The Laplace inverse of ${L}^{-1}\left[\frac{s}{{s}^{2}+{5}^{2}}\right]$ is

$a)\mathrm{cos}(5t)$

$b)\mathrm{sin}h(5t)$

$c)\mathrm{sin}(5t)$

$d)\mathrm{cos}h(5t)$

Armorikam
2020-12-30
Answered

The Laplace inverse of ${L}^{-1}\left[\frac{s}{{s}^{2}+{5}^{2}}\right]$ is

$a)\mathrm{cos}(5t)$

$b)\mathrm{sin}h(5t)$

$c)\mathrm{sin}(5t)$

$d)\mathrm{cos}h(5t)$

You can still ask an expert for help

oppturf

Answered 2020-12-31
Author has **94** answers

Step 1

To find the Laplace transformation of${L}^{-1}\left[\frac{s}{{s}^{2}+{5}^{2}}\right]$

Solution:

let$\phi (s)$ be the function whose Laplace transformation will be $\frac{s}{{s}^{2}+{5}^{2}}$

So,

$L(\phi (s))=\frac{s}{{s}^{2}+{5}^{2}}$

Applying inverse we get,

$\phi (s)={L}^{-1}\left[\frac{s}{{s}^{2}+{5}^{2}}\right]\dots (1)$

Step 2

since, we know that

$L(\mathrm{cos}as)=\frac{s}{{s}^{2}+{a}^{2}}$

$\text{and}(\mathrm{cos}as)={L}^{-1}\left(\frac{s}{{s}^{2}+{a}^{2}}\right)$

From equation (1),

$\mathrm{cos}(5s)=\frac{s}{{s}^{2}+{5}^{2}}$

Hence, the inverse Laplace for given expression is cos5t.

To find the Laplace transformation of

Solution:

let

So,

Applying inverse we get,

Step 2

since, we know that

From equation (1),

Hence, the inverse Laplace for given expression is cos5t.

asked 2021-02-12

Solve $y"+4y=f(t)\text{subject to}y(0)=0,{y}^{\prime}(0)=2\text{and}$

$f(t)=\{\begin{array}{ll}0& t<\pi \\ 1& t\ge \pi \end{array}$

asked 2020-11-14

Solve differential equation

asked 2021-01-05

Linear equations of second order with constant coefficients. Find all solutions on
$(-\mathrm{\infty},+\mathrm{\infty}).y+4y=0$

asked 2021-12-28

Find the inverse Laplace transform f(t) of the following transforms F(s). Match each of the following to the value of f(1.234) to two decimal places.

asked 2020-12-05

Use Theorem 7.4.2 to evaluate the given Laplace transform. Do not evaluate the convolution integral before transforming. (Write your answer as a function of s.)

$L\{{t}^{2}\cdot t{e}^{t}\}$

asked 2022-01-22

Important

Solve the bernoullis

Solve the bernoullis

asked 2022-07-17

Solve this separable differential equation

How would I go about solving the following separable differential equation?

$\frac{dx(t)}{dt}=8-3x$ with $x(0)=4$?

My solution thus far is the following:

$\int \frac{dx(t)}{8-3x}=\int dt+C$

$\Rightarrow \text{ln}|8-3x(t)|=-3(t+C)$

Now using the fact that $x(0)=4$ we can solve for $C=-\frac{1}{3}\text{ln}|4|$. From this, I would seem to get

$|8-3x(t)|={e}^{-3t-\frac{1}{3}\text{ln}4}$

but something seems to go wrong. Am I making a mistake somewhere and how could I solve this entirely?

How would I go about solving the following separable differential equation?

$\frac{dx(t)}{dt}=8-3x$ with $x(0)=4$?

My solution thus far is the following:

$\int \frac{dx(t)}{8-3x}=\int dt+C$

$\Rightarrow \text{ln}|8-3x(t)|=-3(t+C)$

Now using the fact that $x(0)=4$ we can solve for $C=-\frac{1}{3}\text{ln}|4|$. From this, I would seem to get

$|8-3x(t)|={e}^{-3t-\frac{1}{3}\text{ln}4}$

but something seems to go wrong. Am I making a mistake somewhere and how could I solve this entirely?