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Mrac [35]
2 years ago
11

Determine the longest interval in which the given initialvalue problemis certainto have a unique twicedifferentiable solution. D

o not attempt to find the solutionty'' + 3y = t, y(1) = 1, y'(1) = 9
Mathematics
1 answer:
AlekseyPX2 years ago
6 0

Answer:

t_o = 3, so solution exists on (0,4).  

Step-by-step explanation:

Use Theorem

Divide equation with t(t — 4).

y''+[3/(t-4)]*y'+ [4/t(t-4)]*y=2/t(t-4)

p(t)=3/t-4—> continuous on (-∞, 4) and (4,∞)

q(t) = 4/t(t-4) —> continuous on (-∞,0), (0,4) and (4, ∞)

g(t) = 2/t(t-4)—> continuous on (-∞, 0), (0,4) and (4,∞)

t_o = 3, so solution exists on (0,4).  

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Bas_tet [7]

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11.2

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6 0
2 years ago
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jek_recluse [69]

Step-by-step explanation:

this sequence is geometric not arithmetic

HOw we know that ??

when we get a common difference  that must Be equal  

d=6-2=4  not equal to  d=18-6=12

So it is not arithmetic

but when we get the common ratio that also must be equal

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By using this equation:

a(n)=a(1)*r^(n-1)

and we have a(1)=2 , r=3    

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a(n-1) ⇒ priviuse term

SO:  a(n)= 3 * a(n-1)

For example:

a(3)= 3 * 6 =18

<em>I really hope this helps <3</em>

7 0
3 years ago
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kakasveta [241]

Answer:

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Step-by-step explanation:

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Answer:

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Step-by-step explanation:

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Step-by-step explanation:

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