New answers tagged

0

The Laplace transform of the term $-\frac{4}{e^x}=-4e^{-x}$ is $-\frac{4}{s+1}$ by the following rule: $$L\{e^{ax}\}=\frac{1}{s-a}$$ Hence $$Y(s)=\frac{4+(s+5)(s+1)}{(s^2+4)(s+1)}$$ You will then need to use partial fractions to separate the fraction.


2

The Laplace transform is useful because it transforms an ODE into an algebraic equation in the transformed variable (or a PDE into an ODE), and it includes the initial conditions as part of the algebraic equation. The usefulness of this LT of course depends on the ability to find the inverse transform of the solution to the algebraic equation. In ...


0

There are in fact many other integral transforms with specific kernels. The truth is that there is virtually no advantage of using the Laplace transform to solve differential equations, in comparison to other methods. Indeed, the method requires that you are able to compute explicitly the Laplace transform of any functions that are involved, which is really ...


0

The concept of uniform integrability gives a generalization of the conditions of the dominated convergence theorem. Uniform integrability of a sequence $\left(Y_n\right)_{n\geqslant 1}$ means that $$\lim_{x\to +\infty}\sup_{n\geqslant 1}\mathbb E\left[\left|Y_n\right|\mathbf 1\left\{|Y_n|\gt x\right\}\right]=0.$$ We can prove that a uniformly integrable ...


0

$$x''(t)+x'(t)=-2x(t)+u\Longleftrightarrow$$ $$\mathcal{L}_t\left[x''(t)+x'(t)\right]_{(s)}=\mathcal{L}_t\left[-2x(t)+u\right]_{(s)}\Longleftrightarrow$$ $$\mathcal{L}_t\left[x''(t)\right]_{(s)}+\mathcal{L}_t\left[x'(t)\right]_{(s)}=-2\mathcal{L}_t\left[x(t)\right]_{(s)}+\mathcal{L}_t\left[u(t)\right]_{(s)}\Longleftrightarrow$$ ...


1

Notice: Laplace Transform of $f(t)$, assume $\Re(s)>0$: $$\mathcal{L}_t\left[f(t)\right]_{(s)}=\int_{0}^{\infty}f(t)e^{-st}\space\text{d}t$$ $$\mathcal{L}_t\left[t^n\right]_{(s)}=\int_{0}^{\infty}t^ne^{-st}\space\text{d}t=\lim_{x\to\infty}\int_{0}^{x}t^ne^{-st}\space\text{d}t=\frac{n!}{s^{n+1}}$$ ...


0

Directly: \begin{eqnarray*} I(a) &=&\int_{0}^{\infty }dte^{-at}=\left[ \frac{e^{-at}}{-a}\right] _{0}^{\infty }=\frac{1}{a} \\ \partial _{a}^{3}I(a) &=&(-1)^{3}\int_{0}^{\infty }dtt^{3}e^{-at}=\partial _{a}^{2}\frac{-1}{a^{2}}=\partial _{a}\frac{2}{a^{3}}=\frac{-6}{a^{4}} \\ \int_{0}^{\infty }dtt^{3}e^{-at} &=&\frac{6}{a^{4}} \\ a ...


0

$\int_0^\infty t^3 e^{-3t}dt$ is not $L(t^3 e^{-3t})$. Laplace transform of $f$ is defined by \begin{equation} L(f)=\int_0^\infty e^{-st}f(t)dt. \end{equation} Therefore, \begin{equation} \int_0^\infty t^3 e^{-3t}dt=L(t^3)|_{s=3}. \end{equation}


2

Write $$\frac1{t} = \int_0^{\infty} dp \, e^{-p t}$$ Then the integral is $$\begin{align}\int_0^{\infty} dt \, \frac{\cos{a t}-\cos{b t}}{t} &= \int_0^{\infty} dt \, (\cos{a t} - \cos{b t})\int_0^{\infty} dp \, e^{-p t} \\ &= \int_0^{\infty} dp \, \left (\int_0^{\infty} dt \, \cos{a t} \, e^{- p t} - \int_0^{\infty} dt \, \cos{b t} \, e^{- p ...


0

Have you heard about Frullani's Integrals? Otherwise you can see here for different techniques.


0

This initial value problem is tricky because the ODE has a singularity at $x=0$ (the coefficient of the highest order term $f''$ vanishes there), which is exactly where the initial conditions are given. This is the underlying reason for the unusual pattern in computations. In more concrete terms, you haven't really used the condition $f'(0)=3$ when ...


1

The Fourier transform of $f(x)$ doesn't exist in the usual sense, but since $f$ can be viewed as a tempered distribution, we can interpret the Fourier transform in that setting. (I'm using the normalization $\hat f(\omega) = \int_{-\infty}^\infty e^{-i\omega t}f(t)\,dt$. If you're using something else, the answer is a little different.) First of all, ...


4

One can say without any problem that almost no equation can be solved using Laplace transforms. There are two main reasons for this. One is that in general Laplace transforms cannot be computed, unless for functions with some prescribed growth. Otherwise nothing can be done. Even worse, one often writes a symbol for the Laplace transform of the solution ...


1

Consider a function without a Laplace transform, such as $e^{e^x}$. Then: $$y'(x)=e^{e^x}$$ can't be solved by taking Laplace transforms.


0

Ok, as it stands the integral is not well defined. To give it a meaning we somehow need to regularize it. One way of doing that is to define the value of the integral as ($\delta\rightarrow 0_+$) $$ I\equiv\frac{1}{2}\left(\int_0^{\infty+i\delta}e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt+\int_{\infty-i\delta}^0e^{-t/g}\frac{1}{\sqrt{1-2 t x}}dt\right) $$ This ...


1

If you do not know how to differentiate, where did you get this problem? Or is it just the notation that is bothering you? The "Cauchy-Riemann" equations are $\frac{\partial u(x,y)}{\partial x}= \frac{\partial vx,y)}{\partial y}$ and $\frac{\partial u(x,y)}{\partial y}= -\frac{\partial v(x,y)}{\partial x}$ (I have simply changed the letters "$\sigma$" and ...


0

They are trying to confuse you by changing the variables. If in $(11)$ you replace all the $\sigma$'s with $x$ and all the $w$'s with $y$ are you able to continue? The meaning of the equations remains unchanged.


1

Just for convenience I will rename the variables $\sigma \mapsto x$ and $\omega \mapsto y$ in the CR-equations. Then I'll show it for $$\frac{\partial^2 u(x,y)}{\partial x^2} + \frac{\partial^2 u(x,y)}{\partial y^2} = 0.$$ Now note that $$\frac{\partial^2 u(x,y)}{\partial x^2} = \frac{\partial}{\partial x} \frac{\partial u(x,y)}{\partial x} ...


1

I think you can argue that they are still there, just hidden, because you apply the initial conditions in the middle rather than the end. Consider a non-homogenous initial value problem (with constant coefficients) $$\sum_i\alpha_i\frac{d^iy}{dt^i}=\theta(t).$$ Take the Laplace Transform of both sides. Note that the Laplace transform of a derivative, ...


0

I have dug up some book titles that i found useful, so i am writing my own answer. Namely: -Mathematical methods for physics and engineering by Riley, Hobson, Bence. -Many books or chapters from books of the author Vladimir Arnold. -Strang's Calculus -Hildebrand's Methods of Applied Mathematics -Churchill's Fourier Series and Boundary Value Problems The ...


2

Laplace transform of the convolution of two functions is the product of the Laplace transforms of the functions. Thus, take the LT of both sides: Let $$\int_0^{\infty} dt \, y(t) e^{-s t} = Y(s)$$ $$\int_0^{\infty} dt \, (1-\cos{t}) e^{-s t} = \frac1{s} - \frac{s}{1+s^2} = \frac1{s(1+s^2)}$$ $$\int_0^{\infty} dt \, t^4 e^{-s t} = \frac{4!}{s^5} $$ Thus, ...


0

I recommend checking out Gilbert Strang's book Introduction to Applied Math. Strang has great intuition and I think he explains things in a very clear, simple, and yet deep way.


0

Notice, solve them seperatally: $$y''(t)+y(t)=0\Longleftrightarrow y(t)=\cos(t)$$ $$y''(t)+y(t)=\sin(t)\Longleftrightarrow y(t)=\frac{\sin(t)-(x-2)\cos(t)}{2}$$


0

It would be much easier if you solve the equations independently and then stitch them together We solve $ y'' + y = 0 $ have a unique solution $ \cos t $ that satisfy the initial conditions. Next, we solve $ y'' + y = \sin t $ have a solution $ A \sin t + B \cos t -\frac{t \cos t}{2} $, in order for this to be continuous with the previous solution, we can ...


0

Impulse response represent the system in time domain and transfer function represent the system in frequency domain. Essentially both are same.


0

Let's show that in general, for $a \in \mathbf{C}$ and a polynomial $p$, that $p(t) \mathrm{e}^{at}$ is of exponential type $\Re(a) + \varepsilon$ for any $\varepsilon > 0$. Recall that a function $f\colon [0, \infty) \to \mathbf{C}$ is of exponential type $A$ if there exists a constant $C$ such that $$|f(t)| \le C\mathrm{e}^{At}$$ for all $t \ge 0$. ...


1

That doesn't look right. For $a \geq 0$, the basic identity is $L(u(t-a) f(t-a))=e^{-as} L(f(t))$. From this identity it follows that $L(u(t-a) f(t)) = e^{-as} L(f(t+a))$ (the more commonly used version of the formula when transforming forward). Just $L(f(t-a))$ has no simple formula in terms of $L(f(t))$, because it involves the values of $f$ on $[-a,0)$, ...


1

You are almost there. Let $Y(s)$ be the Laplace transform of $y(t)$. The last step should actually read: $20sY(s) - 20y(0) + Y(s) + 4s^2Y(s) -4sy(0) - 4y'(0) = \frac{20}{s}$ Now lets find an expresion for $Y(s)$: $Y(s)(4s^2 + 20s +1) = \frac{20}{s} + 20y(0) + 4sy(0) + 4y'(0)$ Simplify: $Y(s) = \frac{20 +198s+40s^2}{s(4s^2 + 20s +1)} = \frac{40s}{4s^2 + ...


0

F(s)=log[(s+1)/(s+2)(s+3)] f(s)=log(s+1)-log(s+2)-log(s+3) differentiating with respect to s f'(s)= (1/s+1)-(1/s+2)-(1/s+3) taking inverse laplace -t*f(t)= (e^-1)-(e^-2)-(e^-3) f(t)= -[(e^-1)-(e^-2)-(e^-3)]/t


0

Managed to find my mistake, I solved as if s² wasn't part of the derivative at $-\frac{d(s²L(y))}{ds}$ $-\frac{d(s²L(y))}{ds} = -2sL(y) - \frac{s²d(L(y))}{ds}$ $-2sL(y) - \frac{s²d(L(y))}{ds} + sL(y) -\frac{4dL(y)}{ds} =0$ $ - \frac{s²d(L(y))}{ds} - sL(y) -\frac{4dL(y)}{ds} =0$ $ \frac{s²d(L(y))}{ds} + sL(y) +\frac{4dL(y)}{ds} =0$ $ \frac{d(L(y))(s²+4)}{ds} ...


0

Yes, you can move the 10 out, but no, you cannot proceed as you do. Independent of your definition of Laplace transform (whole $\mathbb R$ or just $\mathbb R^+$ as domain of integration in the first step), you will find $$ (\mathcal L f)(s)=\int 10e^{-200t}u(t)e^{-st}\,dt=10\int_0^{+\infty}e^{-(200+s)t}\,dt=\frac{10}{s+200}. $$ As an alternative, you could ...


0

This the double-sided Laplace transform of a Gumbel probability density function (pdf): $$\int_{-\infty}^\infty dt\ e^{-st} e^{-t} e^{-e^{-t}}\ . $$ With the substitution $e^{-t}=\tau$ it becomes $$ \int_0^\infty \frac{d\tau}{\tau}\tau^{s+1}e^{-\tau}=\Gamma(s+1),\qquad\mbox{for}\quad s>-1\ . $$



Top 50 recent answers are included