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3 years ago

Solve the initial-value problem. x' + 2tx = 5t, x(0) = 8 x(t) =

Mathematics

1 answer:

11111nata11111 [884]3 years ago

8 0

Multiply both sides of the ODE

$x'+2tx+5t$

by $e^{t^2}$ :

$e^{t^2}x'+2te^{t^2}x=5te^{t^2}$

Now the left side can be condensed as the derivative of a product:

$\left(e^{t^2}x\right)'=5te^{t^2}$

Integrate both sides, then solve for <em>x</em> :

$e^{t^2}x=\dfrac52e^{t^2}+C$

$\implies x(t)=\dfrac52+Ce^{-t^2}$

Given that <em>x</em>(0) = 8, we find

$8=\dfrac52+Ce^0\implies C=\dfrac{11}2$

so that the particular solution to this IVP is

$\boxed{x(t)=\dfrac{5+11e^{-t^2}}2}$

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Alice and Bob have two positive integers, x and y respectively, glued to their foreheads, so that each can read the other’s numb

vazorg [7]

Step-by-step explanation: |x − y| = 1, ok lets play as Alice, my number is y, and the bob number is x.

the condition says that x-y = 1 or x-y = -1.

so, if you know x, then y = 1 +y or y = y - 1. so you have two possibilities.

let's see two cases : first, let's suppose there are no code in the conversation. Then the only way of being shure of your number, is if one of them have the lowest positive number, so the other should have the next one. So if Bob have the number one, Alice knows for shure that she has the 2. Bob knows that she has a 2, but that means he could have a 1 or a 3, but when he sees that Alice is shure about her number, he knows that his number is the 1.

the second case is where the conversation may be a sort of code, saying a phrase x times and changing when x = the number of the other person, in this case, bob will have the 201 and alice the 202.

7 0

2 years ago

If Carol hangs a picture

Alina [70]

Answer:

3 - 11 = 8 8 - 8 = 0

i think the answer is 8

5 0

2 years ago

Read 2 more answers

HELP?? A lunch combo has the following options: Entree: Hamburger, Hotdog, Chicken sandwich Side: Apples, Salad Drink: Milk, Wat

murzikaleks [220]

Using the Fundamental Counting Theorem, the sample size of these outcomes is of 12.

<h3>What is the Fundamental Counting Theorem?</h3>

It is a theorem that states that if there are n things, each with $n_1, n_2, \cdots, n_n$ ways to be done, each thing independent of the other, the number of ways they can be done is:

$N = n_1 \times n_2 \times \cdots \times n_n$

Considering the number of options for Entree, Side and Drink, the parameters are:

n1 = 3, n2 = 2, n3 = 2.

Hence the sample size of outcomes is:

N = 3 x 2 x 2 = 12.

More can be learned about the Fundamental Counting Theorem at brainly.com/question/24314866

#SPJ1

3 0

1 year ago

If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by

Vitek1552 [10]

Answer:

a) Average velocity at 0.1 s is 696 ft/s.

b) Average velocity at 0.01 s is 7536 ft/s.

c) Average velocity at 0.001 s is 75936 ft/s.

Step-by-step explanation:

Given : If a ball is thrown straight up into the air with an initial velocity of 70 ft/s, its height in feet after t seconds is given by $y = 70t-16t^2$ .

To find : The average velocity for the time period beginning when t = 2 and lasting. a. 0.1 s. , b. 0.01 s. , c. 0.001 s.

Solution :

a) The average velocity for the time period beginning when t = 2 and lasting 0.1 s.

$(\text{Average velocity})_{0.1\ s}=\frac{\text{Change in height}}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{h_{2.1}-h_2}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{(70(2.1)-16(2.1)^2)-(70(0.1)-16(0.1)^2)}{0.1}$

$(\text{Average velocity})_{0.1\ s}=\frac{69.6}{0.1}$

$(\text{Average velocity})_{0.1\ s}=696\ ft/s$

b) The average velocity for the time period beginning when t = 2 and lasting 0.01 s.

$(\text{Average velocity})_{0.01\ s}=\frac{\text{Change in height}}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{h_{2.01}-h_2}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{(70(2.01)-16(2.01)^2)-(70(0.01)-16(0.01)^2)}{0.01}$

$(\text{Average velocity})_{0.01\ s}=\frac{75.36}{0.01}$

$(\text{Average velocity})_{0.01\ s}=7536\ ft/s$

c) The average velocity for the time period beginning when t = 2 and lasting 0.001 s.

$(\text{Average velocity})_{0.001\ s}=\frac{\text{Change in height}}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{h_{2.001}-h_2}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{(70(2.001)-16(2.001)^2)-(70(0.001)-16(0.001)^2)}{0.001}$

$(\text{Average velocity})_{0.001\ s}=\frac{75.936}{0.001}$

$(\text{Average velocity})_{0.001\ s}=75936\ ft/s$

5 0

3 years ago

Please help with homework

Sedbober [7]

Ashley should get her seeds at garden place as she gets the exact amount of seeds and spends the least amount of money.
first of all I divided 75 by the amount of seeds you get it a packet. then if that was a decimal I rounded it to the larger number (eg. 7.15 rounded to 8)
then I multiplied the cost by that number to get the amount of money itd cost.
so I got in order of the boxes(s=seeds)
75s=$18
80s=$18.4
84s=$21.84
80s=$20
so the best deal is Garden place

7 0

2 years ago