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Colt1911 [192]
3 years ago
10

Consider the differential equation

B2%7D%20%2B6xy" id="TexFormula1" title=" x^{2} \frac{dy}{dx} =6 y^{2} +6xy" alt=" x^{2} \frac{dy}{dx} =6 y^{2} +6xy" align="absmiddle" class="latex-formula"> which may be considered either as a homogenous equation or as a Bernoulli equation.
If we make the substitution y(x)=xv(x) relevant to homogenous equations, we obtain \frac{dv}{dx}=
If we make the substitution z(x)= (y(x))^{-1} relevant to homogenous equations, we obtain \frac{dz}{dx}=
Using either (or both) of these methods, solve the initial value problem for the above equation where y(3)=6. Find the interval of validity of this solution.
Mathematics
1 answer:
Nutka1998 [239]3 years ago
8 0
As a Bernoulli equation:

x^2\dfrac{\mathrm dy}{\mathrm dx}=6y^2+6xy\iff x^2y^{-2}\dfrac{\mathrm dy}{\mathrm dx}-6xy^{-1}=6

Let z=y^{-1}\implies\dfrac{\mathrm dz}{\mathrm dx}=-y^{-2}\dfrac{\mathrm dy}{\mathrm dx}. The ODE becomes

-x^2\dfrac{\mathrm dz}{\mathrm dx}-6xz=6
x^6\dfrac{\mathrm dz}{\mathrm dx}+6x^5z=-6x^4
\dfrac{\mathrm d}{\mathrm dx}[x^6z]=-6x^4
x^6z=-6\displaystyle\int x^4\,\mathrm dx
x^6z=-\dfrac65x^5+C
z=-\dfrac6{5x}+\dfrac C{x^6}
y^{-1}=-\dfrac6{5x}+\dfrac C{x^6}
y=\dfrac1{\frac C{x^6}-\frac6{5x}}
y=\dfrac{5x^6}{C-6x^5}

With y(3)=6, we get

6=\dfrac{5(3)^6}{C-6(3)^5}\implies C=\dfrac{4131}2

so the solution is

y=\dfrac{5x^6}{\frac{4131}2-6x^5}=\dfrac{10x^6}{4131-12x^5}

which is valid as long as the denominator is not zero, which is the case for all x\neq\sqrt[5]{\dfrac{4131}{12}}.
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Occasionally an airline will lose a bag. Suppose a small airline has found it can reasonably model the number of bags lost each
julia-pushkina [17]

Answer:

a) The probability that the airline will lose no bags next monday is 0.1108

b) The probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227

c) I would recommend taking a Poisson model with mean 4.4 instead of a Poisson model with mean 2.2

Step-by-step explanation:

The probability mass function of X, for which we denote the amount of bags lost next monday is given by this formula

P(X=k) = \frac{e^{-2.2} * {2.2}^k }{k!}

a)

P(X=0) = \frac{e^{-2.2} * {2.2}^0 }{0!} = 0.1108

The probability that the airline will lose no bags next monday is 0.1108.

b) Note that P(X \in \{0,1,2\} = P(X=0) + P(X=1) + P(X=2) . And

P(X=0)+P(X=1)+P(X=2) = e^{-2.2} * (1 + 2.2 + 2.2^2/2) = 0.6227

Therefore, the probability that the airline will lose 0,1, or 2 bags next Monday is 0.6227.

c) If the double of flights are taken, then you at least should expect to loose a similar proportion in bags, because you will have more chances for a bag to be lost. WIth this in mind, we can correctly think that the average amount of bags that will be lost each day will double. Thus, i would double the mean of the Poisson model, in other words, i would take a Poisson model with mean 4.4, instead of 2.2.

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3 years ago
Find the unknown angle A<br>​
uranmaximum [27]
40! When you add all four angles of a quadrilateral together, you will always end up with 360. So, 360-75-120-125=40 :) have a great day!
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3 years ago
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swat32

Answer:

D

Step-by-step explanation:

(3^2x5)^3=91125

3^6x5^3=91125

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3 years ago
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A line passes through the points (-3,-4) and (6,2)
Zanzabum

Answer: x - int : (3,0)

Step-by-step explanation:

point-slope form: y - y1 = m(x - x1)

m = y2-y1/x2-x1

   = -4-2/-3-6

   =-6/-9

   = 2/3

y1 = 2

x1 =6

y - 2 = 2/3(x-6)  

x -int: when y = 0, what is x?

0-2 = 2/3(x-6)

-2 = 2/3x - 4

2 = 2/3x

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~Hope this helps!~

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Answer:

2. 7.6

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Step-by-step explanation:

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