ODE RSP3

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Using Frobenius' Theorem find two linearly independent solutions to the following equation. x=0 is a regular singular point.

\displaystyle 2x^{{2}}y''-xy'+\left(x^{{2}}+1\right)y=0



Problem Setup

Substitute:

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}}

Which yields:

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

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

Which yields:

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

Notice that in the third term, the index of summation, the index for the constant, and exponent of x all changed. 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 each sum at k=0,1,... until each index is the same. In this example we need to have the k=0 indexes match the k=2 index so we list the terms at k=0 and k=1. In order to reduce the amount of work required we will add together the sums where k=0 and then factor out the C_{k} term. At this point we can also factor out the x^{r} term.


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

From the sum:

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

The first two terms are:

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

and

\displaystyle \left[2\left(1+r\right)\left(1+r-1\right)x^{{1}}-(1+r)x^{{1}}+x^{{1}}\right]C_{{1}}=\left(2r^{{2}}+r\right)C_{{1}}x

Which finally gives us:

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

(2) \displaystyle \left(2r^{{2}}+r\right)C_{{1}}x=0 hence \displaystyle C_{1}=0

(3) \displaystyle \left[2\left(k+r\right)\left(k+r-1\right)x^{{k}}-\left(k+r\right)x^{{k}}+x^{{k}}\right]C_{{k}}+C_{{k-2}}x^{{k}}=0\;,\;k=2,3,4,\ldots

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

We can now substitute the values for r that satisfy (1) into (3) to yield two different recurrence relations:

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

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

And then using our values for k\;,\;k=2,3,4,\ldots we can find the series solutions for the differential equation.

\displaystyle k (4) \displaystyle \;\;C_{{k}}={\frac  {-C_{{k-2}}}{k\left(2k-1\right)}} (5) \displaystyle \;\;C_{{k}}={\frac  {-C_{{k-2}}}{k\left(2k+1\right)}}
\displaystyle 2 \displaystyle C_{{2}}={\frac  {-C_{{0}}}{6}} \displaystyle C_{{2}}={\frac  {-C_{{0}}}{10}}
\displaystyle 3 \displaystyle C_{{3}}=0 \displaystyle C_{{3}}=0
\displaystyle 4 \displaystyle C_{{4}}={\frac  {-C_{{2}}}{28}}={\frac  {C_{0}}{168}} \displaystyle C_{{4}}={\frac  {-C_{{2}}}{36}}={\frac  {C_{0}}{360}}
\displaystyle 5 \displaystyle C_{{5}}=0 \displaystyle C{5}=0
\displaystyle 6 \displaystyle C_{{6}}={\frac  {-C_{{4}}}{66}}={\frac  {-C_{0}}{11088}} \displaystyle C_{{6}}={\frac  {-C_{{4}}}{78}}={\frac  {-C_{0}}{28080}}


And so on...

This gives us the terms for our infinite sum and from that we have the solutions to the differential equation.

For (4) where \displaystyle r={\frac  {1}{2}} let \displaystyle C_{0}=1 so the sum is \displaystyle y_{{1}}(x)=x^{{{\frac  {1}{2}}}}\left(1-{\frac  {1}{6}}x^{{2}}+{\frac  {1}{168}}x^{{4}}-{\frac  {1}{11088}}x^{{6}}+\ldots \right)

For (5) where \displaystyle r=1 let \displaystyle C_{0}=1 so the sum is \displaystyle y_{{2}}(x)=x\left(1-{\frac  {1}{10}}x^{{2}}+{\frac  {1}{360}}x^{{4}}-{\frac  {1}{28080}}x^{{6}}+\ldots \right)

Which means the general solution \displaystyle y=C_{{1}}y_{{1}}(x)+C_{{2}}y_{{2}}(x) is:

\displaystyle y=C_{{1}}x^{{{\frac  {1}{2}}}}\left(1-{\frac  {1}{6}}x^{{2}}+{\frac  {1}{168}}x^{{4}}-{\frac  {1}{11088}}x^{{6}}+\ldots \right)+C_{{2}}x\left(1-{\frac  {1}{10}}x^{{2}}+{\frac  {1}{360}}x^{{4}}-{\frac  {1}{28080}}x^{{6}}+\ldots \right)



Ordinary Differential Equations

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