MATHEMATICS COLLEGE

Find a solution to the following initial-value problem: dy dx = y(y − 2)e x , y (0) = 1.

Answers

Answer 1

Answer:

This equation is separable, as

$(\mathrm dy)/(\mathrm dx)=y(y-2)e^x\implies(\mathrm dy)/(y(y-2))=e^x\,\mathrm dx$

Integrate both sides; on the left, expand the fraction as

$\frac1{y(y-2)}=\frac12\left(\frac1{y-2}-\frac1y\right)$

Then

$\displaystyle\int(\mathrm dy)/(y(y-2))=\int e^x\,\mathrm dx\implies\frac12(\ln|y-2|-\ln|y|)=e^x+C$

$\implies\frac12\ln\left|\frac{y-2}y\right|=e^x+C$

Since $y(0)=1$ , we get

$\frac12\ln\left|\frac{1-2}1\right|=e^0+C\implies C=-1$

so that the particular solution is

$\frac12\ln\left|\frac{y-2}y\right|=e^x-1\implies\boxed{y=\frac2{1-e^(2e^x-2)}}$

Related Questions

Morningstar tracks the total return for a large number of mutual funds. The following tableshows the total return and the number of funds for four categories of mutual funds(Morningstar Funds500, 2008).Type of FundDomestic EquityInternational EquitySpecialty StockHybrid Number of Funds9191262114192900Total Return (%)4.6518.1511.366.75a. Using the number of funds as weights, compute the weighted average total return forthe mutual funds covered by Morningstar.b. Is there any difficulty associated with using the "number of funds" as the weights incomputing the weighted average total return for Morningstar in part (a)? Discuss. Whatelse might be used for weights?c. Suppose you had invested $10,000 in mutual funds at the beginning of 2007 anddiversified the investment by placing $2000 in Domestic Equity funds, $4000 in International Equity funds, $3000 in Specialty Stock funds, and $1000 in Hybridfunds. What is the expected return on the portfolio?
PLEASE HELP ME OUT M
Which best describes how to find an equation of the line shown?A. The slope m is rise over run, so y over x =m. Solve for Y to get y=mx.B. The slope m is run over rise, so x over y=m. Solve for x=ym.C. The slope m is run over rise, so x over y=m. Solve for y to get y= x over m.D. The slope m is rise over run, so x over y=m. Solve for x to get x=ym.
Carl has three lengths of cable, 3/6 yard long, 1/4 yard long, and 1/3 yard long, which two pieces together make a length of 20/24 yard?
The sum of three consecutive odd integers is -381

An image has coordinates: A (-2, 4), B (4, 6), C (3, -4). Given a scale factor of 3, what would be the new coordinates?Group of answer choices

A' (6, -12), B' (-12, 18), C' (9, -12)

A' (1/2, 1), B' (3/2, 2), C' (3, 9/2)

A' (-6, 12), B' (-12, -18), C' (9, -12)

A' (-6, 12), B' (12, 18), C' (9, -12)

Answers

A’(-6,12) B’(12,18) C’(9,-12) because all numbers must be multiplied by your scale factor in this case 3

Which of the following is a linear function

Answers

Answer: 5/6x=y-4

Step-by-step explanation:

A linear function makes a straight line. When graphed, this makes a straight line

Answer:

Step-by-step explanation:

D would probably be the correct answer

Use induction to prove the following formula is true for all integers n where n greaterthanorequalto 1. 1 + 4 + 9 + .. + n^2 = n(n + 1)(2n + 1)/6

Answers

Answer with Step-by-step explanation:

Since we have given that

1+4+9+........................+n² = $(n(n+1)(2n+1))/(6)$

We will show it using induction on n:

Let n = 1

L.H.S. :1 = R.H.S. : $(1* 2* 3)/(6)=(6)/(6)=1$

So, P(n) is true for n = 1

Now, we suppose that P(n) is true for n = k.

$1+4+9+...................+k^2=(k(k+1)(2k+1))/(6)$

Now, we will show that P(n) is true for n = k+1.

So, it L.H.S. becomes,

$1+4+9+......................+(k+1)^2$

and R.H.S. becomes,

$((k+1)(k+2)(2k+3))/(6)$

Consider, L.H.S.,

$1+4+9+..+k^2+(k+1)^2\n\n=(k(k+1)(2k+1))/(6)+(k+1)^2\n\n=k+1[(k(2k+1))/(6)+(k+1)]\n\n=(k+1)[(2k^2+k+6k+6)/(6)]\n\n=(k+1)(2k^2+7k+6)/(6)]\n\n=(k+1)(2k^2+4k+3k+6)/(6)]\n\n=(k+1)[(2k(k+2)+3(k+2))/(6)]\n\n=((k+1)(2k+3)(k+2))/(6)$

So, L.H.S. = R.H.S.

Hence, P(n) is true for all integers n.

During a nine-hour snowstorm, it snows at a rate of 2 inches per hour for the first 3 hours, at a rate of 3 inches per hour for the next 5 hours, and at a rate of 0.75 inch per hour for the final hour.How many inches of snow accumulated from the storm?

Answers

Answer:

use f(x)=y=mx+b

let snow = S, time = t instead of y and x

S(t)=mt+b

The rate of inches per hour represents the slope of the graph, m.

The y-variable would be the amount of snow, S.

The x-variable would be the time, t, in hours.

The function has three pieces:

i) S(t)= 2t (slope = 2)

ii) S(t) = 3t (slope = 3)

iii) S(t) = 0.75t (slope = 0.75)

For the first piece, i), t=3, so the amount of snow is 6 inches.

For the second piece, ii) t=5, so the amount of snow is 15 inches.

For the third piece, iii) t=1, so the amount of snow is 0.75 inch.

In total, it snowed 21.75 inches.

total snow

Final answer:

To find the total accumulation of snow during the nine-hour snowstorm, we calculate the snow accumulation for each hour and then sum them up. The total accumulation of snow from the storm is 21.75 inches.

Explanation:

To find the total accumulation of snow during the nine-hour snowstorm, we need to calculate the amount of snow that fell during each hour and then sum them up. First, we calculate the snow accumulation for each hour:

For the first 3 hours, it snowed at a rate of 2 inches per hour, so the accumulation is 3 * 2 = 6 inches.
For the next 5 hours, it snowed at a rate of 3 inches per hour, so the accumulation is 5 * 3 = 15 inches.
For the final hour, it snowed at a rate of 0.75 inch per hour, so the accumulation is 1 * 0.75 = 0.75 inches.

Finally, we sum up the accumulations for each hour: 6 + 15 + 0.75 = 21.75 inches. Therefore, the total accumulation of snow from the storm is 21.75 inches.

Learn more about Snow accumulation here:

brainly.com/question/33175695

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W divided by 3 minus 5 equals 7

what is w ?

Answers

Answer:

w=36

Step-by-step explanation:

Your welcome

A line segment is dilated by a scale factor of 2 centered at a point not on the line segment. Which statement regarding the relationshipbetween the given line segment and its image is true?
A The line segments are parallel, and the image is twice the length of the given line segment.
B. The line segments are parallel, and the image is one-half of the length of the given line segment.
C. The line segments are perpendicular, and the image is twice the length of the given line segment.
DD The line segments are perpendicular, and the image is one-half of the length of the given line segment.

Answers

9514 1404 393

Answer:

A The line segments are parallel, and the image is twice the length of the given line segment.

Step-by-step explanation:

Dilation by a factor of 2 means any measure of the image is 2 times the corresponding measure of the original.

Dilation does not change any orientations, so the image will have the same orientation with respect to the origin, axes, or any other line segments. That means the dilated segment is parallel to the original. (If the center of dilation is on the original line segment, the dilated segment will overlay the original segment. That is specifically not the case here.)

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