Can anyone give me a hint to solve laplace transform $$\mathcal{L}\left({\int}_{0}^{t}\frac{{e}^{-\tau}-1}{\tau}d\tau \right)$$

beobachtereb
2022-09-26
Answered

Can anyone give me a hint to solve laplace transform $$\mathcal{L}\left({\int}_{0}^{t}\frac{{e}^{-\tau}-1}{\tau}d\tau \right)$$

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AKPerqk

Answered 2022-09-27
Author has **9** answers

First try finding the laplace transform of this function:

$$\frac{{e}^{-t}-1}{t}$$

Using the rule:

$$L\left(\frac{f(t)}{t}\right)={\int}_{s}^{\mathrm{\infty}}g(s)$$

where

$$L(f(t))=g(s)$$

then try using the rule

$$L({\int}_{0}^{t}f(t)dt)=\frac{g(s)}{s}$$

$$\frac{{e}^{-t}-1}{t}$$

Using the rule:

$$L\left(\frac{f(t)}{t}\right)={\int}_{s}^{\mathrm{\infty}}g(s)$$

where

$$L(f(t))=g(s)$$

then try using the rule

$$L({\int}_{0}^{t}f(t)dt)=\frac{g(s)}{s}$$

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 2022-10-10

Finding the inverse laplace ${\mathcal{L}}^{-1}\{\frac{1}{-{s}^{2}-2s+37}\}=?$

asked 2022-09-10

I've been working to find inverse Laplace transform for the following :

$$\frac{A}{(s-a)(s-{r}_{1})(s-{r}_{2})}$$

However, I'm getting stuck on the partial fraction decomposition. When I run the decomposition in Wolfram Alpha, it comes back as

$$-\frac{A}{(s-{r}_{1})(a-{r}_{1})({r}_{1}-{r}_{2})}-\frac{A}{(s-{r}_{2})(a-{r}_{2})({r}_{2}-{r}_{1})}+\frac{A}{(a-{r}_{1})(a-{r}_{2})(s-a)}$$

Any thoughts on how this decomposition works?

$$\frac{A}{(s-a)(s-{r}_{1})(s-{r}_{2})}$$

However, I'm getting stuck on the partial fraction decomposition. When I run the decomposition in Wolfram Alpha, it comes back as

$$-\frac{A}{(s-{r}_{1})(a-{r}_{1})({r}_{1}-{r}_{2})}-\frac{A}{(s-{r}_{2})(a-{r}_{2})({r}_{2}-{r}_{1})}+\frac{A}{(a-{r}_{1})(a-{r}_{2})(s-a)}$$

Any thoughts on how this decomposition works?

asked 2022-10-12

Find the inverse Laplace transform of $$\frac{1}{{({s}^{2}+1)}^{2}}$$

asked 2022-11-05

Find

$$1\ast 1\ast 1\ast \cdots \ast 1\phantom{\rule{1em}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{factors}$$

that is, a function f(t)=1 convolution with itself for a total of n factors.

$$1\ast 1\ast 1\ast \cdots \ast 1\phantom{\rule{1em}{0ex}}n\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\text{factors}$$

that is, a function f(t)=1 convolution with itself for a total of n factors.

asked 2021-12-28

What is the general solution for the high-order differential equations ${y}^{5}+6{y}^{4}+12y+8y=0$

asked 2021-12-29

Auxiliary Equation:

Solve the following:

$(3{D}^{3}-22{D}^{2}+7D)y=0$

Solve the following: