Carol Gates
2021-02-02
Answered

Solve the following differential equations using the Laplace transform and the unit step function

$y"+4y=g(t)$

$y(0)=-1$

${y}^{\prime}(0)=0,\text{where}g(t)=\{\begin{array}{ll}t& ,t\le 2\\ 5& ,t2\end{array}$

$y"-y=g(t)$

$y(0)=1$

${y}^{\prime}(0)=2,\text{where}g(t)=\{\begin{array}{ll}1& ,t\le 3\\ t& ,t3\end{array}$

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asked 2021-09-21

The Laplace transform function for the output voltage of a network is expressed in the following form:

$V}_{0}\left(s\right)=\frac{12(s+2)}{s(s+1)(s+3)(s+4)$

Determine the final value of this voltage. that is,

${\upsilon}_{0}\left(t\right)$ as $t\to \mathrm{\infty}$

a) 6V

b) 2V

c) 12V

d) 4V

Determine the final value of this voltage. that is,

a) 6V

b) 2V

c) 12V

d) 4V

asked 2021-01-23

Laplace transform of $\mathrm{sin}(2t)+{e}^{t}\mathrm{sin}(t)$

asked 2020-10-18

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

where we assume s is a positive real number. For example, to find the Laplace transform of

Verify the following Laplace transforms, where u is a real number.

asked 2021-02-19

Solve the differential equation using Laplace transform of

${y}^{\u2033}-3{y}^{\prime}+2y={e}^{3t}$

when y(0)=0 and y'(0)=0

when y(0)=0 and y'(0)=0

asked 2022-09-11

Try to find the inverse Laplace transform for:

$$\frac{1}{s+1}{e}^{-s}$$

The answer says that the inverse Laplace transform is:

$${\mathcal{L}}^{-1}\left(\frac{1}{s+1}{e}^{-s}\right)={e}^{t-1}u(t-1)$$

I'm aware that the Heaviside function's transform is:

$$\begin{array}{rl}\mathcal{L}(u(t-a))& =\frac{1}{s}{e}^{-as}\\ \mathcal{L}(f(t-a)u(t-a))& ={e}^{-as}F(s)\end{array}$$

but I'm having trouble figuring out how the inverse transform was derived.

$$\frac{1}{s+1}{e}^{-s}$$

The answer says that the inverse Laplace transform is:

$${\mathcal{L}}^{-1}\left(\frac{1}{s+1}{e}^{-s}\right)={e}^{t-1}u(t-1)$$

I'm aware that the Heaviside function's transform is:

$$\begin{array}{rl}\mathcal{L}(u(t-a))& =\frac{1}{s}{e}^{-as}\\ \mathcal{L}(f(t-a)u(t-a))& ={e}^{-as}F(s)\end{array}$$

but I'm having trouble figuring out how the inverse transform was derived.

asked 2020-11-29

Find the Laplace Transform of the function

$f(t)={e}^{at}$

asked 2021-09-23

Find the inverse Laplace transform of $F\left(s\right)=\frac{7s-6}{{s}^{2}-4}$