$$f(t)=t{\int}_{0}^{t}\tau {e}^{-\tau}$$

$L(f)(s)$=??

Sara Solomon
2022-09-08
Answered

Find the Laplace transform of

$$f(t)=t{\int}_{0}^{t}\tau {e}^{-\tau}$$

$L(f)(s)$=??

$$f(t)=t{\int}_{0}^{t}\tau {e}^{-\tau}$$

$L(f)(s)$=??

You can still ask an expert for help

Peutiedw

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

First write the function as

$$f(t)=t{\int}_{0}^{t}\tau {e}^{-\tau}=tg(t),$$

then you can use the following properties of the Laplace transform :

1) $$\mathcal{L}({\int}_{0}^{t}h(\tau )d\tau )(s)=\frac{H(s)}{s},$$

where H(s) is the Laplace transform of h.

2) $$\mathcal{L}({t}^{n}g(t))=(-1{)}^{n}{G}^{(n)}(s),$$

where G(s) is the Laplace transform of g(t).

$$f(t)=t{\int}_{0}^{t}\tau {e}^{-\tau}=tg(t),$$

then you can use the following properties of the Laplace transform :

1) $$\mathcal{L}({\int}_{0}^{t}h(\tau )d\tau )(s)=\frac{H(s)}{s},$$

where H(s) is the Laplace transform of h.

2) $$\mathcal{L}({t}^{n}g(t))=(-1{)}^{n}{G}^{(n)}(s),$$

where G(s) is the Laplace transform of g(t).

Dana Russo

Answered 2022-09-10
Author has **2** answers

$${\int}_{0}^{t}d\tau \phantom{\rule{thinmathspace}{0ex}}\tau \phantom{\rule{thinmathspace}{0ex}}{e}^{-\tau}={[-\tau \phantom{\rule{thinmathspace}{0ex}}{e}^{-\tau}]}_{0}^{t}+{\int}_{0}^{t}d\tau \phantom{\rule{thinmathspace}{0ex}}{e}^{-\tau}=1-(1+t){e}^{-t}$$

$$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}f(t)\phantom{\rule{thinmathspace}{0ex}}{e}^{-st}& ={\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}t\phantom{\rule{thinmathspace}{0ex}}{e}^{-st}-{\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}(t+{t}^{2}){e}^{-(s+1)t}\\ & =\frac{1}{{s}^{2}}-\frac{1}{(s+1{)}^{2}}-\frac{2}{(s+1{)}^{3}}\end{array}$$

$$\begin{array}{rl}{\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}f(t)\phantom{\rule{thinmathspace}{0ex}}{e}^{-st}& ={\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}t\phantom{\rule{thinmathspace}{0ex}}{e}^{-st}-{\int}_{0}^{\mathrm{\infty}}dt\phantom{\rule{thinmathspace}{0ex}}(t+{t}^{2}){e}^{-(s+1)t}\\ & =\frac{1}{{s}^{2}}-\frac{1}{(s+1{)}^{2}}-\frac{2}{(s+1{)}^{3}}\end{array}$$

asked 2022-11-05

Using Laplace transforms find the solution to a differential equation.

$${y}^{\prime}+\frac{2}{t}y=t-1+\frac{1}{t}.$$

with initial condition $y(1)=1/2$

I know how to do it using Laplace transformations when initial condition is of y(0)=a where a is a constant but not otherwise because we were taught the Laplace transform of y′ is $sY(s)-y(0).$

I can solve this equation easily with a linear approach.

solution being

$$y(t)=({t}^{2}/4)-(t/3)+(1/2)+(1/(12{t}^{2}))$$

Can someone show me how to solve this initial value problem using Laplace transforms?

$${y}^{\prime}+\frac{2}{t}y=t-1+\frac{1}{t}.$$

with initial condition $y(1)=1/2$

I know how to do it using Laplace transformations when initial condition is of y(0)=a where a is a constant but not otherwise because we were taught the Laplace transform of y′ is $sY(s)-y(0).$

I can solve this equation easily with a linear approach.

solution being

$$y(t)=({t}^{2}/4)-(t/3)+(1/2)+(1/(12{t}^{2}))$$

Can someone show me how to solve this initial value problem using Laplace transforms?

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