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3 years ago

Don’t know how to do this one pls help me if you can I would really appreciate it

Mathematics

1 answer:

KengaRu [80]3 years ago

6 0

Answer:

PRICE IS 23.45

Step-by-step explanation:

Branlyist

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Consider the initial value problem y′+5y=⎧⎩⎨⎪⎪0110 if 0≤t<3 if 3≤t<5 if 5≤t<[infinity],y(0)=4. y′+5y={0 if 0≤t<311 i

rosijanka [135]

It looks like the ODE is

$y'+5y=\begin{cases}0&\text{for }0\le t$

with the initial condition of $y(0)=4$ .

Rewrite the right side in terms of the unit step function,

$u(t-c)=\begin{cases}1&\text{for }t\ge c\\0&\text{for }t$

In this case, we have

$\begin{cases}0&\text{for }0\le t$

The Laplace transform of the step function is easy to compute:

$\displaystyle\int_0^\infty u(t-c)e^{-st}\,\mathrm dt=\int_c^\infty e^{-st}\,\mathrm dt=\frac{e^{-cs}}s$

So, taking the Laplace transform of both sides of the ODE, we get

$sY(s)-y(0)+5Y(s)=\dfrac{e^{-3s}-e^{-5s}}s$

Solve for $Y(s)$ :

$(s+5)Y(s)-4=\dfrac{e^{-3s}-e^{-5s}}s\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}{s(s+5)}+\dfrac4{s+5}$

We can split the first term into partial fractions:

$\dfrac1{s(s+5)}=\dfrac as+\dfrac b{s+5}\implies1=a(s+5)+bs$

If $s=0$ , then $1=5a\implies a=\frac15$ .

If $s=-5$ , then $1=-5b\implies b=-\frac15$ .

$\implies Y(s)=\dfrac{e^{-3s}-e^{-5s}}5\left(\frac1s-\frac1{s+5}\right)+\dfrac4{s+5}$

$\implies Y(s)=\dfrac15\left(\dfrac{e^{-3s}}s-\dfrac{e^{-3s}}{s+5}-\dfrac{e^{-5s}}s+\dfrac{e^{-5s}}{s+5}\right)+\dfrac4{s+5}$

Take the inverse transform of both sides, recalling that

$Y(s)=e^{-cs}F(s)\implies y(t)=u(t-c)f(t-c)$

where $F(s)$ is the Laplace transform of the function $f(t)$ . We have

$F(s)=\dfrac1s\implies f(t)=1$

$F(s)=\dfrac1{s+5}\implies f(t)=e^{-5t}$

We then end up with

$y(t)=\dfrac{u(t-3)(1-e^{-5t})-u(t-5)(1-e^{-5t})}5+5e^{-5t}$

3 0

3 years ago

Li believes that the graph shows a direct variation. Why is Li incorrect in saying that the graph shows a direct variation?

padilas [110]

Answer:

Option 2) When the x-value is 0, the y-value is 1.

Step-by-step explanation:

A relationship between two variables, x, and y, represent a proportional variation if it can be expressed in the form $y/x=k$ or $y=kx$

In a proportional relationship the constant of proportionality k is equal to the slope m of the line <em>and the line passes through the origin </em>

Remember that in a direct variation

For x=0, the value of y is equal to zero too

therefore

The graph is not a direct variation , because

When the x-value is 0, the y-value is 1.

3 0

3 years ago

Read 2 more answers

PLZZZ HELP!!!! (WILL GIVE BRAINLIEST!!!!!!!)

tatiyna

Y = -5x -5

To check, we can insert 0 for y and -1 for x

0 ? (-5X-1) - 5
0 ? 5 - 5
0 = 0

So yes, the equation is correct

4 0

3 years ago

A number is four times larger than the square of a third of the number, THIS QUESTION IS WORTH 15

lawyer [7]

The number can be 0 or1

7 0

3 years ago

Why the product (-6)(-3) is positive

rewona [7]

Answer:

hope this helps

have a good day :)

Step-by-step explanation:

7 0

3 years ago