anonymous
  • anonymous
find two series solution of y''-2y'+y=0
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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jamiebookeater
  • jamiebookeater
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tkhunny
  • tkhunny
Have you considered building the generalized series?
abb0t
  • abb0t
You can solve this without using series to solve...must you use series?
Loser66
  • Loser66
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?

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More answers

Loser66
  • Loser66
just factor x^n out, and let the coefficient =0, you can construct the relationship between \(a_{n+2}, a_{n+1}~~and~~~a_n\)
Loser66
  • Loser66
yes, but it's part of the program. Cannot say no to it
abb0t
  • abb0t
What program?!
Loser66
  • Loser66
ODE
abb0t
  • abb0t
Your solution is: y(x) = c\(_1\)e\(^x\)+c\(_2\)e\(^x\)x
abb0t
  • abb0t
It should match something very similar to that. Best of luck! Cheers.
Loser66
  • Loser66
hehehe... fortunately, it's not mine. and it's not that.
abb0t
  • abb0t
Your series solution will look different, but it should match the series to e\(^x\) and e\(^x\) x
goformit100
  • goformit100
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anonymous
  • anonymous
\[y(x) = c_1 e^x+c_2 e^x x\]
anonymous
  • anonymous
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 \(a_1=C_2\).

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