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oksian1 [2.3K]
2 years ago
5

I need help, can you help me?​

Mathematics
2 answers:
Aneli [31]2 years ago
5 0

Answer:

with what? I can try

Step-by-step explanation:

(◍•ᴗ•◍)✿

GREYUIT [131]2 years ago
3 0

depends....

whats the problem or question?

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Select the correct answer.
Serhud [2]

Answer:

Dude

Step-by-step explanation:

There's nothing here

7 0
2 years ago
y′′ −y = 0, x0 = 0 Seek power series solutions of the given differential equation about the given point x 0; find the recurrence
sukhopar [10]

Let

\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = a_0 + a_1x + a_2x^2 + \cdots

Differentiating twice gives

\displaystyle y'(x) = \sum_{n=1}^\infty na_nx^{n-1} = \sum_{n=0}^\infty (n+1) a_{n+1} x^n = a_1 + 2a_2x + 3a_3x^2 + \cdots

\displaystyle y''(x) = \sum_{n=2}^\infty n (n-1) a_nx^{n-2} = \sum_{n=0}^\infty (n+2) (n+1) a_{n+2} x^n

When x = 0, we observe that y(0) = a₀ and y'(0) = a₁ can act as initial conditions.

Substitute these into the given differential equation:

\displaystyle \sum_{n=0}^\infty (n+2)(n+1) a_{n+2} x^n - \sum_{n=0}^\infty a_nx^n = 0

\displaystyle \sum_{n=0}^\infty \bigg((n+2)(n+1) a_{n+2} - a_n\bigg) x^n = 0

Then the coefficients in the power series solution are governed by the recurrence relation,

\begin{cases}a_0 = y(0) \\ a_1 = y'(0) \\\\ a_{n+2} = \dfrac{a_n}{(n+2)(n+1)} & \text{for }n\ge0\end{cases}

Since the n-th coefficient depends on the (n - 2)-th coefficient, we split n into two cases.

• If n is even, then n = 2k for some integer k ≥ 0. Then

k=0 \implies n=0 \implies a_0 = a_0

k=1 \implies n=2 \implies a_2 = \dfrac{a_0}{2\cdot1}

k=2 \implies n=4 \implies a_4 = \dfrac{a_2}{4\cdot3} = \dfrac{a_0}{4\cdot3\cdot2\cdot1}

k=3 \implies n=6 \implies a_6 = \dfrac{a_4}{6\cdot5} = \dfrac{a_0}{6\cdot5\cdot4\cdot3\cdot2\cdot1}

It should be easy enough to see that

a_{n=2k} = \dfrac{a_0}{(2k)!}

• If n is odd, then n = 2k + 1 for some k ≥ 0. Then

k = 0 \implies n=1 \implies a_1 = a_1

k = 1 \implies n=3 \implies a_3 = \dfrac{a_1}{3\cdot2}

k = 2 \implies n=5 \implies a_5 = \dfrac{a_3}{5\cdot4} = \dfrac{a_1}{5\cdot4\cdot3\cdot2}

k=3 \implies n=7 \implies a_7=\dfrac{a_5}{7\cdot6} = \dfrac{a_1}{7\cdot6\cdot5\cdot4\cdot3\cdot2}

so that

a_{n=2k+1} = \dfrac{a_1}{(2k+1)!}

So, the overall series solution is

\displaystyle y(x) = \sum_{n=0}^\infty a_nx^n = \sum_{k=0}^\infty \left(a_{2k}x^{2k} + a_{2k+1}x^{2k+1}\right)

\boxed{\displaystyle y(x) = a_0 \sum_{k=0}^\infty \frac{x^{2k}}{(2k)!} + a_1 \sum_{k=0}^\infty \frac{x^{2k+1}}{(2k+1)!}}

4 0
3 years ago
-3+x=4 solve this question
user100 [1]
Just add 3 to both sides
x = 7
4 0
3 years ago
Read 2 more answers
00:00
irina [24]

Answer:

the answer is 16

Step-by-step explanation:

its 16 because 2 quart= 8 cups

and 4pints = 8 cups

5 0
3 years ago
Find the first term, a1, of an arithmetic sequence if a12 = 38 and a45 = 170.
snow_lady [41]
Standard formula for arithmetic sequence:
an = a0 + d(n-1)

if we use the two terms given, setting a12 as starting term and a45 as the end term an.

170 = 38 + 33d 
170 - 38 = 33d
132 = 33d

132/33 = d
This is the common difference, use it to find the first term.

38 = a0 + (132/33)(12-1)
38 = a0 + (132/33)(11)
38 = a0 + 132/3
38 - 132/3 = a0
38 - 44 = a0
-6 = a0

The starting term is -6

8 0
3 years ago
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