2.5.2.7 Example 7 \(4xy^{\prime \prime }+3y^{\prime }+3y=\sqrt {x}\)

\[ 4xy^{\prime \prime }+3y^{\prime }+3y=\sqrt {x}\]

Comparing the ode to

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

Hence \(p\left ( x\right ) =\frac {3}{4x},q\left ( x\right ) =\frac {3}{4x}\). Therefore \(p_{0}=\lim _{x\rightarrow 0}xp\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3}{4}=\frac {3}{4}\) and \(q_{0}=\lim _{x\rightarrow 0}x^{2}q\left ( x\right ) =\lim _{x\rightarrow 0}\frac {3x}{4}=0\). Hence \(x=0\) is regular singular point. The indicial equation is

\begin{align*} r\left ( r-1\right ) +p_{0}r+q_{0} & =0\\ r\left ( r-1\right ) +\frac {3}{4}r+0 & =0\\ r\left ( r-1\right ) +\frac {3}{4}r & =0\\ r & =\frac {1}{4},0 \end{align*}

Frobenius is now used. Roots differ by non integer.  First we find \(y_{h}\). Let \(y=\sum _{n=0}^{\infty }a_{n}x^{n+r}\).

\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 homogenous ode becomes

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

Re indexing to lowest powers on \(x\) gives

\[ \sum _{n=0}^{\infty }4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+\sum _{n=0}^{\infty }3\left ( n+r\right ) a_{n}x^{n+r-1}+\sum _{n=1}^{\infty }3a_{n-1}x^{n+r-1}=0 \]

When \(n=0\)

\begin{align*} 4\left ( n+r\right ) \left ( n+r-1\right ) a_{n}x^{n+r-1}+3\left ( n+r\right ) a_{n}x^{n+r-1} & =0\\ 4r\left ( r-1\right ) a_{0}+3ra_{0} & =0\\ \left ( 4r\left ( r-1\right ) +3r\right ) a_{0} & =0 \end{align*}

Since \(a_{0}\neq 0\) then \(4r\left ( r-1\right ) +3r=0\), hence \(r=0,r=\frac {1}{4}\) as was found above. Therefore the homogenous ode satisfies

\[ 4xy^{\prime \prime }+3y^{\prime }+3y=\left ( 4r\left ( r-1\right ) +3r\right ) a_{0}x^{r-1}\]

Hence the balance equation is that we will use to find the particular solution is

\[ \left ( 4m\left ( m-1\right ) +3m\right ) c_{0}x^{m-1}=\sqrt {x}\]

We will get back to the above after finding \(y_{h}\). Going over the same steps as before, we find the recurrence relation

\[ a_{n}=-\frac {3a_{n-1}}{4n^{2}+8nr+4r^{2}-n-r}\]

For \(r=\frac {1}{4},n>0\) and similarly

\[ b_{n}=-\frac {3a_{n-1}}{4n^{2}+8nr+4r^{2}-n-r}\]

For \(r=0,n>0\). Finding few terms using the above gives the solution as

\begin{align*} y_{h} & =c_{1}y_{1}\left ( x\right ) +c_{2}y_{2}\left ( x\right ) \\ & =c_{1}x^{\frac {1}{4}}\left ( 1-\frac {3}{5}x+\frac {1}{10}x^{2}-\frac {1}{130}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {3}{14}x^{2}-\frac {3}{154}x^{3}+\cdots \right ) \end{align*}

Now we need to find \(y_{p}\). From the balance equation

\[ \left ( 4m\left ( m-1\right ) +3m\right ) c_{0}x^{m-1}=\sqrt {x}\]

Hence \(m-1=\frac {1}{2}\) or \(m=\frac {3}{2}\). And \(\left ( 4m\left ( m-1\right ) +3m\right ) c_{0}=1\), hence \(\left ( 4\left ( \frac {3}{2}\right ) \left ( \frac {3}{2}-1\right ) +3\left ( \frac {3}{2}\right ) \right ) c_{0}=1\), which gives \(c_{0}=\frac {2}{15}\). Therefore

\begin{align*} y_{p} & =x^{m}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =x^{\frac {3}{2}}\left ( c_{0}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{\frac {3}{2}}\left ( \frac {2}{15}+c_{1}x+c_{2}x^{2}+\cdots \right ) \end{align*}

We now just need to determined \(c_{n}\) for \(n>0\). For this we use the same recurrence relation as found above. We can use \(a_{n}\) or \(b_{n}\) as they are the same, but change \(a_{n}\) to \(c_{n}\) and \(r\) to \(c\) (so not to confuse notations). This gives

\[ c_{n}=-\frac {3c_{n-1}}{4n^{2}+8nm+4m^{2}-n-m}\]

For \(n>0\) and \(m=\frac {3}{2}\). Hence for \(n=1\) the above gives

\begin{align*} c_{1} & =-\frac {3c_{0}}{4+8\left ( \frac {3}{2}\right ) +4\left ( \frac {3}{2}\right ) ^{2}-1-\frac {3}{2}}\\ & =-\frac {3\left ( \frac {2}{15}\right ) }{4+8\left ( \frac {3}{2}\right ) +4\left ( \frac {3}{2}\right ) ^{2}-1-\frac {3}{2}}\\ & =-\frac {4}{225}\end{align*}

For \(n=2\)

\begin{align*} c_{1} & =-\frac {3c_{1}}{4\left ( 2\right ) ^{2}+8\left ( 2\right ) \left ( \frac {3}{2}\right ) +\left ( \frac {3}{2}\right ) ^{2}-2-\left ( \frac {3}{2}\right ) }\\ & =-\frac {3\left ( -\frac {4}{225}\right ) }{4\left ( 2\right ) ^{2}+8\left ( 2\right ) \left ( \frac {3}{2}\right ) +4\left ( \frac {3}{2}\right ) ^{2}-2-\left ( \frac {3}{2}\right ) }\\ & =\frac {8}{6825}\end{align*}

And so on. Hence

\begin{align*} y_{p} & =x^{\frac {3}{2}}\left ( \frac {2}{15}+c_{1}x+c_{2}x^{2}+\cdots \right ) \\ & =x^{\frac {3}{2}}\left ( \frac {2}{15}-\frac {4}{225}x+\frac {8}{6825}x^{2}-\frac {16}{348075}x^{3}+\cdots \right ) \end{align*}

Hence the complete solution is

\begin{align*} y & =y_{h}+y_{p}\\ & =c_{1}x^{\frac {1}{4}}\left ( 1-\frac {3}{5}x+\frac {1}{10}x^{2}-\frac {1}{130}x^{3}+\cdots \right ) +c_{2}\left ( 1-x+\frac {3}{14}x^{2}-\frac {3}{154}x^{3}+\cdots \right ) +x^{\frac {3}{2}}\left ( \frac {2}{15}-\frac {4}{225}x+\frac {8}{6825}x^{2}-\frac {16}{348075}x^{3}+\cdots \right ) \end{align*}