(1 point) The IVP 8(-6) = 1029 d.-4 has a unique solution defined on the interval

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Question

Answer 1

Answer #1

Any problem in this then comment below. I will help you.

Answer = (-7,7)

Here this problem not defined on p=7 and p=-7 ..

So these two points divide this into 3 interval...

(-inf,-7) , (-7,7) ,(7,inf) ..

Now initial condition is on p=-6 . That lies in 2nd interval..

So unique solution defined in (-7,7) ..

Answer 2

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