You and your friend each start a car-washing service.

) 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

Help me solve asap!!

Dovator [93]

Answer:

2/16

Step-by-step explanation:

The answer is 2/16

brainliest?

4 0

3 years ago

Explain it step by step

NemiM [27]

Answer:

Step-by-step explanation:

Like terms have same variable with same power. Combine like terms

153y³ + 132y² + 6y - 5 - 3y³ - 5y² +4y - 2

= 153y³ - 3y³ + 132y² - 5y² + 6y + 4y -5 - 2

= 150y³ + 127y² + 10y - 7

3 0

2 years ago

Eight children are divided into two teams each containing 4 children. how many different divisions are possible?

Alekssandra [29.7K]

So this is what's known as a combination. a combination is a group where the order does not matter. to solve a combination, you use this equation:
n! / (r! *(n-r)!)
n is the total number of objects being sorted (8)
r is the group you need to sort them into (4)
(i know it's 2 groups of 4 but if you solve this for 1 group, the other group is automatically made of the remaining kids)
definition of !:
a number with a ! means 1*2*3...*that number (so that number multiplied by every number smaller than it all the way to one)

so we put 8 and 4 in this equation:
8! / (4! *(8-4)!)
simplify:
8! / *(4! 4!)
multiply out:
(8*7*6*5*4*3*2*1) / (4*3*2*1 * 4*3*2*1)
and put that in your calculator
you will get 70
your answer is 70

4 0

3 years ago

2. The original cost of a clock is $60 but you

Cerrena [4.2K]

Answer:

The tax will be 3.60

Step-by-step explanation:

60 x 6% (0.06)

5 0

2 years ago