Find the surface area. SHOW WORK

) Use the Laplace transform to solve the following initial value problem: y′′−6y′+9y=0y(0)=4,y′(0)=2 Using Y for the Laplace tra

artcher [175]

Answer:

$y(t)=2e^{3t}(2-5t)$

Step-by-step explanation:

Let Y(s) be the Laplace transform Y=L{y(t)} of y(t)

Applying the Laplace transform to both sides of the differential equation and using the linearity of the transform, we get

L{y'' - 6y' + 9y} = L{0} = 0

(*) L{y''} - 6L{y'} + 9L{y} = 0 ; y(0)=4, y′(0)=2

Using the theorem of the Laplace transform for derivatives, we know that:

$\large\bf L\left\{y''\right\}=s^2Y(s)-sy(0)-y'(0)\\\\L\left\{y'\right\}=sY(s)-y(0)$

Replacing the initial values y(0)=4, y′(0)=2 we obtain

$\large\bf L\left\{y''\right\}=s^2Y(s)-4s-2\\\\L\left\{y'\right\}=sY(s)-4$

and our differential equation (*) gets transformed in the algebraic equation

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0$

Solving for Y(s) we get

$\large\bf s^2Y(s)-4s-2-6(sY(s)-4)+9Y(s)=0\Rightarrow (s^2-6s+9)Y(s)-4s+22=0\Rightarrow\\\\\Rightarrow Y(s)=\frac{4s-22}{s^2-6s+9}$

Now, we brake down the rational expression of Y(s) into partial fractions

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4s-22}{(s-3)^2}=\frac{A}{s-3}+\frac{B}{(s-3)^2}$

The numerator of the addition at the right must be equal to 4s-22, so

A(s - 3) + B = 4s - 22

As - 3A + B = 4s - 22

we deduct from here

A = 4 and -3A + B = -22, so

A = 4 and B = -22 + 12 = -10

It means that

$\large\bf \frac{4s-22}{s^2-6s+9}=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

and

$\large\bf Y(s)=\frac{4}{s-3}-\frac{10}{(s-3)^2}$

By taking the inverse Laplace transform on both sides and using the linearity of the inverse:

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}$

we know that

$\large\bf L^{-1}\left\{\frac{1}{s-3}\right\}=e^{3t}$

and for the first translation property of the inverse Laplace transform

$\large\bf L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=e^{3t}L^{-1}\left\{\frac{1}{s^2}\right\}=e^{3t}t=te^{3t}$

and the solution of our differential equation is

$\large\bf y(t)=L^{-1}\left\{Y(s)\right\}=4L^{-1}\left\{\frac{1}{s-3}\right\}-10L^{-1}\left\{\frac{1}{(s-3)^2}\right\}=\\\\4e^{3t}-10te^{3t}=2e^{3t}(2-5t)\\\\\boxed{y(t)=2e^{3t}(2-5t)}$

5 0

3 years ago

A club is making posters to raise money. The printer charges a base fee of $270 and $2 per poster for supplies. A person sells e

sesenic [268]

Answer:

He must sell 190 posters make a profit of $300.

Explanation:

Given:

Base fee= $270

Fee for supplies= $2

Cost of poster=$5

To find:

The number of posters to be sold to get a profit of $300=?

Solution:

Let the number of posters sold be n.

Now, we know that,

profit = selling price – cost price

$ 300 = $ 5 for each poster – ( $ 270 base fee + $ 2 for each poster)

300 = 5 x n – (270 + 2 x n)

300 = 5n – 270 – 2n

5n – 2n = 300 + 270

3n = 570

$n=\frac{570}{3}$

n = 190

6 0

3 years ago

The percent of working students increased 8.1 to 40.5 what was the present prior to increase

Dmitrij [34]

Answer:

32.4

Step-by-step explanation:

prior + 8.1 = 40.5 . . . . . . seems to model the problem statement

prior = 32.4 . . . . . . . subtract 8.1 from both sides

Prior to the increase the percent was 32.4.

_____

Comment on the problem statement

When you're talking about a percentage increase in a percentage, it is almost never clear whether you're talking about the percentage of the underlying number, or the percentage of the percentage.

Here, we assume the 8.1 is a percentage of working students, not a percentage of the percentage of workings students. If you actually intend the latter, the percentage before the increase was about 37.465%.

6 0

3 years ago

Y=x+2

pickupchik [31]

Answer:

Noiccce

Step-by-step explanation:

Noiccce

8 0

3 years ago

Read 2 more answers

Gloria collected 23 faintail and comet goldfish. There were 11 fewer fantails than comets. How many comets did Gloria have?

likoan [24]

There’s 23 total of both faintail and comets. But there’s 11 Fewer faintails, so 23-11= 12

5 0

3 years ago