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3

First of all, $\log\dfrac{ab}c=\log a+\log b-\log c$ $$L^{-1}\log(s-a)=-\dfrac1t L^{-1}\left(\dfrac{d\{\log(s-a)\}}{ds}\right)=-\dfrac1tL^{-1}\left(\dfrac1{s-a}\right)$$ $$L^{-1}\left(\dfrac1{s-a}\right)=e^{at}$$

1

\begin{align} \mathcal{L}\Big(\sinh(ct)\int_a^te^{au}\sinh(bu)\,du\Big)&=\mathcal{L}\Big(\sinh(ct)\frac{e^{a^2} (b \cosh (a b)-a \sinh (a b))+e^{a t} (a \sinh (b t)-b \cosh (b t))}{(a-b) (a+b)}\Big)\\ &=\mathcal{L}\Big(-\frac{a e^{a^2} \sinh (a b) \sinh (c t)}{(a-b) (a+b)}+\frac{e^{a^2} b \cosh (a b) \sinh (c t)}{(a-b) (a+b)}\\ &+\frac{a e^{a t} ...

1

If $F(s)=\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)$ is the Laplace Transform of $f(t)$, then the Inverse Laplace Transform is given by $$f(t)=\frac{1}{2\pi i}\int_{\sigma -i\infty}^{\sigma +i\infty}e^{st}\left(\frac{1}{s}+\frac{6}{s^2}-\frac{10}{s^3}+\frac{1}{s^5}\right)\,ds \tag 1$$ where here $\sigma >0$. For $t>0$, we ...

0

Let $f(z)$ be the ordinary generating function and $\mathcal{L}$ be the Laplace transform. Then the EGF $g(z)$ has the form $$g(z)=\mathcal{L}^{-1}\left(\frac{1}{z} f(z)\Big |_{z=\frac{1}{z}}\right)$$

0

I found the error I made, it was when I used the Residual calculation. $S = -1$ $$Res_{(s \longrightarrow -1 )} Y(s) e^{st} = lim_{s \longrightarrow -1} \frac{se^{st}}{(1-s)} = - \frac{e^{-t}}{2}$$ $S = 1$ Here it should be a minus sign, viola! $$Res_{(s \longrightarrow 1 )} Y(s) e^{st} = lim_{s \longrightarrow 1} \frac{se^{st}(s-1)}{(1+s)(1-s)} = ... 0 In your example, if you interchange two first colums you'll get$$ \left[\matrix{ x_1 & x_2 & x_3 \\ 2 & 1 & 3 \\ 4 & 4 & 2 \\ 1 & 1 & 4}\right]\quad\sim\quad \left[\matrix{ x_2 & x_1 & x_3 \\ 1 & 2 & 3 \\ 4 & 4 & 2 \\ 1 & 1 & 4}\right]. $$What happened in the top most row is that the roles ... 0 Manipulating rows is like manipulating the equations of the system but without changing the order of the unknowns, so interchanging the rows has no effect on the resolution of the system. Interchanging columns however is equivalent to changing the order of the variables, so the only way to obtain the same result as before, you'll have to adjust b ... 1 Keep in mind that y(t)=\mathcal{L}^{-1}(Y(s))= \color{red}{\sum}\text{Res}(Y(s)e^{st}). Plus, in your case, you have two poles both of order 1, so the way to find the residue is :$$\text{Res}_{s_0}\left(\frac{P(s)e^{st}}{Q(s)}\right)=\left(\frac{P(s)e^{st}}{Q'(s)}\right)_{s=s_0}$$where s_0 denotes a pole. Hence, we end up with:$$y(t)=-\frac 1 ...

1

You can separate the dimensions. You start by trying to find a function $G(x,t)$ such that $G(x,t)$ satisfies the homogeneous heat equation in all of space and time, except formally at $x=t=0$; At $t=0$, $G(x,t)$ is zero forall non-zero values of $x$; For any nonzero real $(a,b)$, $\int_{-|a|}^{|b} G(x,0) dx = 1$. I'm not saying yet that this function ...

0

If you're not familiar with the integral approach for the inverse transform, you can refer to a table. Completing the square in the denominator is a good way to start:$$\frac{s+4}{s^2+4s+1}=\frac{s+2}{(s+2)^2-3}+\frac{2}{(s+2)^2-3}$$ Next, recall that $$\mathcal{L}\{\sinh at\}=\frac{a}{s^2-a^2},\quad\mathcal{L}\{\cosh at\}=\frac{s}{s^2-a^2}$$and$$... 2 The Convolution Theorem gives$$\mathscr{L}\left((f*g)(t)\right)(s)=\mathscr{L}\left(f(t)\right)(s)\mathscr{L}\left(g(t)\right)(s) \tag 1$$where \mathscr{L}(f(t))(s) is the Laplace Transform of f and$$(f*g)(t)=\int_{-\infty}^{\infty}f(t')g(t-t')dt'$$is the convolution (integral) of f and g. Now, we let f(t)=\cos (t)\,u(t) and g(t)=u(t), ... 1$$y(t)=\frac{1}{2i\pi}\int_\gamma g(s,t)ds=\frac{1}{2i\pi}\int_\gamma \frac{s+4}{s^2+4s+1}e^{st}ds $$Poles: s=-2 \pm \sqrt 3$$\mathcal{Res}_s(g)=\frac{(s+4)e^{st}}{2s+4}y(t)=\frac{((-2 + \sqrt 3)+4)e^{(-2 + \sqrt 3)t}}{2(-2 + \sqrt 3)+4}+\frac{((-2 - \sqrt 3)+4)e^{(-2 - \sqrt 3)t}}{2(-2 - \sqrt 3)+4}$$The rest is up to you: use hyperbolic ... 0 Since the denominator is not easily factorable, we avoid partial fractions and instead complete the square. Recall the following inverse Laplace transforms:$$ Y(s) = \frac{b}{(s - a)^2 + b^2} \implies y(t) = e^{at}\sin bt \\ Y(s) = \frac{s - a}{(s - a)^2 + b^2} \implies y(t) = e^{at}\cos bt Indeed, observe that: \begin{align*} \frac{s + 1}{s^2 - s ... 0 EDIT: As @georg stated in his comment there is a slight mistake, I've fixed the following accordingly:Y(s)=\frac{s+\dfrac 1 2 + \dfrac 1 2}{(s+\dfrac 1 2)^2+\frac 3 4}Y(s)=\frac{s+\dfrac 1 2}{(s+\dfrac 1 2)^2+\frac 3 4} +\frac 1 {\sqrt 3} \frac{\dfrac {\sqrt 3} 2}{(s+\dfrac 1 2)^2+\frac 3 4}$$And keep in mind:$$\mathcal L (\cos(\omega t))=\frac s ...

1

$$\mathfrak{L}(f(at))=\int_0^\infty f(at)e^{-st}dt$$ Let $u=at$ then $dt=du/a$, so: $$\mathfrak{L}(f(at))=\int_0^\infty f(u)e^{-(s/a)u}\frac {du} a=\frac1 a \hat{F}(\frac s a)$$

2

Hint: From the definition, $\mathcal{L}\{f(at)\}$ is an improper integral, try this with the sustitution $r=at$, then $$\int_0^{\infty}e^{-st}f(at)\,dt=\int_0^{\infty}e^{-\frac{s}{a}r}f(r)\frac{1}{a}dr$$

3

Make a change of variables in the integral. Scale with $a$: $$\mathcal L (f(at))(s)=\int_0^{+\infty}f(at)e^{-st}\,dt=\{u=at\}=\frac{1}{a}\int_0^{+\infty}f(u)e^{-(s/a)u}\,du=\frac{1}{a}\mathcal L(f(t))(s/a).$$

5

I'm not sure about the Laplace transform but in Joel L. Schiff's "The Laplace Transform: Theory and Applications" on page 13 the author proves that a large class of functions has a Laplace transform. I am not sure how to describe nicely the result in terms of domain and range of the operator, buy maybe that helps. As for the Fourier transform, you first ...

2

Notes: The first part of this solution is for an equation that contained an error. It remains here as a demonstration. The corrcted equation is the second half of this solution. The process of the proposer is correct in process but does need some guidance in finding a nice result. Consider the differential equation $$x'' + 2 x' + 5 x = 0, \quad x(0) ... 0 We can write f(t)=e^{At} as$$e^{At}=\sum_{n=0}^{\infty}\frac{t^nA^n}{n!}$$Then, assuming that ||A||<s, the Laplace Transform of e^{At} becomes$$\begin{align} F(s)&=\mathscr{L}\left(e^{At}\right)(s)\\\\ &=\int_0^{\infty} e^{At}e^{-st}dt\\\\ &=\sum_{n=0}^{\infty}\frac{A^n}{n!}\int_0^{\infty}t^ne^{-st}dt\\\\ ...

0

Let's take the scalar case for simplicity. $f(t)=e^{at}$ and $a<0$, so $$F(s)=\int_0^\infty e^{at}e^{-st}\,dt=\left[\frac{e^{(a-s)t}}{a-s}\right]_0^\infty=\frac{1}{s-a}, \qquad \text{Re}\,s\ge 0.$$ Now $$sF(s)=\frac{s}{s-a}=\frac{s-a+a}{s-a}=1+\frac{a}{s-a}=1+aF(s).$$ What this $1$ came from? It comes from $f(0)$: $L(\dot f)=sF(s)-f(0)$. For matrices ...

0

Starting from $$H(z)=\frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})}$$ then \begin{align} H(z) &= \frac{z^{3}}{z^{3}} \cdot \frac{1-z^{-1}+z^{-2}-3z^{-3}}{z^{-2}(1-z^{-1})} \\ &= \frac{z^{3} - z^{2} + z - 3}{z-1} = \frac{z^{2} (z-1) + (z-1) -2}{ z-1} \\ &= 1 + z^{2} + \frac{2}{1-z} = 1 + z^{2} + 2 \, \sum_{n=0}^{\infty} z^{n} \\ &= 3 + 2 z + ...

3

In the form it is in right now, I do not think the Laplace transform is of use. However, you can rewrite it in such a way to use the Laplace transform: $$-y\sin x + y'\cos x = \cos^2 x \Longrightarrow \frac{d}{dx}(y\cos x) = \cos^2 x.$$ From here you can use a Laplace transform. You might want to use a double angle identity to simplify $\cos^2 x$. That ...

0

Your function is $f(t)=t$ but not $f(t)=1$. You need to use the formula $$\mathcal{L}\{u_c(t) f(t-c)\} = e^{-cs}*F(s)$$ 3 times with $c=1,2, 3$ to get the answer. The answer derived by Leucippus looks correct.

0

It might be helpful to consider it as three separate parts: $$h(t) = \left\{\begin{array}{l} 0,\, 0\leq t<\pi \\ 1,\, \pi\leq t<2\pi\\ 0, t\geq2\pi \end{array}\right.$$ The most systematic way I find to turn these into heaviside functions is to start with the top-most, then add heaviside times the function on the next step minus that of the ...

1

The inverse Laplace transform considered is $$F(s)=e^{-3s}\frac{1+s}{s^3+2s^2+2s}$$ and can be reduced as follows. First notice that \begin{align} F(s) = \frac{e^{-3 s}}{s} \cdot \frac{s+1}{(s+1)^{2} + 1} = f_{1}(s) \cdot f_{2}(s) \end{align} where \begin{align} f_{1}(s) = \frac{e^{-3 s}}{s} \hspace{10mm} f_{2}(s) = \frac{s+1}{(s+1)^{2} + 1}. \end{align} ...

1

Using the notation $u_{c}(t) = H(t-c)$, where $H(t)$ is the Heaviside step function, then $$g(t)= (t-1) u_1(t) - 2(t-2) u_2(t) + (t-3) u_3(t)$$ is evaluated as follows. Let $g(s) = \mathcal{L}\{g(t)\}$ such that: \begin{align} g(s) &= \int_{0}^{\infty} e^{-s t} \left[ (t-1) \, H(t-1) - 2 \, (t-2) \, H(t-2) + (t-3) \, H(t-3) \right] \, dt \\ &= ...

0

HINT: $$\mathscr{L}(tu(t))(s)=\int_0^{\infty}te^{-st}dt=\frac1{s^2}$$ Thus, $$\mathscr{L}((t-c)u(t-c))(s)=\int_c^{\infty}(t-c)e^{-st}dt=e^{-sc}\frac1{s^2}$$ NOTE: We derive the result for the Laplace Transform of $t$. To that end, we write $$\mathscr{L}(t)(s)=\int_0^{\infty}te^{-st}dt$$ Now, integrate by parts with $u=t$ and $v=-e^{-st}/s$. Then, ...

2

You can reduce this problem to a more familiar one with the following trick. Consider the problem $$q_t(x,t) = -au_x(x,t) + \frac{D}{2}q_{xx}(x,t) \quad (0 < t, 0 < x < 2L)\\ q(x,0) = 0, \quad aq(0,t) - \frac{D}{2}q_x(0,t) = f(t) = -aq(2L,t) - \frac{D}{2}q_x(2L,t)$$ and solve it with a standard method. The solution will have a specific symmetry ...

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