#### Didn’t find what you are looking for? Laplace transform ### Determine $$L^{-1}\left[\frac{(s-4)e^{-3s}}{s^2-4s+5}\right]$$

Laplace transform ### Obtain the Laplace Transform of $$L\left\{e^{-2x}+4e^{-3x}\right\}$$

Laplace transform ### $$\displaystyle{y}'={6}\frac{{x}^{{2}}}{{{2}{y}+{\cos{{y}}}}}{)}$$

Laplace transform ### Find the Laplace transforms of the given functions. $$f{{\left({t}\right)}}={6}{e}^{{-{5}{t}}}+{e}^{{{3}{t}}}+{5}{t}^{{{3}}}-{9}$$

Laplace transform ### $$L^{-1}\bigg(\frac{5}{s^{2}(s^{2}-4s+5)}\bigg)=5\int_{0}^{\infty}(t-T)e^{2T}\sin 2T\ dT$$ Select one: True or False

Laplace transform ### Solve the following IVP using Laplace Transform $$y′′+3y′+2y=e^(-t), y(0)=0 y′(0)=0$$

Laplace transform ### How to solve for third order differential equation of $$y"'-7y'+6y =2 \sin (t)$$ using Method of Laplace Transform when $$y(0)=0, y'(0)=0, y"(0)=0$$? Step by step

Laplace transform ### Use the appropriate algebra and Table of Laplace's Transform to find the given inverse Laplace transform. $$L^{-1}\left\{\frac{1}{(s-1)^2}-\frac{120}{(s+3)^6}\right\}$$

Laplace transform ### Determine the Laplace transform of the given function f. $$f(t)=(t -1)^2 u_2(t)$$

Laplace transform ### use the Laplace transform to solve the initial value problem. $$y"-3y'+2y=\begin{cases}0&0\leq t<1\\1&1\leq t<2\\ -1&t\geq2\end{cases}$$ $$y(0)=-3$$ $$y'(0)=1$$

Laplace transform ### Find the laplace transform by definition. a) $$\displaystyle{L}{\left\lbrace{2}\right\rbrace}$$ b) $$\displaystyle{L}{\left\lbrace{e}^{{{2}{t}}}\right\rbrace}$$ c) $$\displaystyle{L}{\left[{e}^{{-{3}{t}}}\right]}$$

Laplace transform ### Find the solution of the initial value problem given below by Laplace transform $$y'-y=t e^t \sin t$$ $$y(0)=0$$

Laplace transform ### Find the laplace transform of the following: $$a) t^2 \sin kt$$ $$b) t\sin kt$$

Laplace transform ### Find the inverse Laplace Tranformation by using convolution theorem for the function $$\frac{1}{s^3(s-5)}$$

Laplace transform ### use properties of the Laplace transform and the table of Laplace transforms to determine L[f] $$f(t)=e^{3t}\cos5t-e^{-t}\sin2t$$

Laplace transform ### use the Laplace transform to solve the given initial-value problem. $$y"+y=f(t)$$ $$y(0)=0 , y'(0)=1$$ where $$\displaystyle f{{\left({t}\right)}}={\left\lbrace\begin{matrix}{1}&{0}\le{t}<\frac{\pi}{{2}}\\{0}&{t}\ge\frac{\pi}{{2}}\end{matrix}\right.}$$

Laplace transform ### Find the Laplace transformation (evaluating the improper integral that defines this transformation) of the real valued function f(t) of the real variable t>0. (Assume the parameter s appearing in the Laplace transformation, as a real variable). $$f{{\left({t}\right)}}={2}{t}^{2}-{4} \cosh{{\left({3}{t}\right)}}+{e}^{{{t}^{2}}}$$

Laplace transform ### use properties of the Laplace transform and the table of Laplace transforms to determine L[f] $$f(t)=\int_0^t (t-w)\cos(2w)dw$$

Laplace transform ### $$\text{Laplace transforms A powerful tool in solving problems in engineering and physics is the Laplace transform. Given a function f(t), the Laplace transform is a new function F(s) defined by }$$ $$F(s)=\int_0^\infty e^{-st} f(t)dt \(\text{where we assume s is a positive real number. For example, to find the Laplace transform of } f(t)=e^{-t} \text{ , the following improper integral is evaluated using integration by parts:} \(F(s)=\int_0^\infty e^{-st}e^{-t}dt=\int_0^\infty e^{-(s+1)t}dt=\frac{1}{s+1}$$ $$\text{ Verify the following Laplace transforms, where u is a real number. }$$ $$f(t)=t \rightarrow F(s)=\frac{1}{s^2}$$ 