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3 years ago

10

Use the point and the slope to graph each line. Write the equation of the line. The line contains point (2,-2) and is perpendicu

lar to a line with slope 1

1 answer:

lisabon 2012 [21]3 years ago

4 0

Answer:

x + y = 0

Step-by-step explanation:

Slope 1

b = -2 - (-1)(2)

b = -2 +2

b = 0

y = -x

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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

Compare and contrast the difference between "simplify a fraction" and "simplify a radical."

Ne4ueva [31]

Simplifying a fraction means most likely having it remain as a fraction and not a whole number and it can be converted as a decimal.

on the other hand, simplifying a radical, you would have no remainders, fractions, or decimals after simplifying.

thats the difference.
im not sure about similarities

8 0

3 years ago

What is the sum of the next two terms in the sequence: -3, -1, 3, 9, 17

Korvikt [17]

Answer: 66 because you are adding 2 then 4 then 6 then 8 then 10 then 12 therefore 17+10= 27, and 27+12= 39 and 39+27 equals 66

6 0

3 years ago

Zaria wants you to solve this puzzle: “I am thinking of a number.

pshichka [43]

Answer:

first what you do is divide two and then for and then after you divide them together you can get your number so I'm going to show you example how to do it

Step-by-step explanation:

first divide 4 / 2 and then you can get your answer after you divide that

5 0

3 years ago

What else would need to be congruent to show that A ABC= AXYZ by ASA?

fredd [130]

Answer:

D. AC ≅ XZ

Step-by-step explanation:

To prove that two triangles are congruent by the Angle-Side-Angle theorem, both triangles must have two corresponding angles that are congruent to each other in each triangle, and also a corresponding included side in each triangle that are congruent.

Thus, we are given that,

<X ≅ <A and <Z ≅ <C, therefore, what is needed is a corresponding included angle in each triangle that are congruent to each other, which are,

AC and XZ

Therefore, what is needed is:

AC ≅ XZ

7 0

3 years ago