Evaluate the indefinite integral as an infinite series. int frac(cos x-1)(x) dx

shadsiei 2021-09-20 Answered
Evaluate the indefinite integral as an infinite series. cosx1xdx
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Caren
Answered 2021-09-21 Author has 96 answers

We evaluate the given indefinite integral as an expression of an infinite series.
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A car is travelling at 100 km/h on a level road when it runs out of fuel. Its speed v (in km/h) starts to decrease according to the formula
d v d t = k v ( 1 )
where k is constant. One kilometre after running out of fuel its speed has fallen to 50 km/h. Use the chain rule substitution
d v d t = d v d s d s d t = d v d s v
to solve the differential equation.
Note: Although I haven't solved it yet, the answers say that this isn't a reasonable model as the velocity is always positive; I didn't make a typo in the question.
What I'm trying to do is solve velocity as a function of displacement (s, in km), velocity as a function of time (t, in hours), and displacement as a function of time (I need these functions for later parts of the question).
So far I've found velocity as a function of displacement (v(s)):
d v d t = k d s d t (from (1))
d v d t d t = k d s d t d t
v ( s ) = k s + C
v ( 0 ) = 100 C = 100 ,   v ( 1 ) = 50 k = 50
v ( s ) = 50 s + 100
Then I've tried to find velocity as a function of time (v(t)), but I've got stuck. I can't find any differential equation I can use to get this, or to get displacement as a function of time (s(t)).
The answer key says v ( t ) = 100 e 50 t and s ( t ) = 2 ( 1 e 50 t )
I've solved such questions many times before, but it's been a while so I'm a bit rusty. So, even a hint might be enough for me to realise what to do.
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d P 0 d t = C 1 λ P 0 P 1 + C 1 2 P 1 2 d P 1 d t = C 2 P 1 + C 1 λ P 0 P 1 1 2 ( 1 + λ ) C 1 P 1 2 + C 1 P 1 P 2 d P 2 d t = C 2 P 1 + C 1 2 λ P 1 2 C 1 P 1 P 2
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Below is a problem I did. However, it did not match the back of the book. I would like to know where I went wrong.
Problem:
Solve the following differential equation.
y = y x x
Answer:
d y d x = y x x x = y x 1 y = x v d y d x = x d v d x + v x d v d x + v = v 1 x d v d x = 1 d v = d x x v = ln x + c y x = ln | x | + c y = x ln | x | + c x
The book's answer is:
y = x ln | k x |
Where did I go wrong?
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