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0

Letting capital letters denote the laplace transform and taking the laplace transform of both equations, we get $$ sX(s)-x(0) = 2X(s) + 3Y(s) $$ $$ sY(s)-y(0) = 3X(s) + 2Y(s) $$ Collecting like terms we get $$(s-2)X(s)-3Y(s)-x(0)=0$$ $$(s-2)Y(s)-3X(s)-y(0)=0$$ Using elimination, we can get $$-9Y(s)-3x(0)+(s-2)^2Y(s)-(s-2)y(0)=0$$ Moving things around a ...


1

Let's write down the expression for the inverse transform: $$\frac1{i 2 \pi} \int_{c-i \infty}^{c+i \infty} ds \, \log{\left (1+\frac{w^2}{s^2} \right )} e^{s t} $$ where $c$ is greater than the largest real part of all the singularities of the integrand, if any. To compute the inverse transform above, consider the contour integral for $t \gt 0$: ...


1

$$\ln\left( 1+\frac{w^2}{s^2}\right)=-\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}}{k s^{2k}} $$ The inverse Laplace transform of $\frac{1}{s^{2k}}$ is $\frac{t^{2k-1}}{(2k-1)!}$ So, the inverse laplace transform of $\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}}{k s^{2k}} $ is $-\displaystyle\sum_{k=1}^\infty \frac{(-1)^k w^{2k}t^{2k-1}}{k ...


0

You can write ${1 \over s(s^2+3s)} = {1 \over 9(s+3)} - { 1 \over 9s} + {1 \over 3 s^2}$ and ${1 \over s^2(s^2+3s)} = -{1 \over 27(s+3)} + { 1 \over 27s} - {1 \over 9 s^2}+ {1 \over 3 s^3}$.


3

There may be some cases where complex variables, the residue theorem and the residue at infinity are helpful. Suppose your OGF is $f(z)$ and the desired EGF is $g(w).$ Then we have $$g(w) = \sum_{n\ge 0} \frac{w^n}{n!} \frac{1}{2\pi i} \int_{|z|=\epsilon} \frac{1}{z^{n+1}} f(z) \; dz.$$ This will simplify together with some conditions on convergence ...


1

Both of your methods of computing $\mathcal{L}[r(t-1)u(t-1)]$ will work, but your 1st method needs fixing. 1st way (Linearity): You tried to write $\mathcal{L}[r(t-1)]=\mathcal{L}[t-1]=\mathcal{L}[t]-\mathcal{L}[u(t)]=\frac{1}{s^2}-\frac{1}{s}$. This is wrong because $r(t-1)u(t-1)\neq t - u(t)$. To fix this, use the correct expression $r(t-1)u(t-1) = t ...


1

The Laplace Transform of $r(t-1)$ is $$\begin{align} \mathscr{L}\{r(t-1)\}(s)&=\int_0^{\infty}r(t-1)e^{-st}dt\\\\ &=\int_0^{\infty}(t-1)u(t-1)e^{-st}dt\\\\ &=\int_1^{\infty}(t-1)e^{-st}dt \tag1\\\\ &=e^{-s}\int_0^{\infty}te^{-st}dt \tag 2\\\\ &=e^{-s}\mathscr{L}\{r(t)\}(s)\\\\ &=e^{-s}\frac1{s^2} \end{align}$$ In going from $(1)$ to ...


3

Hint: Set $$f(t):=\frac{1-\cos t}{t}.$$ We have $$\mathcal{L}\{1-\cos t \}(s)=\mathcal{L} \{tf(t) \}(s)=-F'(s), $$ where $F(s)$ is the Laplace transform of $f(t)$.


1

Let continue your calculus : $$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {e^{iut}}{iu+1}du$$ $$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {(\cos(ut)+i \sin(ut))(-iu+1)}{u^2+1}du$$ $$\frac{1}{2\pi}\int ^{\infty}_{-\infty}\frac {\cos(ut)+u \sin(ut)}{u^2+1}du + \frac{i}{2\pi}\int ^{\infty}_{-\infty}\frac {-u\cos(ut)+ \sin(ut)}{u^2+1}du$$ $\int ...


0

I will solve this using convolution but I would LOVE to see someone do it using contour integration. Using the fact that: $\frac{1}{\sqrt t}\rightarrow \sqrt \frac{\pi}{s}$ which means $\frac{1}{\sqrt s}\rightarrow \sqrt \frac{1}{\pi t}$ $\frac{1}{s-a}\rightarrow e^{at}u(t)$, where $u(t)$ is the unit step (which means the exponential is zero for $t<0$, ...


0

Hint: $\dfrac{1}{|s|^\alpha+1}=\dfrac{1}{|s|^\alpha\left(1+\dfrac{1}{|s|^\alpha}\right)}=\dfrac{1}{|s|^\alpha}\sum\limits_{n=0}^\infty(-1)^n|s|^{-\alpha n}=\sum\limits_{n=0}^\infty(-1)^n|s|^{-\alpha(n+1)}$


0

A good example is $\ln x$ . Its laplace transform is $-\dfrac{\gamma+\ln s}{s}$ .


0

...or, consider your integral being the squared integral of (exp(-r^2)dr, from 0 to Infinity) in polar coordinates which can be solved easely. The result is sqrt{1/2*sqrt(pi)*erf (Infinity)}. erf(infinity) is a fancy way of saying "1".


1

Use change of order of the double integral (which can be checked to be valid by Fubini's theorem) in the LHS (i.e. the Laplace transform integral) to get $$\int_{0}^{\infty} e^{-st-t^2/2}\left(\int_{0}^t e^{q^2/2}dq\right)dt=\int_{0}^\infty e^{q^2/2}\left(\int_{q}^\infty e^{-st-t^2/2}dt\right)dq$$ I assume $s$ to be real. Then, $$\int_{q}^\infty ...


0

Note that the inverse Fourier transform of $F(\omega)$ is $$f(t)=\frac14 e^{-2|t|}\tag{1}$$ You can obtain this result via the inverse Laplace transform by using $s=j\omega$: $$F(s)=\frac{1}{4-s^2}=\frac14\frac{1}{2-s}+\frac14\frac{1}{2+s}\tag{2}$$ Note that you also need to specify the region of convergence (ROC) in order to find the correct time ...


0

Division by $(s^3+s-1)$ is allowed whenever none of the factors of the polynomial in $s$ are 0. Then factor, do a partial fraction decomposition and identify each term from a table. If you want all solutions you will need to check what you miss when doing the division. There can be "hidden" functions (distributions) which you need to check for. The ...


0

Let's look at this from a conceptual point of view rather than a numerical one. The equation $s^3+s-1=0$ has three roots. A quick plot (by hand!, try $s^3=1-s$) shows that there is one real root $s_0$ and two complex conjugate roots $s_r \pm i s_i$. Knowing this, then, we can simply write down the inverse LT by the residue theorem: $$f(t) = e^t + ...


1

Assuming that there is no error in the original question, one possibility would be to solve the equation $s^3+s-1=p$ (assuming $p>0$) for $s$, selecting the only real root of this equation. This is doable either using Cardano's formulae, or faster using Mathematica, but the result is very ugly. One might then continue plugging this value of $s$ in $\frac ...


1

The Laplace transform is $L(p)= \int \limits _0 ^\infty \mathbb e ^{-pt} \int \limits _0 ^t \frac {\sin x} x \mathbb d x \mathbb d t$. Compute the outer integral by parts (integrating the exponential): $\frac {\mathbb e ^{-pt}} {-p} \int \limits _0 ^t \frac {\sin x} x \mathbb d x \Big| _0 ^\infty - \int \limits _0 ^\infty \frac {\mathbb e ^{-pt}} {-p} \frac ...


2

Integrate by parts: $$ \int_0^{\infty} e^{-sx} \left( \int_0^x \frac{\sin{t}}{t} \, dt \right) \, dx = \left[ -\frac{e^{-sx}}{s}\left( \int_0^x \frac{\sin{t}}{t} \, dt \right) \right]_{0}^{\infty} + \frac{1}{s}\int_0^{\infty} e^{-sx}\frac{\sin{x}}{x} \, dx $$ The first term is obviously zero since the integral converges. Now consider $$ A(s) = ...


3

We need $$F(s) = \int_0^{\infty} e^{-st} \int_0^t \dfrac{\sin(x)}xdx dt = \int_0^{\infty}\int_x^{\infty} e^{-st} \dfrac{\sin(x)}xdt dx = \int_0^{\infty} \dfrac{\sin(x)}x\cdot \dfrac{e^{-sx}}sdx$$ We have $$F'(s) = \int_0^{\infty} \dfrac{\sin(x)}x\cdot \dfrac{-xe^{-sx}}sdx - \int_0^{\infty}\dfrac{\sin(x)}{x}\cdot \dfrac{e^{-sx}}{s^2}dx$$ This gives us ...


1

Are w, y and z functions of x? If so, hint: $w'' + y' = \cos x$ from eq 1 $\to w'' + e^x + z = \cos x$ $\to w''' + e^x + z' = -\sin x$ $\to w''' + e^x + (1-w-y) = -\sin x$ $\to w''' - w = - \sin x - e^x - 1 + y$ $\to w''' - w = - \sin x - e^x - 1 + (\sin(x) - w')$ $\to w''' + w' - w = - \sin x - e^x - 1 + \sin(x)$ $\to w''' + w' - w =- e^x - 1$ Can ...


0

Note that the laplace transform of $\sin{c t}{c}$ is $\frac{1}{s^2+c^2}$. Now, integrate this with respect to $c$ from $0$ to $1$ to get the laplace transform of the sine integral, as desired. (normally, you get the laplace transform of sinc using the rule for derivatives and laplace transforms from this)


1

You argument is that $$ F(\alpha) = \int_{0}^{\infty}e^{-\alpha t}f(t)dt \\ \int_{0}^{\infty}e^{-\alpha t}\alpha F(\alpha)dt = F(\alpha)=\int_{0}^{\infty}e^{-\alpha t}f(t)dt \\ \implies \alpha F(\alpha)=f(t) $$ Does that help you spot your error?


1

Let $f$ be the density of $X$. The joint density of $(X,Y)$ is $$ f_{X,Y}(x,y)=f_{Y|X}(y|x)f(x) = \begin{cases} \frac1xf(x), &\text{if $x>0$ and $0<y<x$}\\ 0, &\text{otherwise.} \end{cases} $$ Put $Z:=X-Y$. Apply the Jacobian formula to find the joint density of $(Z,Y)$: $$ f_{Z,Y}(z,y)=f_{X,Y}(z+y,y)= \begin{cases} \frac1{z+y}f(z+y), ...


2

There is another way of looking at this that removes the mystery around the delta function, i.e., why do we need to introduce it in the first place? This reason may not be all that obvious to those working from tables. The clue is that $s/(s-1)$ doesn't vanish as $|s| \to \infty$. That alone should put the problem solver on some alert regarding the ...


0

Here's an proof why is $\ L \left\{\delta (t)\right\} =1 $ Defined as $\begin{cases} \delta(t) = \infty &when \; t=0 \\\delta(t)=0 &when \;t\neq 0 \end{cases}$ Now, you must be thinking this is indeed a bizarre looking function, infact it's in exact, is not even a function. :) Now how we will take the laplace transform of this function even? Let ...


0

I guess $H$ is the Heaviside function, i.e. $H=1_{[0,\infty)}$. If so then the Laplace transform $\mathscr{L}(f)$ of $f(t)=H(\pi-t)\sin^2t$ is: \begin{eqnarray} \mathscr{L}(f)(s)&=&\int_0^\infty e^{-st}f(t)\,dt=\int_0^\pi e^{-st}\sin^2t\,dt=\frac12\int_0^\pi e^{-st}(1-\cos2t)\,dt\\ ...


1

Assuming the unilateral Laplace transform, you can see that ${\cal L} (t \mapsto H(\pi-t)) = {\cal L} (1_{[0,\pi]})$. So you need to compute the Laplace transform of a pulse of length $\pi$. Similarly, you could note that ${\cal L} (t \mapsto H(\pi-t)) = {\cal L} (t \mapsto H(t)-H(t-\pi))$.


2

The Heaviside function $H$ is defined as $1$ when its argument is positive and zero otherwise. Thus, for $H(\pi-t)$, the function is $1$ when $t<\pi$ and $0$ otherwise. Thus, $\int_0^{\infty}H(\pi-t)e^{-st}dt=\int_0^{\pi}e^{-st}dt=\frac1 s (1-e^{-\pi t})$ Thus, the result is identical to that from the sum $H(t)-H(t-\pi)$.


0

The continuous Fourier transform is equivalent to evaluating the bilateral Laplace transform with imaginary argument s = iω or s = 2πfi, per this wikipedia page: http://en.wikipedia.org/wiki/Laplace_transform#Fourier_transform So you can perform the Laplace Transform, then make the imaginary substitution in the result.


1

hint: $s^2 = \dfrac{2s^2}{2} = \dfrac{1}{2}\left((s^2-4)+(s^2+4)\right)$


0

You are making this harder than it needs to be, since working with the velocity gives a tracable equation right away. Let $v = \frac{dx}{dt}$, then $$\frac{dv}{dt} = \frac{c - bv}{a} \\ dt = \frac{a}{c - bv} dv\\ t+k = -\frac{a}{b} \ln (c - bv)\\ v=\frac{c-\kappa e^{-\frac{tb}{a}}}{b} $$ Now with $v(0) = \frac12$ we have $$ c-\kappa = \frac{b}{2} \\ \kappa ...


0

Does anyone know how to evaluate the following integral ? No. No one knows how to evaluate that integral. The proof is by reduction to the absurd: If anyone would have known how to express that integral in closed form, then the much simpler case with $q=0,~\alpha=1,$ and only one $\Gamma(s)$ in the denominator would also have been known to possess a ...


0

Suppose $f(z) = g(z)/h(z)$ where $g(z)$ and $h(z)$ are analytic in a neighbourhood of $z=a$, and both $g(a) = h(a) = 0$ but $h$ is not identically $0$. Thus $a$ is a zero of $h$, let's say of multiplicity $m$, and of $g$, say of multiplicity $n$. Thus $a$ is an isolated singularity of $f$. Then if $n \ge m$, $a$ is a removable singularity of $f$ (thus not a ...


0

I think you meant $y'' + 3 t y' - 6 y = 1$. Then you should get $$ s^2 Y(s) - 9 Y(s) - 3 s Y'(s) = \dfrac{1}{s}$$


1

Assume first that $t \gt 0$. To use Cauchy's theorem, close the integration contour with a circular arc to the left of radius $R$ as follows. ($D$ is opposite $C$ along the imaginary axis.) Now consider $$\oint_{ABCDA} dz \, F(z) e^{z t} $$ where $$\int_{AB} dz \, F(z) e^{z t}= \int_{x-i \sqrt{R^2-x^2}}^{x+i \sqrt{R^2-x^2}} ds \, F(s) e^{s t} $$ ...


1

Hint: $$\mathscr{L}\left(\int_0^t f(t')dt'\right)(s)=\int_0^{\infty}e^{-st}\left(\int_0^t f(t')dt'\right)dt=\frac{1}{s}\,\mathscr{L}\{f\}(s)$$ So, the Laplace Transform of the integral of a function $f$ is equal to $1/s$ times the Laplace Transform of $f$.


-1

If you let k=3, then you can use that fact that L.T. t*sin(kt) = 2ks/(s^2+k^2)^2 or in your case, 2*3*s/(s^2+3^2)^2 -> 6s/(s^2+3^2)^2 which if you invert should lead to (1/6)t*sint(3t). Try that.


0

EDIT: (weeks later) I have found an even better approach. The equation becomes much more trivial when fourier transform is used. It is possible to identify, immediately, that the fourier transform of a gaussian is also a gaussian, which gives one of the two solutions, which in return makes it very easy to find the second one. I wasn't satisfied with ...


1

Changing the order of integration gives the integral $$ \int_a^{\infty} e^{-xt} \sin t\,dt $$ which is equivalent to $$ \int_0^\infty e^{-xt} \sin t\,\operatorname{u}(t-a)\,dt = \mathcal{L}\left\{\sin t \operatorname{u}(t-a) \right\}(x) $$ where $\operatorname{u}(t)$ is the Heaviside step function $$\operatorname{u}(t-a) = \left\{ \begin{matrix} 0, & ...


4

Hint: your D.E is equivalent to ( by taking the integratition for both sides) $$ y'+yx=C_1$$ by using the boundary condition, we will get $$C_1=-1$$


1

Here is another way to show equality. Let $$G(a)=\int_a^{\infty}\int_0^{\infty} e^{-xt}\sin t \,dx\,dt$$ and $$H(a) = \int_0^{\infty} e^{-ax}\frac{\cos a+x\sin a}{1+x^2} \,dx.$$ Taking the derivatives of $G$ and $H$ reveals that $$\begin{align} G'(a)&=-\int_0^{\infty} e^{-ax}\sin a \,dx\\\\ &=-\sin a\int_0^{\infty} e^{-ax}\,dx \end{align}$$ ...


0

Change the order of integration, and do $$ \int_a^{\infty} e^{-xt} \sin{t} \, dt, $$ which, as you can see, is close to, but not quite, the Laplace transform of $\sin{t}$. What you need to do is one of: integrating by parts twice, or switching the sine into complex form and doing the integral directly. It should then fall out without much fuss.


0

Short answer: yes Long answer: For any 2 functions $f(t)$ and $g(t)$, the Laplace tranform of the convolution is the product of Laplace transforms $$\mathcal{L} \{ f(t) * g(t) \} = \mathcal{L}\{f(t)\} \mathcal{L}\{g(t)\}$$ It makes sense that this would hold no matter how many times we do the convolution


0

$$\begin{align} F(s)&=\frac{s}{(s - \frac{1}{2})^{2} + 4}\\ &=\frac{\frac{1}{2}}{(s - \frac{1}{2})^{2} + 4} + \frac{s-\frac{1}{2}}{(s - \frac{1}{2})^{2}+ 4}\\ \\ f(t) &= (1/4)e^{t/2} (\sin(2 t) + 4\cos(2 t)) \end{align} $$ Credit to Chappers, Oliver, and Mattos for guiding me to the above solution


-1

This can be used to solve differential equations having a derivative term multiplied by $t$. e.g. : $$t\frac{\mathrm dy}{\mathrm dt}=t.$$


1

The transform is equal to $$\frac{2}{(s-1)^2+\sqrt{7}^2}+\frac{2(s-1)}{(s-1)^2+\sqrt{7}^2}$$ which turn into cosine and sine transforms combined with the first shift theorem.



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