 0$ otherwise, then by definition of the Laplace transform $$\mathcal{L}\{ u(-t)\}(s) = \int_0^\infty u(-t)e^{-st} \ dt = \int_0^\infty 0 \ dt = 0$$ as for all $t \in [0,\infty)$, $u(-t) = 0$. It is however true by the same sort of calculation that ...


0

One of the things to remember is that MATLAB does the analytical Laplace transform, so it will detect that you are providing an explicit unit step function. Normally the symbolic functions take the "normal" MATLAB functions as arguments, but internally change them over to the analytical versions. I think this is probably the same with the MATLAB unit step ...


1

Use #28 from your table instead of #27. Then $c=2$ and $$g(t)=t^2\implies g(t+c)=g(t+2)=(t+2)^2=t^2+4t+4$$ so $$\mathscr{L}\{g(t+c)\}=\mathscr{L}\{t^2+4t+4\}={2\over s^3}+{4\over s^2}+{4\over s}.$$ Thus, from #28 in your table, $$ \mathscr{L}\{t^2u(t-2)\}=e^{-2s}\mathscr{L}\{g(t+2)\}=e^{-2s}\left({2\over s^3}+{4\over s^2}+{4\over s}\right).$$


2

The Heaviside function effective changes the lower limit of integration so the LT is $$\int_2^{\infty} dt \, t^2 \, e^{-s t} = \frac{d^2}{d s^2} \int_2^{\infty} dt \, e^{-s t} = \frac{d^2}{d s^2} \frac{e^{-2 s}}{s} $$ Taking the derivative, the LT takes the form $$ -\frac{d}{ds} \left [ \left ( \frac{2}{s} + \frac1{s^2} \right ) e^{-2 s} \right ] = \left ...


1

Define $$y(x,s) = \int_0^{\infty} dt \, Y(x,t) \, e^{-s t}$$ Then, integrating by parts: $$\int_0^{\infty} dt \, Y_t(x,t) \, e^{-s t} = -Y(x,0) + s y(x,s)$$ $$\int_0^{\infty} dt \, Y_{tt}(x,t) \, e^{-s t} = -Y_t(x,0) + s Y(x,0) + s^2 y(x,s)$$ Then using the initial conditions $Y(x,0)=Y_y(x,0)=0$, the PDE becomes the following ODE: $$y''-2 s y' + s^2 y ...


1

This looks like a good candidate for the convolution theorem, which states that the inverse transform of a product of two transforms is the convolution of the inverse transforms. In this case, the inverse transform of $$F(s) = \frac1{a s^2+b s+c} \implies f(t) = \frac{e^{-b t/(2 a)}}{\sqrt{b^2-4 a c}} \left (e^{\frac{\sqrt{b^2-4 a c}}{2 a} ...


0

Suppose $f$ is $T$-periodic. Then $({\cal L} f)(s) = \int_0^\infty f(t) e^{-st} dt = \sum_{n=0}^\infty \int_{nT}^{(n+1)T} f(t) e^{-st} dt$, and using the substitution $\tau = t-nT$ and the fact that $f$ is $T$-periodic, we get $\int_{nT}^{(n+1)T} f(t) e^{-st} dt = \int_0^T f(\tau)e^{-s(\tau + nT)} d \tau = e^{-s nT} \int_0^T f(\tau)e^{-s\tau} d \tau $, and ...


-1

You can always verify your results by taking the inverse laplace transform of your function. Here: $$\mathcal{L}^{-1}(e^{-sc})=\delta(t-c)\text{ where }\delta\text{ is Dirac's Delta Function}$$


0

Yes your analysis is correct. Notice that $X$ is a discrete random variable where $E(e^{X})=\sum_x e^{x}P(X=x)$ where we have $P(X=c)=1$



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