Problem 3
1 answer:
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Answer:
The answer to the question is
The longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution is (-∞, 4)
Step-by-step explanation:
To apply look for the interval, we divide the ordinary differential equation by (t-4) to
y'' + y' +
y =
Using theorem 3.2.1 we have p(t) = , q(t) =
, g(t) =
Which are undefined at 4. Therefore the longest interval in which the given initial value problem is certain to have a unique twice-differentiable solution, that is where p, q and g are continuous and defined is (-∞, 4) whereby theorem 3.2.1 guarantees unique solution satisfying the initial value problem in this interval.
There are 12 such shapes. One can start with two squares, adding squares around the edge and rejecting shapes that have been previously seen. Repeating until you have 5 squares results in 12 distinct shapes.
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2 shapes are possible with 3 squares: L, I
5 shapes are possible with 4 squares: L, I, N, T, O
