How do I find it?

Noe Cowan
2022-11-19
Answered

The Laplace transform of $\mathcal{L}(t{e}^{t}\mathrm{cos}t)$

How do I find it?

How do I find it?

You can still ask an expert for help

metodikkf6z

Answered 2022-11-20
Author has **14** answers

$\text{Another approach:}\mathcal{L}({e}^{t}tcost)=F(s-1)$

$\mathcal{L}(tcost)=-\frac{d}{ds}(\frac{s}{{s}^{2}+1})=\frac{{s}^{2}-1}{({s}^{2}+1{)}^{2}}$

$\text{so the final answer is:}$

$\text{}F(s-1)=\frac{(s-1{)}^{2}-1}{[(s-1{)}^{2}+1{]}^{2}}$

$\mathcal{L}(tcost)=-\frac{d}{ds}(\frac{s}{{s}^{2}+1})=\frac{{s}^{2}-1}{({s}^{2}+1{)}^{2}}$

$\text{so the final answer is:}$

$\text{}F(s-1)=\frac{(s-1{)}^{2}-1}{[(s-1{)}^{2}+1{]}^{2}}$

Annie French

Answered 2022-11-21
Author has **4** answers

You need the relation

$\mathcal{L}\{tf(t)\}\u27fa-{F}^{\prime}(s)$

i.e. multiplication in the time domain corresponds to differentiation in the s-domain (and a negative sign). Since you know F(s), you can easily derive the result.

$\mathcal{L}\{tf(t)\}\u27fa-{F}^{\prime}(s)$

i.e. multiplication in the time domain corresponds to differentiation in the s-domain (and a negative sign). Since you know F(s), you can easily derive the result.

asked 2021-12-31

Find the solution of the following Differential Equations by Exact $(2y+xy)dx+2xdy=0,y\left(3\right)=\sqrt{2}$

asked 2021-02-25

Use properties of the Laplace transform to answer the following

(a) If$f(t)=(t+5{)}^{2}+{t}^{2}{e}^{5t}$ , find the Laplace transform,$L[f(t)]=F(s)$ .

(b) If$f(t)=2{e}^{-t}\mathrm{cos}(3t+\frac{\pi}{4})$ , find the Laplace transform, $L[f(t)]=F(s)$ . HINT:

$\mathrm{cos}(\alpha +\beta )=\mathrm{cos}(\alpha )\mathrm{cos}(\beta )-\mathrm{sin}(\alpha )\mathrm{sin}(\beta )$

(c) If$F(s)=\frac{7{s}^{2}-37s+64}{s({s}^{2}-8s+16)}$ find the inverse Laplace transform, ${L}^{-1}|F(s)|=f(t)$

(d) If$F(s)={e}^{-7s}(\frac{1}{s}+\frac{s}{{s}^{2}+1})$ , find the inverse Laplace transform, ${L}^{-1}[F(s)]=f(t)$

(a) If

(b) If

(c) If

(d) If

asked 2021-12-29

A metal is heated up to a temperature of ${500}^{\circ}C$ . It is then exposed to a temperature of ${38}^{\circ}C$ . After 2 minutes, the temperature of the metal becomes ${190}^{\circ}C$ . When will the temperature be ${100}^{\circ}C$ ?

asked 2022-01-20

The author of my textbook asks to verify that the function:

$y=\sqrt{\frac{2}{3}\mathrm{ln}(1+{x}^{2})+C}$

solves the differential equation

$\frac{dy}{dx}=\frac{{x}^{3}}{y+y{x}^{3}}$

However, this is an error and this y does not solve the differential equation. Is there a simple typo that makes the problem workable?

solves the differential equation

However, this is an error and this y does not solve the differential equation. Is there a simple typo that makes the problem workable?

asked 2021-06-24

In many physical applications, the nonhomogeneous term F(x) is specified by different formulas in different intervals of x. (a) Find a general solution of the equation

asked 2022-09-26

I calculate Laplace transform.

$$f\left(t\right)={\displaystyle \frac{{e}^{{\lambda}_{1}t}-{e}^{{\lambda}_{2}t}}{t}}$$

$${\lambda}_{1},{\lambda}_{2}>0$$

I think

$${\int}_{0}^{\mathrm{\infty}}f\left(t\right){e}^{-st}dt={\int}_{0}^{\mathrm{\infty}}{\displaystyle \frac{{e}^{{\lambda}_{1}t}-{e}^{{\lambda}_{2}t}}{t}}{e}^{-st}dt$$

$$={\int}_{0}^{\mathrm{\infty}}{\displaystyle \frac{{e}^{({\lambda}_{1}-s)t}}{t}}dt-{\int}_{0}^{\mathrm{\infty}}{\displaystyle \frac{{e}^{({\lambda}_{2}-s)t}}{t}}dt$$

I don't know what to do next.

$$f\left(t\right)={\displaystyle \frac{{e}^{{\lambda}_{1}t}-{e}^{{\lambda}_{2}t}}{t}}$$

$${\lambda}_{1},{\lambda}_{2}>0$$

I think

$${\int}_{0}^{\mathrm{\infty}}f\left(t\right){e}^{-st}dt={\int}_{0}^{\mathrm{\infty}}{\displaystyle \frac{{e}^{{\lambda}_{1}t}-{e}^{{\lambda}_{2}t}}{t}}{e}^{-st}dt$$

$$={\int}_{0}^{\mathrm{\infty}}{\displaystyle \frac{{e}^{({\lambda}_{1}-s)t}}{t}}dt-{\int}_{0}^{\mathrm{\infty}}{\displaystyle \frac{{e}^{({\lambda}_{2}-s)t}}{t}}dt$$

I don't know what to do next.

asked 2022-01-20

I was reading some of my notes and I was not sure how the following works:

$\frac{{d}^{2}y}{{dx}^{2}}+\frac{dy}{dx}=0$

Solve the above with the condition$y\left(0\right)=0$

$\Rightarrow y\left(x\right)=A(1-{e}^{-x})$ with A an arbitrary constant.

I was just wondering how do you solve the above to get

$y\left(x\right)=A(1-{e}^{-x})$ ? Because when I tried it, I got:

$y=A+B{e}^{-x}$ , and using the condition $y\left(0\right)=0$ , I got $A+B=0$ .

Solve the above with the condition

I was just wondering how do you solve the above to get