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mash [69]
3 years ago
5

Consider the initial value problem 2ty' = 6y, y(1) =-2. Find the value of the constant C and the exponent r so that y = Ctr is t

he solution of this initial value problem. y = help (formulas) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution. help (inequalities) What is the actual interval of existence for the solution (from part a)? help (inequalities)
Mathematics
1 answer:
ELEN [110]3 years ago
5 0

The correct question is:

Consider the initial value problem

2ty' = 6y, y(1) = -2

(a) Find the value of the constant C and the exponent r such that y = Ct^r is the solution of this initial value problem.

b) Determine the largest interval of the form a < t < b on which the existence and uniqueness theorem for first order linear differential equations guarantees the existence of a unique solution.

c) What is the actual interval of existence for the solution obtained in part (a) ?

Step-by-step explanation:

Given the differential equation

2ty' = 6y

a) We need to find the value of the constant C and r, such that y = Ct^r is a solution to the differential equation together with the initial condition y(1) = -2

Since Ct^r is a solution to the initial value problem, it means that y = Ct^r satisfies the said problem. That is

2tdy/dt - 6y = 0

Implies

2td(Ct^r)/dt - 6(Ct^r) = 0

2tCrt^(r - 1) - 6Ct^r = 0

2Crt^r - 6Ct^r = 0

(2r - 6)Ct^r = 0

But Ct^r ≠ 0

=> 2r - 6 = 0 or r = 6/2 = 3

Now, we have r = 3, which implies that

y = Ct^3

Applying the initial condition y(1) = -2, we put y = -2 when t = 1

-2 = C(1)^3

C = -2

So, y = -2t^3

b) Let y = F(x,y)................(1)

Suppose F(x, y) is continuous on some region, R = {(x, y) : x_0 − δ < x < x_0 + δ, y_0 −ę < y < y_0 + ę} containing the point (x_0, y_0). Then there exists a number δ1 (possibly smaller than δ) so that a solution y = f(x) to (1) is defined for x_0 − δ1 < x < x_0 + δ1.

Now, suppose that both F(x, y)

and ∂F/∂y are continuous functions defined on a region R. Then there exists a number δ2

(possibly smaller than δ1) so that the solution y = f(x) to (1) is

the unique solution to (1) for x_0 − δ2 < x < x_0 + δ2.

c) Firstly, we write the differential equation 2ty' = 6y in standard form as

y' - (3/t)y = 0

0 is always continuous, but -3/t has discontinuity at t = 0

So, the solution to differential equation exists everywhere, apart from t = 0.

The interval is (-infinity, 0) n (0, infinity)

n - means intersection.

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An examination of the records for a random sample of 16 motor vehicles in a large fleet resulted in the sample mean operating co
levacccp [35]

Answer:

1. The 95% confidence interval would be given by (24.8190;27.8010)  

2. 4.7048 \leq \sigma^2 \leq 16.1961

3. t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

t_{crit}=1.753

Since our calculated value it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

4. t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

\chi^2 =24.9958

Since our calculated value is less than the critical value we don't hav enough evidence to reject the null hypothesis at the significance level provided.

Step-by-step explanation:

Previous concepts

\bar X=26.31 represent the sample mean for the sample  

\mu population mean (variable of interest)

s=2.8 represent the sample standard deviation

n=16 represent the sample size

Part 1

The confidence interval for the mean is given by the following formula:

\bar X \pm t_{\alpha/2}\frac{s}{\sqrt{n}}   (1)

In order to calculate the critical value t_{\alpha/2} we need to find first the degrees of freedom, given by:

df=n-1=16-1=15

Since the Confidence is 0.95 or 95%, the value of \alpha=0.05 and \alpha/2 =0.025, and we can use excel, a calculator or a tabel to find the critical value. The excel command would be: "=-T.INV(0.025,15)".And we see that t_{\alpha/2}=2.13

Now we have everything in order to replace into formula (1):

26.31-2.13\frac{2.8}{\sqrt{16}}=24.819    

26.31+2.13\frac{2.8}{\sqrt{16}}=27.801

So on this case the 95% confidence interval would be given by (24.8190;27.8010)  

Part 2

The confidence interval for the population variance is given by the following formula:

\frac{(n-1)s^2}{\chi^2_{\alpha/2}} \leq \sigma^2 \leq \frac{(n-1)s^2}{\chi^2_{1-\alpha/2}}

The next step would be calculate the critical values. First we need to calculate the degrees of freedom given by:

df=n-1=16-1=15

Since the Confidence is 0.90 or 90%, the value of \alpha=0.1 and \alpha/2 =0.05, and we can use excel, a calculator or a table to find the critical values.  

The excel commands would be: "=CHISQ.INV(0.05,15)" "=CHISQ.INV(0.95,15)". so for this case the critical values are:

\chi^2_{\alpha/2}=24.996

\chi^2_{1- \alpha/2}=7.261

And replacing into the formula for the interval we got:

\frac{(15)(2.8)^2}{24.996} \leq \sigma^2 \leq \frac{(15)(2.8)^2}{7.261}

4.7048 \leq \sigma^2 \leq 16.1961

Part 3

We need to conduct a hypothesis in order to determine if actual mean operating cost is at most 25 cents per mile , the system of hypothesis would be:    

Null hypothesis:\mu \leq 25      

Alternative hypothesis:\mu > 25      

We don't know the population deviation, so for this case is better apply a t test to compare the actual mean to the reference value, and the statistic is given by:      

t=\frac{\bar X-\mu_o}{\frac{s}{\sqrt{n}}} (1)      

Calculate the statistic      

We can replace in formula (1) the info given like this:      

t=\frac{26.31-25}{\frac{2.8}{\sqrt{16}}}=1.871    

Critical value  

On this case we need a critical value on th t distribution with 15 degrees of freedom that accumulates 0.05 of th area on the right and 0.95 of the area on the left. We can calculate this value with the following excel code:"=T.INV(0.95,15)" and we got t_{crit}=1.753

Conclusion      

Since our calculated valu it's higher than the critical value we have enough evidence at 5% of significance to reject th null hypothesis and conclude that the true mean is higher than 25 cents per mile.

Part 4

State the null and alternative hypothesis

On this case we want to check if the population standard deviation is more than 2.3, so the system of hypothesis are:

H0: \sigma \leq 2.3

H1: \sigma >2.3

In order to check the hypothesis we need to calculate the statistic given by the following formula:

t=(n-1) [\frac{s}{\sigma_o}]^2

This statistic have a Chi Square distribution distribution with n-1 degrees of freedom.

What is the value of your test statistic?

Now we have everything to replace into the formula for the statistic and we got:

t=(16-1) [\frac{2.8}{2.3}]^2 =22.2306

What is the critical value for the test statistic at an α = 0.05 significance level?

Since is a right tailed test the critical zone it's on the right tail of the distribution. On this case we need a quantile on the chi square distribution with 15 degrees of freedom that accumulates 0.05 of the area on the right tail and 0.95 on the left tail.  

We can calculate the critical value in excel with the following code: "=CHISQ.INV(0.95,15)". And our critical value would be \chi^2 =24.9958

Since our calculated value is less than the critical value we FAIL to reject the null hypothesis.

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