Example 1. homogeneous ode x2y′′(x) + xy′(x) + x2y(x) = 0

1.1 Example 1. homogeneous ode \(x^{2}y^{\prime \prime }\left ( x\right ) +xy^{\prime }\left ( x\right ) +x^{2}y\left ( x\right ) =0\)

Solve

\begin{equation} x^{2}y^{\prime \prime }\left ( x\right ) +xy^{\prime }\left ( x\right ) +x^{2}y\left ( x\right ) =0 \tag {1}\end{equation}

Using power series method by expanding around \(x=0\). Writing the ode as

\[ y^{\prime \prime }\left ( x\right ) +\frac {1}{x}y^{\prime }\left ( x\right ) +y\left ( x\right ) =0 \]

Shows that \(x=0\) is a singular point. But \(\lim _{x\rightarrow 0}x\frac {1}{x}=1\). Hence the singularity is removable. This means \(x=0\) is a regular singular point. In this case the Frobenius power series will be used instead of the standard power series. Let

\[ y\left ( x\right ) =\sum _{n=0}^{\infty }a_{n}x^{n+r}\]

Where \(r\) is to be determined. It is the root of the indicial equation.

\begin{align*} y^{\prime }\left ( x\right ) & =\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}\\ y^{\prime \prime }\left ( x\right ) & =\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}\end{align*}

Substituting the above in (1) gives

\begin{align} x^{2}\sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-2}+x\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r-1}+x^{2}\sum _{n=0}^{\infty }a_{n}x^{n+r} & =0\nonumber \\ \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }a_{n}x^{n+r+2} & =0 \tag {2}\end{align}

Here, we need to make all powers on \(x\) the same, without making the sums start below zero. This can be done by adjusting the last term above as follows

\[ \sum _{n=0}^{\infty }a_{n}x^{n+r+2}=\sum _{n=2}^{\infty }a_{n-2}x^{n+r}\]

Eq (2) becomes

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r}+\sum _{n=0}^{\infty }\left ( n+r\right ) a_{n}x^{n+r}+\sum _{n=2}^{\infty }a_{n-2}x^{n+r}=0 \tag {3}\end{equation}

Eq (3) is the base equation which is used to ﬁnd roots of indicial equation. For \(n=0\) the above gives

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{0}x^{r}+\left ( n+r\right ) a_{0}x^{r} & =0\nonumber \\ \left ( r\right ) \left ( r-1\right ) a_{0}x^{r}+\left ( r\right ) a_{0}x^{r} & =0\nonumber \\ \left ( r\left ( r-1\right ) a_{0}+ra_{0}\right ) x^{r} & =0\nonumber \\ \left ( r\left ( r-1\right ) +r\right ) a_{0}x^{r} & =0 \tag {4}\end{align}

Since \(a_{0}\neq 0\) then (4) gives

\begin{align*} r\left ( r-1\right ) +r & =0\\ r^{2}-r+r & =0\\ r^{2} & =0 \end{align*}

Hence the roots of the indicial equation are \(r=0\) which is a double root. Hence \(r_{1}=r_{2}=0\). This is the case when roots of indicial equation are repeated. In this case the solution \(y_{h}\left ( x\right ) \) is given by

\[ y_{h}\left ( x\right ) =c_{1}y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \]

Where \(y_{1}\left ( x\right ) \) is the ﬁrst solution, which is assumed to be

\begin{equation} y_{1}\left ( x\right ) =\sum _{n=0}^{\infty }a_{n}x^{n+r} \tag {5}\end{equation}

This is the standard Frobenius power series, just like we did to ﬁnd the indicial equation, the only diﬀerence is that now we use \(r=r_{1}\), and hence it is a known value. Once we ﬁnd \(y_{1}\left ( x\right ) \), then the second solution is

\begin{equation} y_{2}\left ( x\right ) =y_{1}\left ( x\right ) \ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r} \tag {6}\end{equation}

Something important to notice. In the sum above, it starts from \(1\) and not from \(0\). The main issue is how to ﬁnd \(b_{n}\). Since that is the only thing we need to be able to complete the solution as \(y_{1}\left ( x\right ) \) is easily found. It turns out that there is a relation between the \(b_{n}\) and the \(a_{n}\). The \(b_{n}\) can be found by taking just derivative of \(a_{n}\) as function of \(r\) for each \(n\) and then evaluating the result at \(r=r_{1}\). How this is done will be shown below.

First we need to ﬁnd \(y_{1}\left ( x\right ) \). Substituting (5) in the ODE gives (3) again (but now with \(r\) having speciﬁc value \(r_{1}\)).

Now we are ready to ﬁnd \(a_{n}\) for \(n>0\). We skip \(n=0\) since that was used to obtain the indicial equation.

For \(n=1\). Eq (7) gives

\begin{align*} \left ( n+r\right ) \left ( n+r-1\right ) a_{1}+\left ( n+r\right ) a_{1} & =0\\ \left ( 1+r\right ) \left ( 1+r-1\right ) a_{1}+\left ( 1+r\right ) a_{1} & =0\\ \left ( \left ( 1+r\right ) \left ( 1+r-1\right ) +\left ( 1+r\right ) \right ) a_{1} & =0\\ \left ( r+1\right ) ^{2}a_{1} & =0 \end{align*}

But \(r=r_{1}=0\). The above becomes \(a_{1}=0\). It is a good idea to use a table to keep record of the \(a_{n}\) values as function of \(r\), since this will be used later to ﬁnd \(b_{n}\).


\(n\)	\(a_{n}\)	\(a_{n}\left ( r=r_{1}\right ) \)

\(0\)	\(a_{0}\)	\(a_{0}\)

\(1\)	\(0\)	\(0\)

For \(n\geq 2\) now we obtain the recursive equation. Notice that the recursive equation starts from \(n\) which is the largest lower summation index. In this case it is \(n=2\). For all lower index, we have to ﬁnd \(a_{n}\) without the use of recursive equation as we did above for \(a_{1}\). Using (7), the recursive equation is

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r\right ) a_{n}+a_{n-2} & =0\nonumber \\ a_{n} & =-\frac {a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) } \tag {8}\end{align}

The above EQ. (8) is very important, since we will use it to ﬁnd all \(a_{n}\). It is valid only for \(n\geq 2\). Now we ﬁnd few more \(a_{n}\) terms. From above and for \(n=2\)

\begin{align*} a_{2} & =-\frac {a_{0}}{\left ( 2+r\right ) \left ( 2+r-1\right ) +\left ( 2+r\right ) }\\ & =-\frac {a_{0}}{\left ( r+2\right ) ^{2}}\end{align*}

and \(r=r_{1}=0\) then the above becomes

\[ a_{2}=-\frac {a_{0}}{\left ( 2\right ) ^{2}}=-\frac {a_{0}}{4}\]

The table now becomes


\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=r_{1}\right ) \)

\(0\)	\(a_{0}\)	\(1\)

\(1\)	\(0\)	\(0\)

\(2\)	\(-\frac {a_{0}}{\left ( r+2\right ) ^{2}}\)	\(-\frac {a_{0}}{4}\)

And for \(n=3\)

\[ a_{3}=-\frac {a_{1}}{\left ( 3+r\right ) \left ( 3+r-1\right ) +\left ( 3+r\right ) }\]

But \(a_{1}=0\). Then \(a_{3}=0\). The table becomes


\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=r_{1}\right ) \)

\(0\)	\(a_{0}\)	\(a_{0}\)

\(1\)	\(0\)	\(0\)

\(2\)	\(-\frac {a_{0}}{\left ( r+2\right ) ^{2}}\)	\(-\frac {a_{0}}{4}\)

\(3\)	\(0\)	\(0\)

For \(n=4\) Eq (8) gives

\[ a_{4}=-\frac {a_{2}}{\left ( 4+r\right ) \left ( 4+r-1\right ) +\left ( 4+r\right ) }\]

But \(a_{2}\) from the table is \(-\frac {1}{\left ( r+2\right ) ^{2}}\). Hence

\[ a_{4}\left ( r\right ) =-\frac {-\frac {a_{0}}{\left ( r+2\right ) ^{2}}}{\left ( 4+r\right ) \left ( 4+r-1\right ) +\left ( 4+r\right ) }=\frac {a_{0}}{\left ( r^{2}+6r+8\right ) ^{2}}\]

The above becomes at \(r=r_{1}=0\)

\[ a_{4}=\frac {a_{0}}{\left ( 8\right ) ^{2}}=\frac {a_{0}}{64}\]

The Table now becomes


\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=r_{1}\right ) \)

\(0\)	\(a_{0}\)	\(a_{0}\)

\(1\)	\(0\)	\(0\)

\(2\)	\(-\frac {a_{0}}{\left ( r+2\right ) ^{2}}\)	\(-\frac {a_{0}}{4}\)

\(3\)	\(0\)	\(0\)

\(4\)	\(\frac {a_{0}}{\left ( r^{2}+6r+8\right ) ^{2}}\)	\(\frac {a_{0}}{64}\)

For \(n=5\) Eq (8) gives

\[ a_{5}=-\frac {a_{3}}{\left ( n+r\right ) \left ( n+r-1\right ) +\left ( n+r\right ) }\]

But \(a_{3}=0\), hence \(a_{5}=0\). The table becomes


\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=r_{1}\right ) \)

\(0\)	\(a_{0}\)	\(a_{0}\)

\(1\)	\(0\)	\(0\)

\(2\)	\(-\frac {a_{0}}{\left ( r+2\right ) ^{2}}\)	\(-\frac {a_{0}}{4}\)

\(3\)	\(0\)	\(0\)

\(4\)	\(\frac {a_{0}}{\left ( r^{2}+6r+8\right ) ^{2}}\)	\(\frac {a_{0}}{64}\)

\(5\)	\(0\)	\(0\)

For \(n=6\) Eq (8) gives

\[ a_{6}\left ( r\right ) =-\frac {a_{4}\left ( r\right ) }{\left ( 6+r\right ) \left ( 6+r-1\right ) +\left ( 6+r\right ) }\]

But from the table \(a_{4}=\frac {a_{0}}{\left ( r^{2}+6r+8\right ) ^{2}}\), so the above becomes

\[ a_{6}\left ( r\right ) =-\frac {\frac {a_{0}}{\left ( r^{2}+6r+8\right ) ^{2}}}{\left ( 6+r\right ) \left ( 6+r-1\right ) +\left ( 6+r\right ) }=-\frac {a_{0}}{\left ( r+6\right ) ^{2}\left ( r^{2}+6r+8\right ) ^{2}}\]

At \(r=r_{1}=0\) the above becomes

\[ a_{6}=-\frac {a_{0}}{\left ( 6\right ) ^{2}\left ( 8\right ) ^{2}}=-\frac {a_{0}}{2304}\]

The table becomes


\(n\)	\(a_{n}\)	\(a_{n}\left ( r=r_{1}\right ) \)

\(0\)	\(a_{0}\)	\(a_{0}\)

\(1\)	\(0\)	\(0\)

\(2\)	\(-\frac {a_{0}}{\left ( r+2\right ) ^{2}}\)	\(-\frac {a_{0}}{4}\)

\(3\)	\(0\)	\(0\)

\(4\)	\(\frac {a_{0}}{\left ( r^{2}+6r+8\right ) ^{2}}\)	\(\frac {a_{0}}{64}\)

\(5\)	\(0\)	\(0\)

\(6\)	\(-\frac {a_{0}}{\left ( r+6\right ) ^{2}\left ( r^{2}+6r+8\right ) ^{2}}\)	\(-\frac {a_{0}}{2304}\)

And so on. Hence \(y_{1}\left ( x\right ) \) is

\[ y_{1}\left ( x\right ) =\sum _{n=0}^{\infty }a_{n}x^{n+r}\]

But \(r=r_{1}=0\). Therefore

\begin{align} y_{1}\left ( x\right ) & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+a_{5}x^{5}+a_{6}x^{6}+\cdots \nonumber \\ & =a_{0}\left ( 1-\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {1}{2304}x^{6}+\cdots \right ) \tag {9}\end{align}

We are done ﬁnding \(y_{1}\left ( x\right ) \). This was not bad at all. Now comes the hard part. Which is ﬁnding \(y_{2}\left ( x\right ) \). From (6) it is given by

\[ y_{2}\left ( x\right ) =y_{1}\left ( x\right ) \ln \left ( x\right ) +\sum _{n=1}^{\infty }b_{n}x^{n+r}\]

To ﬁnd \(b_{n}\), we will use the following

\begin{equation} b_{n}=\frac {d}{dr}\left ( a_{n}\right ) \tag {10}\end{equation}

Notice that \(n\) starts from \(1\). Hence

\[ b_{1}\left ( r\right ) =\left . \frac {d}{dr}\left ( a_{1}\left ( r\right ) \right ) \right \vert _{r=r_{1}}\]

What the above says, is that we ﬁrst take derivative of \(a_{n}\) w.r.t. \(r\) and evaluate the result at the root of the indicial equation. Using the table above, we obtain (recalling that \(r_{1}=0\) in this example)


\(n\)	\(a_{n}\)	\(a_{n}\left ( r=r_{1}\right ) \)	\(b_{n}=\frac {d}{dr}\left ( a_{n}\right ) \)	\(b_{n}\left ( r=r_{1}\right ) \)

\(0\)	\(a_{0}\)	\(a_{0}\)	N/A since \(b\) starts from \(n=1\)	N/A

\(1\)	\(0\)	\(0\)	\(0\)	\(0\)

\(2\)	\(-\frac {a_{0}}{\left ( r+2\right ) ^{2}}\)	\(-\frac {a_{0}}{4}\)	\(\frac {d}{dr}\left ( -\frac {a_{0}}{\left ( r+2\right ) ^{2}}\right ) =\frac {2a_{0}}{\left ( r+2\right ) ^{3}}\)	\(\frac {2a_{0}}{\left ( 2\right ) ^{3}}=\) \(\frac {a_{0}}{4}\)

\(3\)	\(0\)	\(0\)	\(0\)	\(0\)

\(4\)	\(\frac {a_{0}}{\left ( r^{2}+6r+8\right ) ^{2}}\)	\(\frac {a_{0}}{64}\)	\(\frac {d}{dr}\left ( \frac {a_{0}}{\left ( r^{2}+6r+8\right ) ^{2}}\right ) =-2a_{0}\frac {2r+6}{\left ( r^{2}+6r+8\right ) ^{3}}\)	\(-2a_{0}\frac {6}{\left ( 8\right ) ^{3}}=-\frac {3a_{0}}{128}\)

\(5\)	\(0\)	\(0\)	\(0\)	\(0\)

\(6\)	\(-\frac {a_{0}}{\left ( r+6\right ) ^{2}\left ( r^{2}+6r+8\right ) ^{2}}\)	\(-\frac {a_{0}}{2304}\)	\(\frac {d}{dr}\left ( -\frac {a_{0}}{\left ( r+6\right ) ^{2}\left ( r^{2}+6r+8\right ) ^{2}}\right ) =2a_{0}\frac {3r^{2}+24r+44}{\left ( r^{3}+12r^{2}+44r+48\right ) ^{3}}\)	\(2a_{0}\frac {44}{\left ( 48\right ) ^{3}}=\frac {11a_{0}}{13\,824}\)

We have found all \(b_{n}\) terms. Hence using (6) and since \(r=r_{1}=0\) then

\[ y_{2}\left ( x\right ) =y_{1}\left ( x\right ) \ln \left ( x\right ) +\left ( b_{1}x+b_{2}x^{2}+b_{3}x^{3}+b_{4}x^{4}+b_{5}x^{5}+b_{6}x^{6}+\cdots \right ) \]

But from the above table, we see that \(b_{1}=0,b_{2}=\frac {a_{0}}{4},b_{3}=0,b_{4}=-\frac {3a_{0}}{128},b_{5}=0,b_{6}=\frac {11a_{0}}{13\,824}\). The above becomes

\begin{equation} y_{2}\left ( x\right ) =y_{1}\left ( x\right ) \ln \left ( x\right ) +a_{0}\left ( \frac {1}{4}x^{2}-\frac {3}{128}x^{4}+\frac {11}{13\,824}x^{6}+O\left ( x^{8}\right ) \right ) \tag {11}\end{equation}

And we know what \(y_{1}\left ( x\right ) \) is from Eq (9). Hence the above becomes

\[ y_{2}\left ( x\right ) =a_{0}\left ( 1-\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {1}{2304}x^{6}+O\left ( x^{8}\right ) \right ) \ln \left ( x\right ) +a_{0}\left ( \frac {1}{4}x^{2}-\frac {3}{128}x^{4}+\frac {11}{13\,824}x^{6}+O\left ( x^{8}\right ) \right ) \]

Therefore the general solution is

\begin{align*} y\left ( x\right ) & =c_{1}y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \\ & =a_{0}c_{1}\left ( 1-\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {1}{2304}x^{6}+O\left ( x^{8}\right ) \right ) \\ & +a_{0}c_{2}\left ( \left ( 1-\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {1}{2304}x^{6}+O\left ( x^{8}\right ) \right ) \ln \left ( x\right ) +\left ( \frac {1}{4}x^{2}-\frac {3}{128}x^{4}+\frac {11}{13\,824}x^{6}+O\left ( x^{8}\right ) \right ) \right ) \end{align*}

We can now absorb \(a_{0}\) into the constants \(c_{1},c_{2}\) and the above becomes

\begin{align*} y\left ( x\right ) & =c_{1}\left ( 1-\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {1}{2304}x^{6}+O\left ( x^{8}\right ) \right ) \\ & +c_{2}\left ( \left ( 1-\frac {1}{4}x^{2}+\frac {1}{64}x^{4}-\frac {1}{2304}x^{6}+O\left ( x^{8}\right ) \right ) \ln \left ( x\right ) +\left ( \frac {1}{4}x^{2}-\frac {3}{128}x^{4}+\frac {11}{13\,824}x^{6}+O\left ( x^{8}\right ) \right ) \right ) \end{align*}

It it easier in implementation to just make \(a_{0}=1\) at the start of this process so we do not have to carry it around, and that is what we will do from now on.

This completes the solution. The only diﬃculty in this method, is to make sure when ﬁnding the \(b_{n}\) is to have access to the \(a_{n}\) with \(r\) being unevaluated form in order to take derivatives correctly. This was done above by keeping a table of these quantities updated.

[next] [front] [up]