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0

For the special case $k=1$ you should have the answer $$ {{\rm e}^{-bt}} \left( i\sqrt {{\frac {b}{t}}} {\rm I_{1} \left(\,2\,\sqrt {-bt}\right)}+\delta \left( t \right) \right) ,$$ where $I_n(x)$ is the modified Bessel function of the first kind and $\delta(x)$ is the dirac delta function.


0

So you need to compute $$ \frac1{2\pi i}\int_{-i\infty}^{i\infty}e^{xs-\left(\tfrac{b}{b+s}\right)^k}ds=e^{-bx} \frac1{2\pi i}\int_{-i\infty}^{i\infty}e^{x(b+s)-\left(\tfrac{b}{b+s}\right)^k}ds. $$ For $x$ positive and $k$ positive integer this is equal to $$ e^{-bx}\operatorname{res}_{z=0}e^{xz-\left(\tfrac{b}{z}\right)^k}. $$ The residue is the ...


0

By the uniqueness of the Laplace transform it suffices to show that: $$\mathcal{L}\left(\int_{0}^t x(u)x(t-u)du\right)=\frac{1}{\lambda^2+1} \tag{1}$$ To do so, insert the sum definition: $$x \left( t \right) =\sum _{n=0}^{\infty }{\frac { \left( -1 \right) ^{ n} \left(\dfrac{t}{2} \right) ^{2\,n}}{ \left( n! \right) ^{2}}} \tag{2}$$ into the integral and ...


1

Here is an approach. The integral in consideration is $$ \int_{0}^{\infty}t^{-3/2} e^{-\frac{a^2}{4t}}e^{-st}dt. $$ Taking the Mellin transform w.r.t. $a$ (see note 1), we need to change the order of integration, gives $$ \int_{0}^{\infty}t^{-3/2}e^{-st} \int_{0}^{\infty} a^{w-1}e^{-\frac{a^2}{4t}} da\,dt = \frac{4^{-w/2}}{2}\,\Gamma\left( ...


0

simplification of the above st = u s dt = du dt = du/s 1/s integral(0- infinity ) e^-u (u/s)^-1/2 du s^1/2 / s^1 = 1/s^1/2 therefore the answer is root pi by s since the multiplying term is s^1/2 again.


0

Consider $f(t) = \sum_{j=0}^\infty X_j(t)$ where $X_j(t) = j!$ for $j < t < j + 1/(j!)^2$ and $0$ otherwise. This is not exponentially bounded: $j! \exp(-sj) \to \infty$ as $j \to \infty$ for every real $s$. But for $s > 0$, ${\mathscr L}(f)(s) \le \sum_{j=0}^\infty \exp(-sj)/j! = \exp(\exp(-s))$ converges.


2

Hint: $\mathscr{L^{-1}}\left(\frac{b}{(s-a)^2+b^2}\right) = e^{at}\sin{(bt)}$; in your case, $a=-1$ and $b=1$. You can prove it by the integral definition of Laplace transformation. More info here. Another hint:, you can also find the inverse of $\frac{s}{((s+1)^2+1)}$ by smart usage of the $\mathscr{L^{-1}}$ of $e^{at}\cos{(bt)}$ in the link above. (That ...


0

Here is one way to think of it: if we multiply throughout by $\frac{2}{2} = 1$ we get $$\frac{1}{2}\cdot\frac{2s^2 + 6s + 6}{2s^2 + 7s + 7}$$ $$= \frac{1}{2}\cdot\frac{(2s^2 + 7s + 7) - s - 1}{2s^2 + 7s + 7}$$ $$= \frac{1}{2}\cdot\left(1 + \frac{-s-1}{2s^2 + 7s + 7}\right)$$ $$= \frac{1}{2}\cdot\left(1 - \frac{s + 1}{2s^2 + 7s + 7}\right)$$ $$= \frac{1}{2} ...


0

Consider the Laplace transform of $e^{x^{2}}$. \begin{align} e^{t^{2}} &\doteqdot \int_{0}^{\infty} e^{-st + t^{2}} dt \\ &\doteqdot \int_{0}^{\infty} e^{(t-s/2)^{2}- s^{2}/4} \ dt \\ &\doteqdot e^{- s^{2}/4} \ \int_{0}^{\infty} e^{(t-s/2)^{2}} \ dt \\ &\doteqdot e^{- s^{2}/4} \ \int_{-s/2}^{\infty} e^{u^{2}} du = e^{-s^{2}/4} \left[ ...


0

$\textbf{Hint:}$ You can start by using $L\big(u(t-\pi)(2\cos t-3\sin 3t)\big)=e^{-\pi s}L\big(2\cos(t+\pi)-3\sin(3(t+\pi)\big)=e^{-\pi s}L\big(-2\cos t+3\sin3t\big)$


0

By the formula $L(u(t-a)f(t))=e^{-as}Lf(t+a)$ we have \begin{aligned} L(U(t-2)(2t^2-6t+5)) &=e^{-2s}L((2(t+2)^2-6(t+2)+5)) \\ &=e^{-2s}(\frac{4}{s^3}+\frac{2}{s^3}+\frac{1}{s}) \\ \end{aligned}


0

Since $F\left( {\alpha ,\beta ,\delta ;t} \right) = \sum {\frac{{\left( \alpha \right)_n \left( \beta \right)_n }}{{n!\left( \delta \right)_n }}t^n } $. Assuming the uniform convergence of the series, then term by term integration yields that \begin{align} \int_0^\infty {e^{ - st} } t^{\gamma - 1} F\left( {\alpha ,\beta ,\delta ;t} \right)dt \\ ...


0

$\mathcal L[g''(r)] = t^2f(t) - t\,g(0) - g'(0) \Rightarrow\\ \mathcal L[r\,g''(r)] = \dfrac{d}{dt}[t^2f(t) - t\,g(0) - g'(0)] = t^2f'(t) + 2t\,f(t) - g(0)$ $\mathcal L[g'(r)] = tf(t) - g(0)$ $\mathcal L[r\,g(r)] = -f'(t), \mathcal L[r^2 g(r)] = f''(t)$ Rewriting the given equation as $r\,g''(r) - (2\mu_3 - 1) g'(r) - \mu_1^2r^2 g(r) - \mu_2^2 r\,g(r) = ...


2

Here is a start. $$ f(t) = 3\int^t_0{\sin{u}(t-u)e^{-(t-u)}du} = 3e^{-t}\int^t_0{\sin{u}(t-u)e^{u}du} $$ $$ = 3te^{-t} \int^t_0{\sin{u}\,e^{u}du} - 3e^{-t} \int^t_0{u\sin{u}\,e^{u}du}. $$ To make it easier to evaluate the last two integrals use the identity $$ \sin u = \frac{e^{iu}-e^{-iu}}{2i} $$ Another approach: We can use the fact $$ ...


0

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0

The inverse Laplace transform in question is a reduction of the more general form given by \begin{align} \mathcal{L}^{-1}\left\{ \frac{1}{(s^{2}+a^{2})^{\nu}} \right\} = \frac{ \sqrt{\pi} t^{\nu - 1/2} J_{\nu-1/2}(at)}{(2a)^{\nu-1/2} \Gamma(\nu)} \end{align} where $J_{\mu}(z)$ is the Bessel function of the first kind. In the case of $a=1$ and $\nu = 3$ it is ...


1

Yes, you have done it correctly. Now, notice that $$ \mathcal{L}[\sin at]=\frac{a}{s^2+a^2} $$ and $$ \mathcal{L}[\sin at-at\cos at]=\frac{2a^3}{(s^2+a^2)^2}, $$ then rewrite $$ \frac{1}{(s^2+1)^3}=\frac12\cdot\frac{1}{s^2+1}\cdot\frac{2}{(s^2+1)^2}. $$ Taking the inverse Laplace transform and using the convolution theorem, we obtain \begin{align} ...


2

$$\cfrac{2s+12}{s^2+9}=2\cfrac{s}{s^2+3^2}+\cfrac{12}{3}\cfrac{3}{s^2+3^2}$$ Here a good table for Laplace transform (note the inverse transform is linear) Solution


2

Hint: $F(s) = \dfrac{s}{s+3} = \dfrac{(s+3)-3}{s+3} = 1-\dfrac{3}{s+3}$. Can you find $1$ and $\dfrac{1}{s+3}$ in your Laplace table?


1

Working intuitively, it seems to me that if $\dfrac{\partial u}{\partial t} - \alpha \nabla^2 u=0, \tag{1}$ or $\dfrac{\partial u}{\partial t} = \alpha \nabla^2 u, \tag{1}$ and we are working in an open set $U \in \Bbb R \times \Bbb R^n$ on which $u(t, \vec x)$, $\vec x = (x_1, x_2, \ldots, x_n)$, and enough of its derivatives are "nice", that is, ...


5

Yes. Writing $\Delta$ instead of $\nabla^2$ one has $$u_{tt}={\partial\over\partial t}(u_t)={\partial\over\partial t}\bigl(\alpha\>\Delta u\bigr)=\alpha\>\Delta(u_t)=\alpha\>\Delta\bigl(\alpha\>\Delta u\bigr)=\alpha^2\>\Delta^2 u\ \ .$$


1

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0

Given \begin{align} \mathcal{L}\{f(t); s\} = \int_{0}^{\infty} e^{-st} f(t) \ dt \end{align} then the inverse Laplace transforms are given by the following \begin{align} \mathcal{L}\{ \frac{2 \sinh(at)}{t}; s\} = \ln\left(\frac{s+a}{s-a}\right) \end{align} and \begin{align} \mathcal{L}\{ \frac{1 - e^{at}}{t}; s\} = \ln\left(\frac{s-a}{s}\right). \end{align} ...


1

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0

Hint: First use partial fraction to get $$ \frac{6}{s} - \frac{2}{\left( s+1/2+i/2 \right)}-\frac{2}{\left( s+1/2-i/2 \right)}\quad i=\sqrt{-1}$$ then you can use the tables.


2

Consider Laplace transform $$ \mathcal{L}\left[f(t)\right]=F(s)=\int_0^\infty f(t)\ e^{-st}\ dt $$ and property of the unilateral Laplace transform $$ \mathcal{L}\left[\frac{f(t)}{t}\right]=\int_s^\infty F(\omega)\ d\omega, $$ where $F(\omega)$ is Laplace transform of $f(t)$. We choose $f(t)=(1-e^{-t})$ and it is easy to show that $$ ...


0

Hint: compute both $$ \frac d{ds} \int_0^\infty \frac{f(t)}{t}e^ {-st} dt \\ \lim _{s\to\infty} \int_0^\infty \frac{f(t)}{t}e^ {-st} dt $$


0

I only know how to solve it directly. You decide if it is useful to you or not. Differentiating (1) wrt $t$ twice and we obtain: $$x^{(3)}(t)+x(t)-1-t^2/2=0$$ The solution (via Mathematica) is given by: $$x(t)=1+t^2/2+c_1e^{-t}+c_2 e^t \sin\left(\frac{\sqrt{3}}{2}t\right)+c_3 e^t \cos\left(\frac{\sqrt{3}}{2}t\right)$$ $$1=x(0)=1+c_1+c_3$$ Therefore ...


4

Hint: $$ \int_X \alpha f + \beta g \, \mathrm{d} \mu = \alpha \int_X f \, \mathrm{d} \mu + \beta \int_X g \, \mathrm{d} \mu $$



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