Let
. Then
and
are two fundamental, linearly independent solution that satisfy
Note that
, so that
. Adding
doesn't change this, since
.
So if we suppose
then substituting
would give
To make sure everything cancels out, multiply the second degree term by
, so that
Then if
, we get
as desired. So one possible ODE would be
(See "Euler-Cauchy equation" for more info)
Let the total toys be x.
The first has (1/10)x = x/10
The third child has one more toys than the first = x/10 + 1
The fourth has double of the third: = 2* (x/10 + 1) = 2x/10 + 2
Since we know that the second child has 12 more toys than the first
Therefore the Second minus First = 12.
First let us find the second.
1st + 2nd + 3rd + 4th = x
x/10 + 2nd + (x/10 + 1) + (2x/10 + 2) = x
2nd + x/10 + x/10 + 1 + 2x/10 + 2 = x
2nd + x/10 + x/10 + 2x/10 + 1 + 2 = x
2nd + 4x/10 + 3 = x
2nd = x - 4x/10 - 3
2nd = 6x/10 - 3
But recall
2nd - 1st = 12
6x/10 - 3 - x/10 = 12
6x/10 - x/10 = 12 + 3
5x/10 = 15
5x = 15 * 10
x = 15*10 /5
x = 30
So there were 30 toys in all.
Answer:
1.5 miles or kilograms
Step-by-step explanation:
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