Question

find two series solution of y''-2y'+y=0

Answer 1

Have you considered building the generalized series?

Answer 2

You can solve this without using series to solve...must you use series?

Answer 3

look for \[y = \sum_{n=0}^\infty a_nx^n\]
\[y'= \sum_{n=0}^\infty na_nx^{n-1}\]
\[y"=\sum_{n=0}^\infty n(n-1)a_nx^{n-2}\]
so, \[yours =\sum_{n=0}^\infty n(n-1)a_nx^{n-2}-2\sum_{n=0}^\infty na_nx^{n-1}+\sum_{n=0}^\infty a_nx^n\]
now make them have the same exponent . Pay attention: the first term, exponent of x is n-2, you want to get n , that means you +2 into n (everywhere you see n, +2), The second term +1, the last term, keep it as it is. so you have
\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n}-2\sum_{n=0}^\infty (n+1)a_{n+1}x^{n}+\sum_{n=0}^\infty a_nx^n\]
now you can put everything under the one sum
\[\sum_{n=0}^\infty (n+2)(n+1)a_{n+2}x^{n}-2(n+1)a_{n+1}x^{n}+a_nx^n\]
can you step up from here?

Answer 4

just factor x^n out, and let the coefficient =0, you can construct the relationship between \(a_{n+2}, a_{n+1}~~and~~~a_n\)

Answer 5

yes, but it's part of the program. Cannot say no to it

Answer 6

What program?!

Answer 7

ODE

Answer 8

Your solution is: y(x) = c\(_1\)e\(^x\)+c\(_2\)e\(^x\)x

Answer 9

It should match something very similar to that. Best of luck! Cheers.

Answer 10

hehehe... fortunately, it's not mine. and it's not that.

Answer 11

Your series solution will look different, but it should match the series to e\(^x\) and e\(^x\) x

Answer 12

@priyeshvs
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Answer 13

\[y(x) = c_1 e^x+c_2 e^x x\]

Answer 14

Solving by series-solution-method will give you something like
\[y=a_0\left(1+x+\frac{1}{2!}x^2+\frac{1}{3!}x^3+\cdots\right)+a_1\left(x+x^2+\frac{1}{2!}x^3+\frac{1}{3!}x^4+\cdots\right)\]
which is the same as
\[y=C1e^x+C_2xe^x\]
where \(a_0=C_1\) and \(a_1=C_2\).