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I have to find the power series solution to this differential equation (y'' − 2xy' + y = 0) 

giving that y(0)=0 and y'(0)=1  

 

Please help if you can, Thanks. 

 Sep 18, 2019
 #1
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I have to find the power series solution to this differential equation (\(y'' - 2xy' + y = 0\))
giving that \(y(0)=0\) and \(y'(0)=1\)  .

 

\(\begin{array}{|lcll|} \hline \text{power series}: \\ \hline y = \sum \limits_{n=0}^{\infty}a_nx^n &=& a_0 +a_1x+a_2x^2+a_3x^3+a_4x^4+a_5x^5+\cdots \\\\ y' = \sum \limits_{n=1}^{\infty}na_nx^{n-1} &=& a_1 +2a_2x+3a_3x^2+4a_4x^3+5a_5x^4+\cdots \\\\ y'' = \sum \limits_{n=2}^{\infty}n(n-1)a_nx^{n-2} &=& 2\cdot 1 \cdot a_2+3\cdot 2 \cdot a_3x+4\cdot 3 \cdot a_4x^2+5\cdot 4 \cdot a_5x^3+\cdots \\ \hline && \begin{array}{|rcll|} \hline y(0) = 0 &=& a_0 +a_1\cdot 0+a_20^2+a_30^3+a_40^4+a_50^5+\cdots \\ 0 &=& a_0 \\ \mathbf{a_0} &=& \mathbf{0} \\ \hline y'(0) = 1 &=& a_1 +2a_2\cdot 0+3a_30^2+4a_40^3+5a_50^4+\cdots \\ 1 &=& a_1 \\ \mathbf{a_1} &=& \mathbf{1} \\ \hline \end{array} \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline y'' - 2xy' + y &=& 0 \\ \sum \limits_{n=2}^{\infty}n(n-1)a_nx^{n-2} - 2x\sum \limits_{n=1}^{\infty}na_nx^{n-1} + \sum \limits_{n=0}^{\infty}a_nx^n &=& 0 \\ \cdots \\ \mathbf{\text{see: https://www.youtube.com/watch?v=SS6bniyB7rw }} \\ \cdots \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline y &=& a_0 \left( 1-\dfrac{1}{2!}x^2 - \sum \limits_{n=2}^{\infty} \dfrac{3\cdot 7 \cdot 11\cdots(4n-5)} {(2n)!} x^{2n} \right) +a_1 \left( x+\sum \limits_{n=1}^{\infty} \dfrac{1\cdot 5 \cdot 9 \cdots(4n-3)} {(2n+1)!} x^{2n+1} \right) \\ \hline \mathbf{a_0} &=& \mathbf{0} \\ \mathbf{a_1} &=& \mathbf{1} \\ \mathbf{y} &=& \mathbf{ x+\sum \limits_{n=1}^{\infty} \dfrac{1\cdot 5 \cdot 9 \cdots(4n-3)} {(2n+1)!} x^{2n+1} } \\ \hline \end{array}\)

 

see: https://www.youtube.com/watch?v=SS6bniyB7rw

 

laugh

 Sep 18, 2019
edited by heureka  Sep 18, 2019
edited by heureka  Sep 18, 2019
edited by heureka  Sep 18, 2019

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