All categories

3 years ago

Consider the system x' = rx + x3 - x5, which exhibits a pitchfork bifurcation.

Mathematics

1 answer:

trapecia [35]3 years ago

5 0

Answer:

rs = -3/16 and the three minima x₁, x ₂ and x₄ fall at the value of integrating curve C

Step-by-step explanation:

The answer is found in the attachment.

You might be interested in

How much of the 8% solution should we use to make 100g of a 3% solution?

sp2606 [1]

Suppose x grams of 8% solution should be mixed to make 100 g of 3% solution.
here we shall have:
the amount of 3% solution in the 100 g will be:
3/100*100=3 grams
the the equation will be, 8% of what quantity becomes 3g?
(8/100)*x=3
hence solving for x we shall have:
0.08x=3
x=3/0.08
x= 37.8 grams
The implication of this is that, when you have 37.5 grams of 8% solution, that will contain 3 of pure chemical and hence when you add more solvent on top of 37.5 g until you reach 100g, that will be 3% solution.

8 0

3 years ago

Read 2 more answers

Suppose a batch of metal shafts produced in a manufacturing company have a standard deviation of 1.5 and a mean diameter of 205

Crank

Answer:

$P(205-0.3=204.7$

$z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778$

$z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778$

So we can find this probability:

$P(-1.778$

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

$P = 1-0.9243 = 0.0757$

Step-by-step explanation:

Let X the random variable that represent the diamters of interest for this case, and for this case we know the following info

Where $\mu=205$ and $\sigma=1.5$

We can begin finding this probability this probability

$P(205-0.3=204.7$

For this case they select a sample of n=79>30, so then we have enough evidence to use the central limit theorem and the distirbution for the sample mean can be approximated with:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

And the best way to solve this problem is using the normal standard distribution and the z score given by:

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

And we can find the z scores for each limit and we got:

$z=\frac{204.7-205}{\frac{1.5}{\sqrt{79}}}=-1.778$

$z=\frac{205.3-205}{\frac{1.5}{\sqrt{79}}}=1.778$

So we can find this probability:

$P(-1.778$

And then since the interest is the probability that the mean diameter of the sample shafts would differ from the population mean by more than 0.3 inches using the complement rule we got:

$P = 1-0.9243 = 0.0757$

6 0

3 years ago

Solve the initial value problem where y′′+4y′−21 y=0, y(1)=1, y′(1)=0 . Use t as the independent variable.

igor_vitrenko [27]

Answer:

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

Step-by-step explanation:

y′′ + 4y′ − 21y = 0

The auxiliary equation is given by

m² + 4m - 21 = 0

We solve this using the quadratic formula. So

$m = \frac{-4 +/- \sqrt{4^{2} - 4 X 1 X (-21))} }{2 X 1}\\ = \frac{-4 +/- \sqrt{16 + 84} }{2}\\= \frac{-4 +/- \sqrt{100} }{2}\\= \frac{-4 +/- 10 }{2}\\= -2 +/- 5\\= -2 + 5 or -2 -5\\= 3 or -7$

So, the solution of the equation is

$y = Ae^{m_{1} t} + Be^{m_{2} t}$

where m₁ = 3 and m₂ = -7.

So,

$y = Ae^{3t} + Be^{-7t}$

Also,

$y' = 3Ae^{3t} - 7e^{-7t}$

Since y(1) = 1 and y'(1) = 0, we substitute them into the equations above. So,

$y(1) = Ae^{3X1} + Be^{-7X1}\\1 = Ae^{3} + Be^{-7}\\Ae^{3} + Be^{-7} = 1 (1)$

$y'(1) = 3Ae^{3X1} - 7Be^{-7X1}\\0 = 3Ae^{3} - 7Be^{-7}\\3Ae^{3} - 7Be^{-7} = 0 \\3Ae^{3} = 7Be^{-7}\\A = \frac{7}{3} Be^{-10}$

Substituting A into (1) above, we have

$\frac{7}{3}B e^{-10}e^{3} + Be^{-7} = 1 \\\frac{7}{3}B e^{-7} + Be^{-7} = 1\\\frac{10}{3}B e^{-7} = 1\\B = \frac{3}{10} e^{7}$

Substituting B into A, we have

$A = \frac{7}{3} \frac{3}{10} e^{7}e^{-10}\\A = \frac{7}{10} e^{-3}$

Substituting A and B into y, we have

$y = Ae^{3t} + Be^{-7t}\\y = \frac{7}{10} e^{-3}e^{3t} + \frac{3}{10} e^{7}e^{-7t}\\y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

So the solution to the differential equation is

$y = \frac{7}{10} e^{3(t - 1)} + \frac{3}{10}e^{-7(t - 1)}$

6 0

3 years ago

I need help I don’t understand

Arte-miy333 [17]

Use mathpapa.com or cymath.com

7 0

3 years ago

What is the work to find out the awnser to find out how many centimeters in a foot

igor_vitrenko [27]

To find how many cm are in a ft, you have to use the method of unit rate:

so we know that 12 in. equals a ft. so,

1 ft
------
12 in.

and 2.25 cm equals an in. so,

1 in.
-------
2.54cm

then we put the rates together and get:

1ft 1 in.
---- X -------................we cancel out the in. units and get:
12 in. 2.54 cm

Solve and get 1 ft = 30.48 cm

4 0

3 years ago

Consider the system x' = rx + x3 - x5​, which exhibits a pitchfork bifurcation.

Consider the system x' = rx + x3 - x5, which exhibits a pitchfork bifurcation.