Safety regulations require that the time between airplane takeoffs (on the same runway) will be at least 2 minutes. When taking
1 answer:
Answer:
1 plane
Step-by-step explanation:
Let's suppose the number of planes waiting or on the runway " P "
the number of planes taking off per hour ' '
the time for waiting and the runway so:
P = x
= 18 airplanes per hour
we know that 1 min = 60s
36s = 36/60 = 0.6 min
Also, 3 min and 30 s = 3 + 30/60 = 3.5 min
Next to find the time for waiting and the runway
∴= 0.6 + 3.5 = 4.1 min/60 (converting into hour)
= 0.068 hour
P = 18x0.068 = 1.23
therefore, there is 1 plane either on the runway or waiting to take off
So, there is 1 plane either on the runway or waiting to take off
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The next step is to solve the recurrence, but let's back up a bit. You should have found that the ODE in terms of the power series expansion for 
which indeed gives the recurrence you found,
but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that
, and substituting this into the recurrence, you find that 
for all 
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Next, the linear term tells you that
, or 
.
Now, if
is the first term in the sequence, then by the recurrence you have
and so on, such that
for all 
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Finally, the quadratic term gives
, or 
. Then by the recurrence,
and so on, such that
for all.
Now, the solution was proposed to be
so the general solution would be
which indeed gives the recurrence you found,
but in order to get anywhere with this, you need at least three initial conditions. The constant term tells you that
Next, the linear term tells you that
Now, if
and so on, such that
Finally, the quadratic term gives
and so on, such that
for all
Now, the solution was proposed to be
so the general solution would be
x*y' + y = 8x
y' + y/x = 8 .... divide everything by x
dy/dx + y/x = 8
dy/dx + (1/x)*y = 8
We have something in the form
y' + P(x)*y = Q(x)
which is a first order ODE
The integrating factor is
Multiply both sides by the integrating factor (x) and we get the following:
dy/dx + (1/x)*y = 8
x*dy/dx + x*(1/x)*y = x*8
x*dy/dx + y = 8x
y + x*dy/dx = 8x
Note the left hand side is the result of using the product rule on xy. We technically didn't need the integrating factor since we already had the original equation in this format, but I wanted to use it anyway (since other ODE problems may not be as simple).
Since (xy)' turns into y + x*dy/dx, and vice versa, this means
y + x*dy/dx = 8x turns into (xy)' = 8x
Integrating both sides with respect to x leads to
xy = 4x^2 + C
y = (4x^2 + C)/x
y = (4x^2)/x + C/x
y = 4x + Cx^(-1)
where C is a constant. In this case, C = -5 leads to a solution
y = 4x - 5x^(-1)
you can check this answer by deriving both sides with respect to x
dy/dx = 4 + 5x^(-2)
Then plugging this along with y = 4x - 5x^(-1) into the ODE given, and you should find it satisfies that equation.