I need help with this equation

Use Laplace transform methods to solve the differential equation: The initial conditions are f(0) = 0 and <img src="https://tex.

Llana [10]

Answer:

Step-by-step explanation:

$\frac{d^2 y}{dt^2} +12\frac{dy}{dt} +32y = 10e^{-2t}$ where f(x) is replaced by y

Take Laplace on both sides

$s^2 Y -s(0)-0 +12s Y-0+32Y = \frac{10}{s+2} \\Y(s^2+12s+32) = \frac{10}{s+2}\\Y = \frac{10}{(s+2)(s+4)(s+8)}$

We can resolve into partial fraction to get Y = L(y)

Let this equals

$\frac{A}{s+2} +\frac{B}{s+4} +\frac{C}{s+8} \\10 = A(s+4)(s+8) + B(s+2)(s+8) +Cs+4)(s+2) \\$

Solving we get

s=-9

24C = 10 or C =5/12

s=-2: A=10/12=5/6

s=-4: B = -5/4

Taking inverse Laplace

$y(t) = \frac{-5e^{-4t} }{4} + \frac{5e^{-8t} }{12} +\frac{-5e^{-2t} }{6}$

6 0

3 years ago

A survey among freshman at a certain university revealed that the number of hours spent studying before final exams was normally

patriot [66]

Answer: 0.0775

Step-by-step explanation:

Given : Mean : $\mu = 25$

Standard deviation : $\sigma =15$

Sample size : $n=36$

Since its normal distribution , then the formula to calculate the z-score is given by :-

$z=\dfrac{x-\mu}{\dfrac{\sigma}{\sqrt{n}}}$

For x= 28.2 hours

$z=\dfrac{28.2-25}{\dfrac{15}{\sqrt{36}}}=1.28$

For x= 30 hours

$z=\dfrac{30-25}{\dfrac{15}{\sqrt{36}}}=2$

The P- value = $P(1.28$

$=P(z$

Hence, the probabiliy that the average time spent stydying for the sampe was between 28.2 and 30 hours = 0.0775

6 0

3 years ago

15 points for free i guess lol

marysya [2.9K]

Answer:

WEEEEEEE!

Step-by-step explanation:

Freeeee? Sorry.

7 0

3 years ago

Read 2 more answers

What is the common ratio for the geometric sequence?<br><br> 18, 12, 8, 16/3,…

alex41 [277]

Answer:

2/3

Step-by-step explanation:

Given : A geometric sequence : 18, 12, 8, 16/3,…

Solution :

Each term is found by multiplying the previous term by a constant in a Geometric Sequence .

In a Geometric Sequence common ratio is denoted by r

$r = \frac{a_{n} }{a_{n-1} }$

for n = 2

$r = \frac{a_{2} }{a_{1} }$

⇒ $r = \frac{12 }{18 }$

⇒ $r = \frac{2 }{3 }$

for n = 3

$r = \frac{a_{3} }{a_{2} }$

⇒ $r = \frac{8 }{12 }$

⇒ $r = \frac{4 }{6 }$

⇒ $r = \frac{2}{3 }$

Thus the common ratio 'r' = 2/3

6 0

3 years ago

Read 2 more answers

you drop a ball from a height of 98 feet. at the same time, your friend throws a ball upward. the polynomials represent the heig

Ostrovityanka [42]

a) $h_0 -u_y t$

b) See interpretation below

Step-by-step explanation:

a)

The motion of both balls is a free-fall motion: it means that the ball is acted upon the force of gravity only.

Therefore, this means that the motion of the ball is a uniformly accelerated motion, with constant acceleration equal to the acceleration of gravity:

$g=32 ft/s^2$

in the downward direction.

For the ball dropped from the initial height of $h_0 = 98 ft$ , the height at time t is given by

$h(t) = h_0 -\frac{1}{2}gt^2$ (1)

The ball which is thrown upward from the ground instead is fired with an initial vertical velocity $u_y$ , and its starting height is zero, so its position at time t is given by

$h'(t)=u_y t - \frac{1}{2}gt^2$ (2)

Therefore, the polynomial that represents the distance between the two balls is:

$h(t)-h'(t)=h_0 - \frac{1}{2}gt^2 - (u_y t - \frac{1}{2}gt^2) = h_0 -u_y t$

b)

Now we interpret this polynomial, which is:

$\Delta h(t) = h_0 -u_y t$

which represents the distance between the two balls at time t.

The interpretation of the two terms is the following:

- The constant term, $h_0$ , is the initial distance between the two balls, at time t=0 (in fact, the first ball is still at the top of the building, while the second ball is on the ground). For this problem, $h_0 = 98 ft$

- The coefficient of the linear term, $u_y$ , is the initial velocity of the second ball; this terms tells us that the distance between the two balls decreases every second by $u_y$ feet.

5 0

3 years ago