Question

solve the de xy''-(2x-1)y'+(x-1)y=0 using frobenius series

Answer 1

Would you get the same result if you don't rewrite the DE

Answer 2

\[xy''-(2x-1)y'+(x-1)y=0\]
sub into DE:
\[=x \sum_{n=0}^{\infty}a_{n}(n+r)(n+r-1)x ^{n+r-2}-2x \sum_{n=0}^{\infty}a_{n}(n+r)x ^{n+r-1}+\sum_{n=0}^{\infty}a_{n}(n+r)x ^{n+r-1}\]
\[ + x \sum_{n=0}^{\infty}a_{n}x ^{n+r} - \sum_{n=0}^{\infty}a_{n}x ^{n+r} =0\]

\[=\sum_{n=0}^{\infty}a_n(n+r)(n+r-1)x^{n+r-1}-2\sum_{n=0}^{\infty}a_n(n+r)x^{n+r}+\sum_{n=0}^{\infty}a_n(n+r)x^{n+r-1}\]
\[+ \sum_{n=0}^{\infty}a_{n}x ^{n+r+1} - \sum_{n=0}^{\infty}a_{n}x ^{n+r} =0\]

This is where am stuck!

Answer 3

I'm not sure... The tutorial I had on this method mentioned that you have to divide through by the leading power of the independent variable, in this case \(x\) because the leading term is \(xy'\). I don't see why it would matter, but I'm sticking to the way I learned it.

Answer 4

so from your solution, how then do you get the indicial equation

Answer 5

I am trying to find two linearly independent solutions to the DE. I was taught that you rewrite the DE when you are finding the second solution(will be using reduction of order).
I just need help splitting the about equation because it is very confusing!

Answer 6

I'm not entirely sure... I've yet to come across any examples that deal with an indicial equation containing more than just \(a_0\) and a single power of \(x\). Might be that I made a mistake somewhere?

Answer 7

Ugh, I did... Let me repost with corrections

Answer 8

\[\begin{align*}
xy''-(2x-1)y'+(x-1)y&=0\\\\
y''-\left(2-\frac{1}{x}\right)y'+\left(1-\frac{1}{x}\right)y&=0
\end{align*}\]
Assume a solution of the form
\[\large\begin{cases}\displaystyle y=\sum_{k=0}^\infty a_kx^{k+r}\\\\
\displaystyle y'=\sum_{k=0}^\infty (k+r)a_kx^{k+r-1}\\\\
\displaystyle y''=\sum_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2}
\end{cases}\]
Substitute into the equation:
\[\large\begin{align*}
\sum_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2}\\-\left(2-\frac{1}{x}\right)\sum_{k=0}^\infty (k+r)a_kx^{k+r-1}\\
\quad\quad+\left(1-\frac{1}{x}\right)\sum_{k=0}^\infty a_kx^{k+r}&=0\\\\

\sum_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2}\\
\color{red}{-2\sum_{k=0}^\infty (k+r)a_kx^{k+r-1}}\\
\color{blue}{+\sum_{k=0}^\infty (k+r)a_kx^{k+r-2}}\\
\color{lightgreen}{+\sum_{k=0}^\infty a_kx^{k+r}}\\
\color{orange}{-\sum_{k=0}^\infty a_kx^{k+r-1}}&=0\\\\

\sum_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2}\\
\color{red}{-2\sum_{k-1=0}^\infty (k-1+r)a_{k-1}x^{k-1+r-1}}\\
\color{blue}{+\sum_{k=0}^\infty (k+r)a_kx^{k+r-2}}\\
\color{lightgreen}{+\sum_{k-2=0}^\infty a_{k-2}x^{k-2+r}}\\
\color{orange}{-\sum_{k-1=0}^\infty a_{k-1}x^{k-1+r-1}}&=0\\\\

\sum_{k=0}^\infty (k+r)(k+r-1)a_kx^{k+r-2}\\
\color{red}{-2\sum_{k=1}^\infty (k+r-1)a_{k-1}x^{k+r-2}}\\
\color{blue}{+\sum_{k=0}^\infty (k+r)a_kx^{k+r-2}}\\
\color{lightgreen}{+\sum_{k=2}^\infty a_{k-2}x^{k+r-2}}\\
\color{orange}{-\sum_{k=1}^\infty a_{k-1}x^{k+r-2}}&=0\\\\

r(r-1)a_0x^{r-2}+r(r+1)a_1x^{r-1}\\
+\sum_{k=2}^\infty (k+r)(k+r-1)a_kx^{k+r-2}\\
\color{red}{-2ra_0x^{r-1}-2\sum_{k=2}^\infty (k+r-1)a_{k-1}x^{k+r-2}}\\
\color{blue}{+ra_0x^{r-2}+(r+1)a_1x^{r-1}+\sum_{k=2}^\infty (k+r)a_kx^{k+r-2}}\\
\color{lightgreen}{+\sum_{k=2}^\infty a_{k-2}x^{k+r-2}}\\
\color{orange}{-a_0x^{r-1}-\sum_{k=2}^\infty a_{k-1}x^{k+r-2}}&=0\\\\

r(r-1)a_0x^{r-2}+r(r+1)a_1x^{r-1}\color{red}{-2ra_0x^{r-1}}\\
\color{blue}{+ra_0x^{r-2}+(r+1)a_1x^{r-1}}\color{orange}{-a_0x^{r-1}}\\
+\sum_{k=2}^\infty\color{red}{\bigg\{}(k+r)(k+r-1)a_k\color{red}{-2(k+r-1)a_{k-1}}\quad\quad\quad\quad\\
\color{blue}{+(k+r)a_k}\color{lightgreen}{+a_{k-2}}\color{orange}{-a_{k-1}}

\color{red}{\bigg\}}x^{k+r-2}&=0
\end{align*}\]
There, that should be right...

Answer 9

Okay, sorting out the indicial equation:
\[\left[r(r-1)+r\right]a_0x^{r-2}-(2r+1)a_0x^{r-1}+[r(r+1)+r+1]a_1x^{r-1}=0\]
The first term \(\left(x^{r-2}\right)\) gives
\[r(r-1)+r=r^2=0~~\implies~~r=0\]
The second term \(\left(x^{r-1}\right)\) is trickier to deal with... I'm not sure if there's a way to reconcile the fact that we have two different unknowns here, but it might be as simple as solving for \(r\) in terms of either/both \(a_0\) and/or \(a_1\).
\[\begin{align*}
-(2r+1)a_0+(r^2+2r+1)a_1&=0\\\\
a_1r^2+(2a_1-2a_0)r+(a_1+a_0)&=0\\\\
r&=\frac{(2a_0-2a_1)\pm\sqrt{(2a_1-2a_0)^2-4a_1(a_1+a_0)}}{2a_1}\\\\
r&=\frac{(a_0-a_1)\pm\sqrt{a_0^2-3a_1a_0}}{a_1}
\end{align*}\]

Answer 10

Okay i got it. Thank you