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Mrac [35]
3 years ago
11

Determine the longest interval in which the given initialvalue problemis certainto have a unique twicedifferentiable solution. D

o not attempt to find the solutionty'' + 3y = t, y(1) = 1, y'(1) = 9
Mathematics
1 answer:
AlekseyPX3 years ago
6 0

Answer:

t_o = 3, so solution exists on (0,4).  

Step-by-step explanation:

Use Theorem

Divide equation with t(t — 4).

y''+[3/(t-4)]*y'+ [4/t(t-4)]*y=2/t(t-4)

p(t)=3/t-4—> continuous on (-∞, 4) and (4,∞)

q(t) = 4/t(t-4) —> continuous on (-∞,0), (0,4) and (4, ∞)

g(t) = 2/t(t-4)—> continuous on (-∞, 0), (0,4) and (4,∞)

t_o = 3, so solution exists on (0,4).  

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A 75-gallon tank is filled with brine (water nearly saturated with salt; used as a preservative) holding 11 pounds of salt in so
Debora [2.8K]

Let A(t) = amount of salt (in pounds) in the tank at time t (in minutes). Then A(0) = 11.

Salt flows in at a rate

\left(0.6\dfrac{\rm lb}{\rm gal}\right) \left(3\dfrac{\rm gal}{\rm min}\right) = \dfrac95 \dfrac{\rm lb}{\rm min}

and flows out at a rate

\left(\dfrac{A(t)\,\rm lb}{75\,\rm gal + \left(3\frac{\rm gal}{\rm min} - 3.25\frac{\rm gal}{\rm min}\right)t}\right) \left(3.25\dfrac{\rm gal}{\rm min}\right) = \dfrac{13A(t)}{300-t} \dfrac{\rm lb}{\rm min}

where 4 quarts = 1 gallon so 13 quarts = 3.25 gallon.

Then the net rate of salt flow is given by the differential equation

\dfrac{dA}{dt} = \dfrac95 - \dfrac{13A}{300-t}

which I'll solve with the integrating factor method.

\dfrac{dA}{dt} + \dfrac{13}{300-t} A = \dfrac95

-\dfrac1{(300-t)^{13}} \dfrac{dA}{dt} - \dfrac{13}{(300-t)^{14}} A = -\dfrac9{5(300-t)^{13}}

\dfrac d{dt} \left(-\dfrac1{(300-t)^{13}} A\right) = -\dfrac9{5(300-t)^{13}}

Integrate both sides. By the fundamental theorem of calculus,

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac1{(300-t)^{13}} A\bigg|_{t=0} - \frac95 \int_0^t \frac{du}{(300-u)^{13}}

\displaystyle -\dfrac1{(300-t)^{13}} A = -\dfrac{11}{300^{13}} - \frac95 \times \dfrac1{12} \left(\frac1{(300-t)^{12}} - \frac1{300^{12}}\right)

\displaystyle -\dfrac1{(300-t)^{13}} A = \dfrac{34}{300^{13}} - \frac3{20}\frac1{(300-t)^{12}}

\displaystyle A = \frac3{20} (300-t) - \dfrac{34}{300^{13}}(300-t)^{13}

\displaystyle A = 45 \left(1 - \frac t{300}\right) - 34 \left(1 - \frac t{300}\right)^{13}

After 1 hour = 60 minutes, the tank will contain

A(60) = 45 \left(1 - \dfrac {60}{300}\right) - 34 \left(1 - \dfrac {60}{300}\right)^{13} = 45\left(\dfrac45\right) - 34 \left(\dfrac45\right)^{13} \approx 34.131

pounds of salt.

7 0
2 years ago
Nationwide, it is estimated that 40% of service stations have gas tanks that leak to some extent. A new program in California is
fredd [130]

Answer:

Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .

Step-by-step explanation:

The sample proportion is p2= 7/27= 0.259

and q2= 0.74

The sample size = n= 27

The population proportion = p1= 0.4

q1= 0.6

We formulate the null and alternate hypotheses that the new program is effective

H0: p2> p1   vs  Ha: p2 ≤ p1

The test statistic is

z= p2- p1/√ p1q1/n

z= 0.259-0.4/ √0.4*0.6/27

z=  -0.141/0.09428

z= -1.496

The significance level ∝ is 0.05

The critical region for one tailed test is z ≤ ± 1.645

Since the calculated value of z= -1.496 does not fall in the critical region z < -1.645 we conclude that the new program is effective. We fail to reject the null hypothesis .

8 0
3 years ago
For the following equations, find:
astraxan [27]

Answer:

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4 0
3 years ago
I need help on this, please help​
AfilCa [17]

Answer:i tihnk football is .45 and tennis is .18?

Step-by-step explanation:

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3 years ago
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maks197457 [2]

Answer:

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