New answers tagged

2

We have: $$G(s)=F(s-2) = \frac{1}{3s-2+\sqrt{s^2-4}} $$ so, in order to tackle this problem like the previous one, we should find the coefficients of the Taylor series at $x=0$ of: $$ H(x) = G\left(\frac{1}{x}\right) = \frac{x}{3-2x+\sqrt{1-4x^2}}=\sum_{n\geq 1}g_n x^n\tag{1} $$ to deduce (have also a look at Ramanujan's master theorem): $$ (\mathcal{L}^{-1}...


4

First, enforcing the substitution $t\to t^2$ in the integral of interest yields $$\begin{align} \int_0^\infty e^{-st}\,\frac{e^{-u^2/4t}}{\sqrt{\pi t}}\,dt&=\frac{2}{\sqrt\pi}\int_0^\infty e^{-s\left(t^2+u^2/4st^2\right)}\,dt\\\\ &=\frac{2e^{-u/\sqrt{s}}}{\sqrt\pi}\int_0^\infty e^{-s\left(t-u/2\sqrt{s}t\right)^2}\,dt \tag 1 \end{align}$$ Second, ...


3

It is probably easier to prove first that if $A,B\in\mathbb{R}^+$ we have: $$ \int_{0}^{+\infty}\exp\left(-A^2 x^2-\frac{B^2}{x^2}\right)\,dx = \frac{\sqrt{\pi}}{2A} e^{-2AB}\tag{1} $$ that is the same as proving that for every $C\in\mathbb{R}^+$ we have: $$ \int_{0}^{+\infty}\exp\left(-x^2-\frac{C^2}{x^2}\right)\,dx = \frac{\sqrt{\pi}}{2}e^{-2C}\tag{2} $$ ...


4

One possibility: $$ \frac{s}{s + \beta} = 1 - \frac{\beta}{s + \beta} $$


2

Hint: the inverse Laplace transform of: $$ f(s) = \frac{1}{s+\sqrt{s^2-4}}=\frac{s-\sqrt{s^2-4}}{4} $$ is given by: $$ (\mathcal{L}^{-1} f)(x) = \color{red}{\frac{I_1(2x)}{2x}}=\frac{1}{2}\sum_{n\geq 0}\frac{x^{2n}}{n!(n+1)!}\tag{1} $$ where $I_1$ is a modified Bessel function of the first kind, and it is probably easier to show that the Laplace ...


0

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1

To avoid using concepts maybe unfamiliar, but also maintain rigor, I'll introduce some lemmas. Let $f: [a,b]\to \mathbb{R}$ be a function. We say that $f$ is uniformly differentiable if for every $\epsilon > 0$ there exists a $\delta > 0$ so that $$ 0 < |x-y| < \delta \qquad\Rightarrow\qquad \left|\frac{f(y)-f(x)}{y-x} - f'(x)\right| <...


2

It suffices that one of the two functions is in $L^1$ and the other is in $L^{\infty} \cap C^1$ and its derivative is in $L^{\infty}$. More generally: If $f \in L^1(\Bbb R^d)$ and $g \in C^k(\Bbb R^d)$, such that $D^{\alpha}g \in L^{\infty}(\Bbb R^d)$ for all $|\alpha| \le k$, then $f \star g \in C^k(\Bbb R^d)$ and: $D^{\alpha}(f \star g) = f \star D^{\...


2

If $f$ and $g$ are $\mathcal{C}^1$ and $f',g'$ are both integrable, then: $$(f\ast g)'=f'\ast g=f\ast g'.$$ Basically, whenever $f'\ast g$ and $f\ast g'$ are well-defined, one has the equality above.


0

We have: $$\begin{eqnarray*}(\mathcal{L}h)(0^+)&=&\lim_{s\to 0^+}\int_{0}^{+\infty}\int_{t}^{+\infty}\frac{1}{e^u \sqrt{u}}\cdot e^{-st}\,du\,dt\\ (u=tv)\quad&=&\lim_{s\to 0^+}\int_{0}^{+\infty}\int_{1}^{+\infty}\frac{t}{e^{tv} \sqrt{tv}}\cdot e^{-st}\,dv\,dt\\(\text{Fubini})\quad&=&\lim_{s\to 0^+}\int_{1}^{+\infty}\int_{0}^{+\infty}\...


0

$\because\mathcal{L}^{-1}\{c\}=c\delta(t)$ $\therefore\mathcal{L}^{-1}\{\lfloor s\rfloor\}=\lfloor t\rfloor\delta(t)$


1

Now, we got: $$ \begin{cases} f(t)=\text{M}_2x''_2(t)+\text{K}_2(x_2(t)-x_1(t))\\ \text{M}_1x''_1(t)+\text{K}_2(x_1(t)-x_2(t))+\text{K}_1x_1(t)+\text{B}x'_1(t)=0 \end{cases} $$ So, we can take the Laplace transform of both these equations: $$\text{F}(s)=\text{M}_2\left(s^2\text{X}_2(s)-sx_2(0)-x_2'(0)\right)+\text{K}_2(\text{X}_2(s)-\text{X}_1(s))$$ ...


-1

Take a look at the The Laplace Transform, Joel L. Schiff 1991 Springer Verlag, ISBN 0-387-98698-7, Examples on p182 onwards.


0

For 21 take a Laplace transform to obtain $$(s^2F(s)-s-1)+(sF(s)-1)+F(s)=\frac2{s^3}$$ which you can rearrange to obtain $$F(s)=\frac{s^4+2s^3+2}{s^5+s^4+s^3}$$ can you take it from here? PS. See this page. Addendum we have $$F(s)=\frac2{s^3}-\frac2{s^2}+\frac{s+\frac12}{s^2+s+1}+\frac72\frac1{s^2+s+1}$$


0

First observe that $$F(s)=\frac{4}{\left[(s+1)^2+2^2\right]^2}$$ Let $\mathcal{L}^{-1}\left\{F(s)\right\}=f(t)$, then from the First Shifting Theorem we have $$\mathcal{L}\left\{e^tf(t)\right\}=F(s-1)=\frac4{\left(s^2+2^2\right)^2}=\left(\frac2{s^2+2^2}\right)^2\tag{1}$$ Applying the Inverse Laplace Transform to the last equation we get, from the ...


2

Laplace-transforming both sides of $y'' + y = 2t$, $$s^2 Y (s) - y_0 s - v_0 + Y (s) = \frac{2}{s^2}$$ Hence, $$Y (s) = y_0 \left(\frac{s}{s^2 + 1}\right) + (v_0 - 2)\left(\frac{1}{s^2 + 1}\right) + \frac{2}{s^2}$$ Taking the inverse Laplace transform, $$y (t) = y_0 \cos (t) + (v_0 - 2) \sin (t) + 2 t$$ From the conditions $y(\pi /4) = \pi / 2 $ and $y'...


0

You did two things wrong. First of all, you have $j$ in your answer because you did not notice that $j^2=1$. But before that, you did the integrals wrong, or much more likely, you did the integrals right but you got wrong the step where $$ \frac{1}{\alpha-i\omega}-\frac{1}{\alpha+i\omega}=\frac{2\alpha}{\alpha^2+\omega^2} $$ Note the denominator.


1

One may observe that, for $\alpha>0$, $w \in \mathbb{R}$, one has $$ \int_0^\infty \cos(wx)\:e^{-\alpha x}dx=\Re \int_0^\infty e^{-(\alpha -wj) x}dx=\Re \:\frac1{\alpha-wj}=\frac{\alpha}{\alpha^2+w^2} \tag1 $$ Then one gets $$ \begin{align} CF(w)=&\alpha/2\int_{0}^{\infty} e^{jwx} e^{-\alpha x} dx + \alpha/2\int_{-\infty}^{0} e^{jwx} e^{\alpha x}...


0

$$(\mathcal{L}^{-1}F)(x)=\sum_{n\geq 0}\mathcal{L}^{-1}(e^{-ns})=\sum_{n\geq 0}\delta(x-n) $$ is a one-sided Dirac comb.


3

I'm not entirely sure which step is confusing you. So starting from: $$=\cfrac{4}{2} e^t + \cfrac{3i+1}{-2-2i} e^{it} + \cfrac{-3i+1}{-2+2i}e^{-it}$$ As you noted, the next step is simply just the division of complex numbers $$\cfrac{3i+1}{-2-2i}=\cfrac{3i+1}{-2-2i}\cfrac{-2+2i}{-2+2i}=\cfrac{-4i-8}{8}=(-1-1/2 i)$$ $$\cfrac{-3i+1}{-2+2i}=\cfrac{-3i+1}{-2+...


0

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1

I assume that the OP actually wants to solve the following ODE $$y'' + y = \delta + \delta'$$ rather than $y'' + y' = \delta + \delta'$. Assuming zero initial conditions and taking the Laplace transform of both sides, $$s^2 Y (s) + Y (s) = 1 + s$$ and, thus, $$Y (s) = \dfrac{s+1}{s^2+1} = \dfrac{s}{s^2+1} + \dfrac{1}{s^2+1}$$ Taking the inverse Laplace ...


1

$$y''(t)+y'(t)=\delta(t)+\delta'(t)$$ $\delta(t)$ is the Dirac function. First integration : $$y'+y=u(t)+\delta(t)+c_1$$ $u(t)$ is the Heaviside function. Seconf integration : The solution of the homogeneous ODE $y'+y=0$ is $y=c\:e^{-t}$ Let $y(t)=f(t)e^{-t}$ $(f'-f)e^{-t}+fe^{-t}=u(t)+\delta(t)+c_1$ $f'=(u(t)+\delta(t)+c_1)e^t$ $f=(e^t-1)u(t)+u(t)+...


0

Just like for any differential equation, you check whether it's a solution by plugging the alleged solution into the equation and see whether the difference of the two sides simplifies to $0$. Here it doesn't, so this is not a solution.


0

$$f(t)=t^2H(t-1)=\left[(t-1)^2+2(t-1)+1\right]H(t-1)$$ $$\mathcal L \{f(t)\}=\mathcal L \left\{\left[(t-1)^2+2(t-1)+1\right]H(t-1)\right\}=\mathrm e^{-s}\left[\frac{2}{s^3}+2\cdot\frac{1}{s^2}+\frac{1}{s}\right]$$


1

You are given: $$ \int_{t=0}^{\infty}f(t)e^{-st}\;dt=\phi (s) $$ and are asked for: $$ \int_{t=0}^{\infty} e^{bt}f(at)e^{-st}\;dt=\int_{t=0}^{\infty} f(at)e^{-(s-b)t}\;dt $$ Now put $t'=at$ when the integral becomes: $$ \int_{t=0}^{\infty} e^{bt}f(at)e^{-st}\;dt=\int_{t'=0}^{\infty}\frac{1}{a}f(t')e^{-\frac{s-b}{a}t'}\;dt'=\frac{1}{a}\phi\left(\frac{s-b}{a}\...


3

Assuming $\text{Re}(s)>0$ we have: $$\begin{eqnarray*}\mathcal{L}\left(\int_{0}^{t}\frac{1-e^{-u}}{u}\,du\right)&=&\int_{0}^{+\infty}\int_{0}^{t}\frac{e^{-st}-e^{-st-u}}{u}\,du\,dt\\(\text{Sub.}\;u=tv)\quad&=&\int_{0}^{+\infty}\int_{0}^{1}\frac{e^{-st}-e^{-(s+v)t}}{v}\,du\,dt\\(\text{Fubini})\quad&=&\int_{0}^{1}\frac{1}{v}\int_{0}^{...


1

If $\mu(\{0\})<1$, then as $e^{-\theta x}<1$ for $\theta,x>0$ we have \begin{align} \hat\mu(\theta) &= \mu(\{0\}) + \int_{(0,\infty)} e^{-\theta x}\ \mathsf d\mu(x)\\ &< \mu(\{0\}) + \int_{(0,\infty)}\ \mathsf d\mu(x)\\ &=\mu(\{0\}) + \mu((0,\infty))\\ &=\mu([0,\infty))\\ &\leqslant 1. \end{align}


1

HINT $$ v'(t)=-g+c u^2(t)-2 c u(t) v (t)+c v^2(t) $$ This is a Riccati's Equation, of the form $$ v'(t) = q_0(t) + q_1(t) \, v(t) + q_2(t) \, v^2(t) $$ where $q_0(t)= -g+c u^2(t)\neq 0$, $q_1(t)=-2 c u(t)$ and $q_2(t)=c\neq 0$.



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