2.3.2.0 Example 3 \(y^{\prime \prime }+y^{\prime }+y=\sqrt {x}\)

\[ y^{\prime \prime }+y^{\prime }+y=\sqrt {x}\]

Let the solution be \(y=y_{h}+y_{p}\). We start by finding \(y_{h}\) which is solution to \(y^{\prime \prime }+y^{\prime }+y=0\). Comparing this ode to

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

Hence \(p\left ( x\right ) =1,q\left ( x\right ) =1\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}x=0\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}x^{2}=0\). Hence the indicial equation is

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

Therefore \(r_{1}=1,r_{2}=0\).  Expansion around \(x=0\). Let

\begin{align*} y & =\sum _{n=0}^{\infty }a_{n}x^{n+r}\\ 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*}

The ode becomes

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

Reindexing to lowest powers on \(x\) gives

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

\(n=0\) gives

\[ r\left ( r-1\right ) a_{0}x^{r-2}=0 \]

Since \(a_{0}\neq 0\), then \(r_{1}=1,r_{2}=0\) as was found above. Since roots differ by integer \(N=r_{1}-r_{2}=1\), then the two linearly independent solutions can be constructed using

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

Where \(C\) above can be zero depending on a condition given below. We start by finding \(y_{1}\). When \(n=1\) then (2) gives

\begin{align} \left ( 1+r\right ) \left ( r\right ) a_{1}+ra_{0} & =0\nonumber \\ a_{1} & =\frac {-a_{0}}{1+r} \tag {3}\end{align}

The recurrence relation is when \(n\geq 2\) from (2) is given by

\begin{align} \left ( n+r\right ) \left ( n+r-1\right ) a_{n}+\left ( n+r-1\right ) a_{n-1}+a_{n-2} & =0\nonumber \\ a_{n} & =\frac {-\left ( n+r-1\right ) a_{n-1}-a_{n-2}}{\left ( n+r\right ) \left ( n+r-1\right ) } \tag {4}\end{align}

Now, using \(r=1\). For \(n=1\) and from (3) and using \(a_{0}=1\) gives

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

From \(n=2\) from (4) and using \(r=1\) it becomes

\begin{equation} a_{2}=\frac {-2a_{1}-a_{0}}{\left ( 2+1\right ) \left ( 2\right ) }=\frac {-2a_{1}-a_{0}}{6}=\frac {-2\left ( \frac {-1}{2}\right ) -1}{6}=0\nonumber \end{equation}

For \(n=3\) then (5) gives

\[ a_{3}=\frac {-\left ( 3\right ) a_{2}-a_{1}}{\left ( 3+1\right ) \left ( 3\right ) }=\frac {-a_{1}}{12}=\frac {-\left ( \frac {-1}{2}\right ) }{12}=\frac {1}{24}\]

And so on. Hence

\begin{align*} y_{1} & =\sum _{n=0}^{\infty }a_{n}x^{n+1}\\ & =x\left ( a_{0}+a_{1}x+a_{2}x^{2}+a_{3}x^{3}+a_{4}x^{4}+\cdots \right ) \\ & =x\left ( 1-\frac {1}{2}x+\frac {1}{24}x^{2}-\frac {1}{120}x^{3}+\cdots \right ) \end{align*}

Now we need to find \(y_{2}\). We first check if \(y_{2}\) can be found using standard method as was done above for \(y_{1}\).  For this we look at

\begin{align*} \lim _{r\rightarrow r_{2}}a_{N}\left ( r\right ) & =\lim _{r\rightarrow 0}a_{1}\left ( r\right ) \\ & =\lim _{r\rightarrow 0}\frac {-a_{0}}{1+r}\\ & =-a_{0}\end{align*}

And see this is defined. Hence \(C=0\) and we can find \(y_{2}\) using same series expansion and using \(b_{0}=1\) (we do not need the log term)

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

For \(n\geq 2\) we have

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

Which for \(r=0\) becomes

\begin{equation} b_{n}=\frac {-\left ( n-1\right ) b_{n-1}-b_{n-2}}{n\left ( n-1\right ) } \tag {5}\end{equation}

For \(n=2\)

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

For \(n=3\)

\begin{align*} b_{3} & =\frac {-\left ( 3-1\right ) b_{2}-b_{1}}{3\left ( 3-1\right ) }\\ & =\frac {1}{6}\end{align*}

For \(n=4\)

\begin{align*} b_{4} & =\frac {-\left ( 3\right ) b_{3}-b_{2}}{4\left ( 3\right ) }\\ & =\frac {-\left ( 3\right ) \left ( \frac {1}{6}\right ) }{4\left ( 3\right ) }\\ & =-\frac {1}{24}\end{align*}

And so on. Hence

\begin{align*} y_{2} & =\sum _{n=0}^{\infty }b_{n}x^{n+0}\\ & =\left ( b_{0}+b_{1}x+b_{2}x^{2}+\cdots \right ) \\ & =1-x+\frac {1}{6}x^{3}-\frac {1}{24}x^{4}+\cdots \end{align*}

Therefore \(y_{h}\)

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

Now we find \(y_{p}\). Let

\[ y_{p}=\sum _{n=0}^{\infty }c_{n}x^{n+r}\]

Substituting into the original ode gives

\begin{equation} \sum _{n=0}^{\infty }\left ( n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-2}+\sum _{n=1}^{\infty }\left ( n+r-1\right ) c_{n-1}x^{n+r-2}+\sum _{n=2}^{\infty }c_{n-2}x^{n+r-2}=\sqrt {x} \tag {6}\end{equation}

For \(n=0\)

\[ r\left ( r-1\right ) c_{0}x^{r-2}=\sqrt {x}\]

For balance we need \(r-2=\frac {1}{2}\) or \(r=\frac {5}{2}\). Hence the coefficient is \(r\left ( r-1\right ) c_{0}=1\) or \(\frac {5}{2}\left ( \frac {5}{2}-1\right ) c_{0}=1\). Solving for \(c_{0}\) gives

\[ c_{0}=\frac {4}{15}\]

Therefore

\[ y_{p}=\sum _{n=0}^{\infty }c_{n}x^{n+\frac {5}{2}}\]

For \(n=1\) EQ. (6) gives (where now we put zero on the right side, since there can be no more balance terms, as it was used for \(n=0\))

\begin{align*} \left ( 1+r\right ) \left ( r\right ) c_{1}+rc_{0} & =0\\ c_{1} & =\frac {-c_{0}}{1+r}\\ & =\frac {-c_{0}}{1+\frac {5}{2}}\\ & =\frac {-\frac {4}{15}}{1+\frac {5}{2}}\\ & =-\frac {8}{105}\end{align*}

Recusive relation is for \(n\geq 2\) from EQ.(6) and remebering to always have zero on right side from now on. This gives

\begin{align*} \left ( n+r\right ) \left ( n+r-1\right ) c_{n}+\left ( n+r-1\right ) c_{n-1}+c_{n-2} & =0\\ c_{n} & =\frac {-c_{n-2}-\left ( n+r-1\right ) c_{n-1}}{\left ( n+r\right ) \left ( n+r-1\right ) }\\ & =\frac {-c_{n-2}-\left ( n+\frac {5}{2}-1\right ) c_{n-1}}{\left ( n+\frac {5}{2}\right ) \left ( n+\frac {5}{2}-1\right ) }\\ & =\frac {-c_{n-2}-\left ( n+\frac {3}{2}\right ) c_{n-1}}{\left ( n+\frac {5}{2}\right ) \left ( n+\frac {3}{2}\right ) }\end{align*}

For \(n=2\)

\begin{align*} c_{2} & =\frac {-c_{n-2}-\left ( 2+\frac {3}{2}\right ) c_{1}}{\left ( 2+\frac {5}{2}\right ) \left ( 2+\frac {3}{2}\right ) }\\ & =\frac {-\frac {4}{15}-\left ( 2+\frac {3}{2}\right ) \left ( -\frac {8}{105}\right ) }{\left ( 2+\frac {5}{2}\right ) \left ( 2+\frac {3}{2}\right ) }\\ & =0 \end{align*}

And so on. Hence

\begin{align*} y_{p} & =\sum _{n=0}^{\infty }c_{n}x^{n+r}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n+\frac {5}{2}}\\ & =x^{\frac {5}{2}}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{\frac {5}{2}}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{\frac {5}{2}}\left ( \frac {4}{15}-\frac {8}{105}x+\frac {32}{10\,395}x^{3}+\cdots \right ) \end{align*}

Hence the final solution

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}x\left ( 1-\frac {1}{2}x+\frac {1}{24}x^{2}-\frac {1}{120}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {1}{6}x^{3}-\frac {1}{24}x^{4}+\cdots \right ) +x^{\frac {5}{2}}\left ( \frac {4}{15}-\frac {8}{105}x+\frac {32}{10\,395}x^{3}+\cdots \right ) \end{align*}