ODE RSP2

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We use the method of Frobenius to obtain two linearly independent series solution. Where x=0 is a regular singular point of the differential equation. The general solution on (0,\infty )

\displaystyle 2xy''-\left(3+2x\right)y'+y=0

Problem Setup

Let :

\displaystyle y=\sum _{{n=0}}^{{\infty }}C_{{n}}x^{{n+r}} , \displaystyle y'=\sum _{{n=0}}^{{\infty }}\left(n+r\right)C_{{n}}x^{{n+r-1}} , and \displaystyle y''=\sum _{{n=0}}^{{\infty }}\left(n+r\right)\left(n+r-1\right)C_{{n}}x^{{n+r-2}}

Substitute into DE:

\displaystyle 2x\sum _{{n=0}}^{{\infty }}\left(n+r\right)\left(n+r-1\right)C_{{n}}x^{{n+r-2}}-3\sum _{{n=0}}^{{\infty }}\left(n+r\right)C_{{n}}x^{{n+r-1}}-2x\sum _{{n=0}}^{{\infty }}\left(n+r\right)C_{{n}}x^{{n+r-1}}+\sum _{{n=0}}^{{\infty }}C_{{n}}x^{{n+r}}=0

Simplify:

\displaystyle \sum _{{n=0}}^{{\infty }}2\left(n+r\right)\left(n+r-1\right)C_{{n}}x^{{n+r-1}}-\sum _{{n=0}}^{{\infty }}3\left(n+r\right)C_{{n}}x^{{n+r-1}}-\sum _{{n=0}}^{{\infty }}2\left(n+r\right)C_{{n}}x^{{n+r}}+\sum _{{n=0}}^{{\infty }}C_{{n}}x^{{n+r}}=0

We want to have all the powers of x match. In order to do this we will substitute k for the value of n:

\displaystyle \underbrace {{\sum _{{n=0}}^{{\infty }}2\left(n+r\right)\left(n+r-1\right)C_{{n}}x^{{n+r-1}}}}_{{k=n-1}}-\underbrace {{\sum _{{n=0}}^{{\infty }}3\left(n+r\right)C_{{n}}x^{{n+r-1}}}}_{{k=n-1}}-\underbrace {{\sum _{{n=0}}^{{\infty }}2\left(n+r\right)C_{{n}}x^{{n+r}}}}_{{k=n}}+\underbrace {{\sum _{{n=0}}^{{\infty }}C_{{n}}x^{{n+r}}}}_{{k=n}}=0

Which yields:

\displaystyle \sum _{{{\color {red}k=-1}}}^{{\infty }}2\left(k+r+1\right)\left(k+r\right)C_{{k+1}}x^{{{\color {red}k+r}}}-\sum _{{{\color {red}k=-1}}}^{{\infty }}3\left(k+r+1\right)C_{{k+1}}x^{{{\color {red}k+r}}}-\sum _{{{\color {red}k=0}}}^{{\infty }}2\left(k+r\right)C_{{k}}x^{{{\color {red}k+r}}}+\sum _{{{\color {red}k=0}}}^{{\infty }}C_{{k}}x^{{{\color {red}k+r}}}=0

Now, in order to add together the sums we need to have the indexes for each sum match. To do this we list the terms for the first and second sum at k=-1. In this example we need to have the k=-1 indexes match the k=0. In order to reduce the amount of work required we will add together the sums where k=-1 and then factor out the C_{k}+1 term. At this point we can also factor out the x^{r} term.

The first two terms are:

\displaystyle \left[2r\left(r-1\right)C_{{0}}x^{{-1}}-3rC_{{0}}x^{{-1}}\right]x^{{r}}=\left[r\left(2r-5\right)C_{{0}}x^{{-1}}\right]x^{{r}}

From the sum we have:

\displaystyle x^{{r}}\left[\sum _{{k=0}}^{{\infty }}\left[2\left(k+r+1\right)\left(k+r\right)C_{{k+1}}-3\left(k+r+1\right)C_{{k+1}}-2\left(k+r\right)C_{{k}}+C_{{k}}\right]x^{k}\right]=x^{{r}}\left[\sum _{{k=0}}^{{\infty }}[C_{{k+1}}\left(k+r+1\right)\left(2k+2r-3\right)-C_{{k}}\left(2k+2r-1\right)]x^{k}\right]

Finally we now have:

\displaystyle x^{{r}}\left[r\left(2r-5\right)C_{{0}}x^{{-1}}+\sum _{{k=0}}^{{\infty }}[C_{{k+1}}\left(k+r+1\right)\left(2k+2r-3\right)-C_{{k}}\left(2k+2r-1\right)]x^{k}\right]=0

Solving

Now we are ready to solve the equation. We start by equating the terms on the left hand side with the terms on the right hand side (which are all equal to zero). Which gives us the following implications:

(1) \displaystyle r\left(2r-5\right)C_{{0}}=0\rightarrow the indicial roots are: \displaystyle \;r=0\; and \displaystyle \;r={\frac  {5}{2}}\;.


(2) \displaystyle C_{{k+1}}\left(k+r+1\right)\left(2k+2r-3\right)-C_{{k}}\left(2k+2r-1\right)=0,\;k=0,1,2,\ldots

The recurrence relation is:

\;\Rightarrow C_{{k+1}}={\frac  {C_{{k}}\left(2k+2r-1\right)}{\left(k+r+1\right)(2k+2r-3)}}\;,\;k=0,1,2,\ldots

We can now substitute the values for \displaystyle r that satisfy (1) into (2) to yield two different recurrence relations:

(4) r=0\;,\;C_{{k+1}}={\frac  {C_{{k}}\left(2k-1\right)}{\left(k+1\right)(2k-3)}}\;,\;k=0,1,2,\ldots

(5) r={\frac  {5}{2}}\;,\;C_{{k+1}}={\frac  {C_{{k}}\left(2k+4\right)}{\left(2k+7\right)(k+1)}}\;,\;k=0,1,2,\ldots


Now we can let \;k=0,1,2,\ldots on equations (4) and (5) for the first 3 terms:

\displaystyle k (4) \displaystyle \;\;C_{{k+1}}={\frac  {C_{{k}}\left(2k-1\right)}{\left(k+1\right)(2k-3)}} (5) \displaystyle \;\;C_{{k+1}}={\frac  {C_{{k}}\left(2k+4\right)}{\left(2k+7\right)(k+1)}}
\displaystyle 0 \displaystyle C_{{1}}={\frac  {C_{{0}}}{3}} \displaystyle C_{{1}}={\frac  {4}{7}}C_{{0}}
\displaystyle 1 \displaystyle C_{{2}}={\frac  {-C_{{0}}}{6}} \displaystyle C_{{2}}={\frac  {4}{21}}C_{{0}}
\displaystyle 2 \displaystyle C_{{3}}={\frac  {-C_{{0}}}{6}} \displaystyle C_{{3}}={\frac  {32}{693}}C_{{0}}


And so on ...

For simplification purposes we let \displaystyle C_{0}=1
so we get the general solution to the differential equation on (0,\infty ) :

\displaystyle y(x)=C_{{1}}\left(1+{\frac  {1}{3}}x-{\frac  {1}{6}}x^{{2}}-{\frac  {1}{6}}x^{{3}}+\ldots \right)+C_{{2}}x^{{{\frac  {5}{2}}}}\left(1+{\frac  {4}{7}}x+{\frac  {4}{21}}x^{{2}}+{\frac  {32}{693}}x^{{3}}+\ldots \right)