Question

@perl

Answer 1

find a series of a soluction y''+y'=0

Answer 2

i can get a0 and a1 but dount know how to summarise like

Answer 3

the file question number 10

Answer 4

ok one sec

Answer 5

$$
\Large {
y = \sum_{n=0}^{\infty}a_n(x-x_0)^n= \sum_{n=0}^{\infty}a_n(x-0)^n
\\ y ~' = \sum_{n=1}^{\infty}a_n \cdot n ~x^{n-1}
\\ y ~'' = \sum_{n=2}^{\infty}a_n \cdot n (n-1)~x^{n-2}

}

$$

Answer 6

are we assuming \( \Large x_0 = 0 \)

Answer 7

x0=\[-\infty\]

Answer 8

can you take a screen shot of question number 10

Answer 9

i only see questions 1 through 9

Answer 10

and am still having problem in 7,8,9 10

Answer 11

@perl

Answer 12

there is a general solution to 'airy's equation' here
http://www.sosmath.com/diffeq/series/series04/series04.html

Answer 13

which of your answer choices match the power series expansion of that

Answer 14

i think a

Answer 15

for
10. the general solution to y ' ' + y = 0 is y = a_1 sin(x)+a_2 cos(x)

now see if you can find a power series for that.

Answer 16

here are the first few power series terms
y = a_1 sin(x)+a_2 cos(x)
=a_2+a_1*x-(1/2)*a_2*x^2-(1/6)*a_1*x^3+(1/24)*a_2*x^4

Answer 17

waw.neva seen that

Answer 18

You can also try reducing the order by replacing \(t=y'\), then you have
\[t'+t=0\]
and find a series solution for \(t\), then integrate to find the solution for \(y\).

Answer 19

ok let me start over. This is not the best approach, but it works (reverse engineering).

10) The solution of \( \large y ' ' + y = 0 \)
http://www.wolframalpha.com/input/?i=solve+y+%27+%27+%2B+y+%3D+0

Now we know that power series (taylor series) for sin x and cos x power series
http://en.wikibooks.org/wiki/Trigonometry/Power_Series_for_Cosine_and_Sine

$$\Large {
\displaystyle \cos(x) = 1 - {x^{2} \over 2!} + {x^{4} \over 4!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!}\\

\sin(x) = x - {x^{3} \over 3!} + {x^{5} \over 5!} - \cdots = \sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}
\\ \therefore \\
y = a_0 \cos x + a_1 \sin x \\
= a_0 \sum_{n=0}^{\infty} \frac{(-1)^nx^{2n}}{(2n)!} + a_1\sum_{n=0}^{\infty} \frac{(-1)^{n}x^{2n+1}}{(2n+1)!}

}
$$