Question

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Answer 1

You could get a textbook, Sears and Zemansky's University Physics.
You could also use some online resources.

Answer 2

y"-xy'-y=0, \(x_0=1\)
\[y=\sum_{n=0}^{\infty}a_nx^n\]
\[y'=\sum_{n=1}^{\infty}a_n n x^{n-1}\\xy'=\sum_{n=1}^{\infty}a_n n x^n\]
\[y"=\sum_{n=2}^{\infty}a_n n (n-1)x ^{n-2}\]
then I don't know how to make them have the same indices and same exponent of x.
@SithsAndGiggles

Answer 3

\[y''-xy'-y=0\]
Since \(x_0=1\), you have
\[y=\sum_{n=0}^\infty a_n(x-1)^n\\
y'=\sum_{n=1}^\infty a_nn(x-1)^{n-1}\\
y''=\sum_{n=2}^\infty a_nn(n-1)(x-1)^{n-2}\]
Note that in the second term of the ODE, you will not be able to distribute the \(x\) readily:
\[xy'=\color{blue}{x}\sum_{n=0}^\infty a_nn\color{red}{(x-1)}^{n-1}\]
(blue and red don't work together)

So, you'll have to rewrite the equation:
\[y''-xy'-y=0\\
y''-xy'+y'-y=y'\\
y''-(x-1)y'-y=y'\]
Now, substitute the series representations:
\[\sum_{n=2}^\infty a_nn(n-1)(x-1)^{n-2}\\
~~~~~~~~-(x-1)\sum_{n=1}^\infty a_nn(x-1)^{n-1}\\
~~~~~~~~~~~~~~~~~~~~~~-\sum_{n=0}^\infty a_n(x-1)^n=\sum_{n=1}^\infty a_nn(x-1)^{n-1}\]
\[\sum_{n=2}^\infty a_nn(n-1)(x-1)^{n-2}\\
~~~~~~~~-\sum_{n=1}^\infty a_nn(x-1)^n\\
~~~~~~~~~~~~~~~~~~~~~~-\sum_{n=0}^\infty a_n(x-1)^n=\sum_{n=1}^\infty a_nn(x-1)^{n-1}\]

(Sorry the equation is so staggered, I couldn't think of another way to fit it all in.)

Answer 4

Since \(\displaystyle \sum_{n=0}^\infty a_n(x-1)^n=a_0+\sum_{n=1}^\infty a_n(x-1)^n\), you have
\[\sum_{n=2}^\infty a_nn(n-1)(x-1)^{n-2}\\
~~~~~~~~-\sum_{n=1}^\infty a_nn(x-1)^n\\
~~~~~~~~~~~~~~~~~~~~~~-\left(a_0+\sum_{n=1}^\infty a_n(x-1)^n\right)=\sum_{n=1}^\infty a_nn(x-1)^{n-1}\]
\[\sum_{n=2}^\infty a_nn(n-1)(x-1)^{n-2}-\sum_{n=1}^\infty a_n(n+1)(x-1)^n-a_0=\sum_{n=1}^\infty a_nn(x-1)^{n-1}\]

Answer 5

After some work, you'll find that
\[\large a_n=\frac{a_{n-1}-a_{n-2}}{n}\]

Answer 6

Thank you very much.