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Nuetrik [128]
3 years ago
9

PLEASE PLEASE HURRY!!!!!!!

Mathematics
1 answer:
zvonat [6]3 years ago
3 0

Answer:x=2 y=3

Step-by-step explanation:

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Step-by-step explanation:

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2 years ago
Find the Laplace Transform of y''+7y'+7y given that y(0)=0 and y'(0)=7
Iteru [2.4K]
Assuming you start with the homogeneous ODE,

y''+7y'+7y=0

upon taking the Laplace transform of both sides, you end up with

\mathcal L\left\{y''+7y'+7y\right\}=\mathcal L\{0\}
\mathcal L\{y''\}+7\mathcal L\{y'\}+7\mathcal L\{y\}=0

since the transform operator is linear, and the transform of 0 is 0.

I'll denote the Laplace transform of a function y(t) into the s-domain by \mathcal L_s\{y(t)\}:=Y(s).

Given the derivative of y(t), its Laplace transform can be found easily from the definition of the transform itself:

Y(s)=\displaystyle\int_0^\infty y(t)e^{-st}\,\mathrm dt
\implies\mathcal L_s\{y'(t)\}=\displaystyle\int_0^\infty y'(t)e^{-st}\,\mathrm dt

Integrate by parts, setting

u=e^{-st}\implies\mathrm du=-se^{-st}\,\mathrm dt
\mathrm dv=y'(t)\,\mathrm dt\implies v=y(t)

so that

\mathcal L_s\{y'(t)\}=y(t)e^{-st}\bigg|_{t=0}^{t\to\infty}+s\displaystyle\int_0^\infty y(t)e^{-st}\,\mathrm dt

The second term is just the transform of the original function, while the first term reduces to y(0) since e^{-st}\to0 as t\to\infty, and e^{-st}\to1 as t\to0. So we have a rule for transforming the first derivative, and by the same process we can generalize it to any order provided that we're given the value of all the preceeding derivatives at t=0.

The general rule gives us

\mathcal L_s\{y(t)\}=Y(s)
\mathcal L_s\{y'(t)\}=sY(s)-y(0)
\mathcal L_s\{y''(t)\}=s^2Y(s)-sy(0)-y'(0)

and so our ODE becomes

\bigg(s^2Y(s)-sy(0)-y'(0)\bigg)+7\bigg(sY(s)-y(0)\bigg)+7Y(s)=0
(s^2+7s+7)Y(s)-7=0
Y(s)=\dfrac7{s^2+7s+7}
Y(s)=\dfrac{14}{\sqrt{21}}\dfrac{\frac{\sqrt{21}}2}{\left(s+\frac72\right)^2-\left(\frac{\sqrt{21}}2\right)^2}

Depending on how you learned about finding inverse transforms, you should either be comfortable with cross-referencing a table and do some pattern-matching, or be able to set up and compute an appropriate contour integral. The former approach seems to be more common, so I'll stick to that.

Recall that

\mathcal L_s\{\sinh(at)\}=\dfrac a{s^2-a^2}

and that given a function y(t) with transform Y(s), the shifted transform Y(s-c) corresponds to the function e^{ct}y(t).

We have

Y(s)=\dfrac{14}{\sqrt{21}}\dfrac{\frac{\sqrt{21}}2}{\left(s+\frac72\right)^2-\left(\frac{\sqrt{21}}2\right)^2}
\implies Y\left(s-\dfrac72\right)=\dfrac{14}{\sqrt{21}}\dfrac{\frac{\sqrt{21}}2}{s^2-\left(\frac{\sqrt{21}}2\right)^2}

and so the inverse transform for our ODE is

\mathcal L^{-1}_t\left\{Y\left(s-\dfrac72\right)\right\}=\dfrac{14}{\sqrt{21}}\mathcal L^{-1}_t\left\{\dfrac{\frac{\sqrt{21}}2}{s^2-\left(\frac{\sqrt{21}}2\right)^2}\right\}
\implies e^{7/2t}y(t)=\dfrac{14}{\sqrt{21}}\sinh\left(\dfrac{\sqrt{21}}2t\right)
\implies y(t)=\dfrac{14}{\sqrt{21}}e^{-7/2t}\sinh\left(\dfrac{\sqrt{21}}2t\right)

and in case you're not familiar with hyperbolic functions, you have

\sinh t=\dfrac{e^t-e^{-t}}2
7 0
2 years ago
D. $3.25 A cell phone at Store A costs $10 less than twice the cost of the same cell phone at Store B. If the phone costs $49.99
Gekata [30.6K]

Answer:

A or $89.98

Step-by-step explanation:

49.99 x 2 = 99.98. 99.98 - 10.00 = 89.98

4 0
3 years ago
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