2.3.2 Example 5. xy′′ + 2y′ + xy = 0,y(0) = 1,y′(0) = 0

\begin{align} xy^{\prime \prime }+2y^{\prime }+xy & =0\tag {1}\\ y\left ( 0\right ) & =1\nonumber \\ y^{\prime }\left ( 0\right ) & =0 \end{align}

Using power series method by expanding around \(x=0\). We see that \(x=0\) is a regular singular point. In this case the Frobenius power series will be used instead of the standard power series. Let

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

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

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

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

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

Since \(a_{0}\neq 0\) then \(r=0\) or \(r=-1\). Roots diﬀer by integer. Let \(r_{1}=0,r_{2}=-1\), hence the solutions are

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+r_{1}}\\ & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ y_{2} & =Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n+r}\\ & =Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n-1}\end{align*}

We do not know yet if \(C\) will be zero or not (it will be). We have to wait after ﬁnding \(y_{1}\) and ﬁnding all \(a_{n}\) to ﬁnd this. \(N=1\) in this case, (which is diﬀerence between roots). So it depends if \(\lim _{r\rightarrow r_{2}}a_{1}\) if this is deﬁned or not. Now we ﬁnd \(y_{1}\). Using (2), for \(n=1\)

But \(r_{1}=0\), hence\(\ 2a_{1}=0\) or \(a_{1}=0\). We setup the table now to help ﬁnd \(C\)


\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=0\right ) \)

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

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

\begin{align*} \left ( 2+r\right ) \left ( 1+r\right ) a_{2}+2\left ( 2+r\right ) a_{2}+a_{0} & =0\\ a_{2} & =\frac {-a_{0}}{\left ( 2+r\right ) \left ( 1+r\right ) +2\left ( 2+r\right ) }\\ & =\frac {-1}{r^{2}+5r+6}\end{align*}


\(n\)	\(a_{n}\left ( r\right ) \)	\(a_{n}\left ( r=0\right ) \)

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

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

\(2\)	\(\frac {-1}{r^{2}+5r+6}\)	\(\frac {-1}{6}\)

If we continue we will ﬁnd that \(a_{3}=0,a_{4}=\frac {1}{120},a_{5}=0,\cdots \). Hence

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n}\\ & =a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+\cdots \\ & =1-\frac {1}{6}x^{2}+\frac {x^{4}}{120}+\cdots \end{align*}

Now that we found \(y_{1}\) we need to decide if \(C=0\) or not. Since \(N=1\) then we need

But \(a_{1}\left ( r\right ) =0\). Does not depend on \(r\) as we see from the above table. Hence \(\lim _{r\rightarrow -1}a_{1}\left ( r\right ) =0\). Since limit exist then \(C=0\). Hence

\begin{align*} y_{2} & =Cy_{1}\ln \left ( x\right ) +\sum _{n=0}^{\infty }b_{n}x^{n+r}\\ & =\sum _{n=0}^{\infty }b_{n}x^{n-1}\end{align*}

We skip \(n=0\) since that was used to ﬁnd roots and let \(b_{0}=1\). To ﬁnd all \(b_{n}\) we can use (2) but replace \(a_{n}\) by \(b_{n}\) and replace \(r\) by \(-1\) which gives

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

Hence \(b_{1}\) can be any value. Let \(b_{1}=0\). Recusive relation for \(n\geq 2\) which becomes

\begin{align*} \left ( n-1\right ) \left ( n-2\right ) b_{n}+2\left ( n-1\right ) b_{n}+b_{n-2} & =0\\ b_{n} & =\frac {-b_{n-2}}{\left ( n-1\right ) \left ( n-2\right ) +2\left ( n-1\right ) }\end{align*}

\begin{align*} b_{2} & =\frac {-b_{0}}{2}\\ & =-\frac {1}{2}\end{align*}

\begin{align*} b_{3} & =\frac {-b_{1}}{\left ( 3-1\right ) \left ( 3-2\right ) +2\left ( 3-1\right ) }\\ & =0 \end{align*}

\begin{align*} b_{4} & =\frac {-b_{2}}{\left ( 4-1\right ) \left ( 4-2\right ) +2\left ( 4-1\right ) }\\ & =\frac {-\left ( -\frac {1}{2}\right ) }{\left ( 4-1\right ) \left ( 4-2\right ) +2\left ( 4-1\right ) }\\ & =\frac {1}{24}\end{align*}

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n-1}\\ & =\frac {1}{x}\sum _{n=0}^{\infty }b_{n}x^{n}\\ & =\frac {1}{x}\left ( b_{0}+b_{1}x+b_{2}x^{2}+b_{3}x^{3}+\cdots \right ) \\ & =\frac {1}{x}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\cdots \right ) \end{align*}

\begin{align} y & =c_{1}y_{1}+c_{2}y_{2}\nonumber \\ & =c_{1}\left ( 1-\frac {1}{6}x^{2}+\frac {x^{4}}{120}+\cdots \right ) +c_{2}\left ( \frac {1}{x}\left ( 1-\frac {1}{2}x^{2}+\frac {1}{24}x^{4}+\cdots \right ) \right ) \nonumber \\ & =c_{1}\left ( 1-\frac {1}{6}x^{2}+\frac {x^{4}}{120}+\cdots \right ) +c_{2}\left ( \frac {1}{x}-\frac {1}{2}x+\frac {1}{24}x^{3}+\cdots \right ) \tag {3}\end{align}

We now need to determine the \(c_{1},c_{2}\) from initial conditions \(y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0\). At \(x=0\) we have

\begin{align*} 1 & =c_{1}\lim _{x\rightarrow 0}\left ( 1-\frac {1}{6}x^{2}+\frac {x^{4}}{120}+\cdots \right ) +\lim _{x\rightarrow 0}c_{2}\left ( \frac {1}{x}-\frac {1}{2}x+\frac {1}{24}x^{3}+\cdots \right ) \\ & =c_{1}+c_{2}\lim _{x\rightarrow 0}\frac {1}{x}\end{align*}

So we need to have \(c_{2}=0\) since\(\ \lim _{x\rightarrow 0}\frac {1}{x}\) is undeﬁned. Hence \(c_{1}=1\) and the solution now becomes

Now we need to verify this solution satisﬁes the second IC \(y^{\prime }\left ( 0\right ) =0\). Taking derivative gives

At \(x=0\), the above gives \(y^{\prime }=0\). Satisﬁed. Hence the solution is

Lets see what happens if the IC was just \(y^{\prime }\left ( 0\right ) =0\). Taking derivative of (3) gives

\[ y^{\prime }=c_{1}\left ( -\frac {1}{3}x+\frac {x^{3}}{30}+\cdots \right ) +c_{2}\left ( -\frac {1}{x^{2}}-\frac {1}{2}+\frac {3}{24}x^{2}+\cdots \right ) \]

\[ 0=c_{1}\left ( 0\right ) +c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{x^{2}}-\frac {1}{2}\right ) \]

Therefore we need to have \(c_{2}=0\) since no limit exist. We see that \(c_{1}\) can be any value. Hence the solution (3) becomes

Lets see what happens if the IC was just \(y^{\prime }\left ( 0\right ) =A\) for \(A\neq 0\). Then taking derivative as above gives

\[ A=c_{1}\left ( 0\right ) +c_{2}\lim _{x\rightarrow 0}\left ( -\frac {1}{x^{2}}-\frac {1}{2}\right ) \]

Which is contradiction. Therefore for the IC \(y^{\prime }\left ( 0\right ) =A\) there is no solution. Finally, lets see what happens when the IC is \(y\left ( 0\right ) =A\) for \(A\neq 0\). From (3)

\[ A=c_{1}\left ( 1\right ) +c_{2}\lim _{x\rightarrow 0}\left ( \frac {1}{x}\right ) \]

Since limit is not deﬁned, we make \(c_{2}=0\) which means \(c_{1}=A\). And the solution (3) becomes

Let verify this satisﬁes the IC \(y\left ( 0\right ) =A\). We see it does. When \(x=0\) the above gives \(A=A\).

2.3.2.0 Example 5. \(xy^{\prime \prime }+2y^{\prime }+xy=0,y\left ( 0\right ) =1,y^{\prime }\left ( 0\right ) =0\)