Answer:
Add 3 each term inside the parentheses, then use inverse operations to solve.
Step-by-step explanation:
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

.
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

.
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


1. The information given in the problem above, is:
- Austin purchases <span>2/3 of a pound of yum yum treats.
- Y</span><span>um yum treats are $6.00 per pound (1 pound=$6.00).
2. Keeping this on mind you have: If 1 pound of </span>yum yum treats is $6.00, how much is 2/3 of a pound?
1 pound------------$6.00
2/3 of a pound----x
x=(2/3)(6.00)/1
x=(2/3)(6.00)
x=12/3
x=$4.00
<span>
Therefore: How much does Austin pay?
</span>
The answer is: Austin pays $4.00.
Answer: 30 feet per second
========================================
x = diameter, y = speed of water
y varies inversely with respect to x, so, y = k/x for some constant k
If x = 0.6 and y = 5 pair up, then y = k/x turns into 5 = k/0.6 which solves to k = 3 when you multiply both sides by 0.6
So the equation y = k/x turns into y = 3/x
Now plug in the diameter x = 0.1 to find the speed to be...
y = 3/x
y = 3/0.1
y = 30 feet per second