\[ x^{-a-1} y(x)^2-x^a+y'(x)=0 \] ✓ Mathematica : cpu = 0.184436 (sec), leaf count = 230
\[\left \{\left \{y(x)\to \frac {x^{a+1} \left (-\frac {1}{2} (-1)^{-a} a x^{-\frac {a}{2}-1} \Gamma (1-a) I_{-a}\left (2 \sqrt {x}\right )+\frac {1}{2} (-1)^{-a} x^{-\frac {a}{2}-\frac {1}{2}} \Gamma (1-a) \left (I_{-a-1}\left (2 \sqrt {x}\right )+I_{1-a}\left (2 \sqrt {x}\right )\right )+c_1 \left (\frac {1}{2} x^{-\frac {a}{2}-\frac {1}{2}} \Gamma (a+1) \left (I_{a-1}\left (2 \sqrt {x}\right )+I_{a+1}\left (2 \sqrt {x}\right )\right )-\frac {1}{2} a x^{-\frac {a}{2}-1} \Gamma (a+1) I_a\left (2 \sqrt {x}\right )\right )\right )}{(-1)^{-a} x^{-a/2} \Gamma (1-a) I_{-a}\left (2 \sqrt {x}\right )+c_1 x^{-a/2} \Gamma (a+1) I_a\left (2 \sqrt {x}\right )}\right \}\right \}\] ✓ Maple : cpu = 0.087 (sec), leaf count = 54
\[y \relax (x ) = \frac {x^{a +1} \left (-\BesselK \left (a +1, 2 \sqrt {x}\right ) c_{1}+\BesselI \left (a +1, 2 \sqrt {x}\right )\right )}{\sqrt {x}\, \left (\BesselK \left (a , 2 \sqrt {x}\right ) c_{1}+\BesselI \left (a , 2 \sqrt {x}\right )\right )}\]
Hand solution
\begin {align} y^{\prime }+x^{-a-1}y^{2}-x^{a} & =0\nonumber \\ y^{\prime } & =x^{a}-x^{-a-1}y^{2}\nonumber \\ & =P\relax (x) +Q\relax (x) y+R\relax (x) y^{2} \tag {1} \end {align}
This is Ricatti first order non-linear ODE. Using standard transformation\[ y=-\frac {u^{\prime }}{uR\relax (x) }=x^{a+1}\frac {u^{\prime }}{u}\]
Hence
\[ y^{\prime }=\left (a+1\right ) x^{a}\frac {u^{\prime }}{u}+x^{a+1}\frac {u^{\prime \prime }}{u}-x^{a+1}\frac {\left (u^{\prime }\right ) ^{2}}{u^{2}}\]
Comparing to (1) gives
\begin {align} x^{a}-x^{-a-1}y^{2} & =\left (a+1\right ) x^{a}\frac {u^{\prime }}{u}+x^{a+1}\frac {u^{\prime \prime }}{u}-x^{a+1}\frac {\left (u^{\prime }\right ) ^{2}}{u^{2}}\nonumber \\ x^{a}-x^{-a-1}\left (x^{a+1}\frac {u^{\prime }}{u}\right ) ^{2} & =\left ( a+1\right ) x^{a}\frac {u^{\prime }}{u}+x^{a+1}\frac {u^{\prime \prime }}{u}-x^{a+1}\frac {\left (u^{\prime }\right ) ^{2}}{u^{2}}\nonumber \\ 1-\frac {x^{-a-1}}{x^{a}}x^{2a+2}\frac {\left (u^{\prime }\right ) ^{2}}{u^{2}} & =\left (a+1\right ) \frac {u^{\prime }}{u}+x\frac {u^{\prime \prime }}{u}-x\frac {\left (u^{\prime }\right ) ^{2}}{u^{2}}\nonumber \\ 1-x\frac {\left (u^{\prime }\right ) ^{2}}{u^{2}} & =\left (a+1\right ) \frac {u^{\prime }}{u}+x\frac {u^{\prime \prime }}{u}-x\frac {\left (u^{\prime }\right ) ^{2}}{u^{2}}\nonumber \\ 1 & =\left (a+1\right ) \frac {u^{\prime }}{u}+x\frac {u^{\prime \prime }}{u}\nonumber \\ xu^{\prime \prime }+\left (1+a\right ) u^{\prime }-u & =0 \tag {2} \end {align}
In standard form \(u^{\prime \prime }+\frac {1}{x}\left (1+a\right ) u^{\prime }-\frac {1}{x}u=0\) or \(u^{\prime \prime }+p\relax (x) \left (1+a\right ) u^{\prime }+q\relax (x) u=0\). We see that \(p\relax (x) \) is not analytic at \(x=0\) (the expansion point). So we can’t use power series solution, and will use Forbenius series. Power series, which is \(u=\sum _{n=0}^{\infty }c_{n}x^{n}\) is used when the expansion point is not singular point. (i.e. \(p\relax (x) \) and \(q\relax (x) \) are analytic there). Forbenius series \(u=x^{r}\sum _{n=0}^{\infty }c_{n}x^{n}\) is used when there is a removable singular point (called also regular singular point), as in this case. Starting with\[ u=x^{r}\sum _{n=0}^{\infty }c_{n}x^{n}=\sum _{n=0}^{\infty }c_{n}x^{n+r}\] Hence \begin {align*} u^{\prime } & =\sum _{n=0}^{\infty }\left (n+r\right ) c_{n}x^{n+r-1}\\ u^{\prime \prime } & =\sum _{n=0}^{\infty }\left (n+r\right ) \left ( n+r-1\right ) c_{n}x^{n+r-2} \end {align*}
Substituting in (2) gives\begin {align*} x\sum _{n=0}^{\infty }\left (n+r\right ) \left (n+r-1\right ) c_{n}x^{n+r-2}+\left (1+a\right ) \sum _{n=0}^{\infty }\left (n+r\right ) c_{n}x^{n+r-1}-\sum _{n=0}^{\infty }c_{n}x^{n+r} & =0\\ \sum _{n=0}^{\infty }\left (n+r\right ) \left (n+r-1\right ) c_{n}x^{n+r-1}+\left (1+a\right ) \sum _{n=0}^{\infty }\left (n+r\right ) c_{n}x^{n+r-1}-\sum _{n=0}^{\infty }c_{n}x^{n+r} & =0 \end {align*}
Dividing out \(x^{r}\)\[ \sum _{n=0}^{\infty }\left (n+r\right ) \left (n+r-1\right ) c_{n}x^{n-1}+\left (1+a\right ) \sum _{n=0}^{\infty }\left (n+r\right ) c_{n}x^{n-1}-\sum _{n=0}^{\infty }c_{n}x^{n}=0 \] Each term should have \(x^{n-1}\) in it. So we adjust the last term\[ \sum _{n=0}^{\infty }\left (n+r\right ) \left (n+r-1\right ) c_{n}x^{n-1}+\left (1+a\right ) \sum _{n=0}^{\infty }\left (n+r\right ) c_{n}x^{n-1}-\sum _{n=1}^{\infty }c_{n-1}x^{n-1}=0 \] Expanding the second term\[ \sum _{n=0}^{\infty }\left (n+r\right ) \left (n+r-1\right ) c_{n}x^{n-1}+\sum _{n=0}^{\infty }\left (n+r\right ) c_{n}x^{n-1}+\sum _{n=0}^{\infty }a\left (n+r\right ) c_{n}x^{n-1}-\sum _{n=1}^{\infty }c_{n-1}x^{n-1}=0 \] Hence for \(n=0\)\begin {align*} \left (n+r\right ) \left (n+r-1\right ) c_{n}x^{n-1}+\left (n+r\right ) c_{n}x^{n-1}+a\left (n+r\right ) c_{n}x^{n-1} & =0\\ r\left (r-1\right ) c_{0}+rc_{0}+arc_{0} & =0 \end {align*}
Since \(c_{0}\neq 0\) then\[ r\left (r-1\right ) +r+ar=0 \] Hence \(r=\) \(-a\) or \(r=0\). Now for \(n\geq 1\)\begin {align*} \ \left (n+r\right ) \left (n+r-1\right ) c_{n}x^{n-1}+\ \left (n+r\right ) c_{n}x^{n-1}+\ a\left (n+r\right ) c_{n}x^{n-1}-\ c_{n-1}x^{n-1} & =0\\ \left (n+r\right ) \left (n+r-1\right ) c_{n}\ +\ \left (n+r\right ) c_{n}\ +\ a\left (n+r\right ) c_{n}\ -\ c_{n-1}\ & =0\\ \left (\left (n+r\right ) \left (n+r-1\right ) \ +\ \left (n+r\right ) \ +\ a\left (n+r\right ) \right ) c_{n}\ \ & =c_{n-1}\\ c_{n}\ \ & =\frac {c_{n-1}}{\left (n+r\right ) \left (n+r-1\right ) \ +\ \left (n+r\right ) \ +\ a\left (n+r\right ) } \end {align*}
For \(r=0\), we obtain\begin {equation} c_{n}\ \ =\frac {c_{n-1}}{n\left (n-1\right ) \ +\ n\ +\ an}\tag {3} \end {equation} For \(r=-a\)\begin {equation} c_{n}\ \ =\frac {c_{n-1}}{\left (n-a\right ) \left (n-a-1\right ) \ +\ \left (n-a\right ) \ +\ a\left (n-a\right ) }\tag {4} \end {equation} There are two solutions. Looking at (3) for now, for \(n=1\)\[ c_{1}=\frac {c_{0}}{\ 1\ +\ a}\] For \(n=2\)\[ c_{2}\ \ =\frac {c_{1}}{4\ +2a}=\frac {c_{0}}{\ 1\ +\ a}\frac {1}{2\left ( 2\ +a\right ) }\] For \(n=3\)\[ c_{3}=\frac {c_{2}}{3\relax (2) \ +\ 3\ +3a}=\frac {c_{2}}{3\left ( 3\ +a\right ) }=\frac {c_{0}}{\ 1\ +\ a}\frac {1}{2\left (2\ +a\right ) }\frac {1}{3\left (3\ +a\right ) }\] And so on. Since the solution is assumed to be \(x^{r}\sum _{n=0}^{\infty }c_{n}x^{n}\) and we are looking at case \(r=0\) then\begin {align} u_{r=0}\relax (x) & =\sum _{n=1}^{\infty }c_{n}x^{n}\nonumber \\ & =c_{0}+c_{1}x+c_{2}x^{2}+\cdots \nonumber \\ & =c_{0}x^{0}+\frac {c_{0}}{\ 1\ +\ a}x+\frac {c_{0}}{\ 1\ +\ a}\frac {1}{2\left (2\ +a\right ) }x^{2}+\frac {c_{0}}{\ 1\ +\ a}\frac {1}{2\left ( 2\ +a\right ) }\frac {1}{3\left (3\ +a\right ) }x^{3}+\cdots \nonumber \\ & =c_{0}\left (x^{0}+\frac {1}{\ 1\ +\ a}x+\frac {1}{\ \left (1\ +\ a\right ) }\frac {1}{2\left (2\ +a\right ) }x^{2}+\frac {1}{\ \left (1\ +\ a\right ) }\frac {1}{2\left (2\ +a\right ) }\frac {1}{3\left (3\ +a\right ) }x^{3}+\cdots \right ) \tag {5} \end {align}
Since \[ \Gamma \relax (n) =\left (n-1\right ) ! \] and \[ a\left (1+a\right ) \left (2+a\right ) \cdots \left (n+a\right ) =\frac {\Gamma \left (a+n+1\right ) }{\Gamma \relax (a) }\] Then\[ \left (1+a\right ) \left (2+a\right ) \cdots \left (n+a\right ) =\frac {\Gamma \left (a+n+1\right ) }{a\Gamma \relax (a) }\] And (5) can now be written as\begin {equation} y_{r=0}\relax (x) =c_{0}\sum _{n=1}^{\infty }\frac {1}{n!}\frac {a\Gamma \relax (a) }{\Gamma \left (a+n+1\right ) }x^{n}\tag {6} \end {equation} But modified Bessel function of first kind is \[ \operatorname {BesselI}\left (a,z\right ) =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left (a+n+1\right ) }\left (\frac {z}{2}\right ) ^{2n+a}\] So if we let \(z=2\sqrt {x}\) we obtain\begin {align*} \operatorname {BesselI}\left (a,2\sqrt {x}\right ) & =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left (a+n+1\right ) }\left (\frac {2\sqrt {x}}{2}\right ) ^{2n+a}\\ & =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left (a+n+1\right ) }\left (\sqrt {x}\right ) ^{2n}\left (\sqrt {x}\right ) ^{a}\\ & =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left (a+n+1\right ) }x^{n}\left (\sqrt {x}\right ) ^{a} \end {align*}
Hence\begin {equation} \frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left (a,2\sqrt {x}\right ) =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left (a+n+1\right ) }x^{n}\tag {7} \end {equation} If we now compare (6) and (7), we see that if we set \(c_{0}\), which is arbitrary, to be \(c_{0}=\frac {1}{a\Gamma \relax (a) }\), then we obtain\begin {align*} u_{r=0}\relax (x) & =\frac {1}{a\Gamma \relax (a) }\sum _{n=0}^{\infty }\frac {1}{n!}\frac {a\Gamma \relax (a) }{\Gamma \left ( a+n+1\right ) }x^{n}\\ & =\sum _{n=0}^{\infty }\frac {1}{n!}\frac {1}{\Gamma \left (a+n+1\right ) }x^{n} \end {align*}
But this is (7). Hence we found the first solution, which is \begin {equation} u_{r=0}\relax (x) =\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) \tag {8} \end {equation}
The above was for \(r=0\). Now we find the second solution for \(r=-a\). From (4)
\[ c_{n}\ \ =\frac {c_{n-1}}{\left (n-a\right ) \left (n-a-1\right ) \ +\ \left (n-a\right ) \ +\ a\left (n-a\right ) }\]
For \(n=1\)
\[ c_{1}\ \ =\frac {c_{0}}{-a\left (1-a\right ) +\ \left (1-a\right ) \ +\ a\left (1-a\right ) }=\frac {c_{0}}{\ \left (1-a\right ) \ }\]
For \(n=2\)
\[ c_{2}\ \ =\frac {c_{1}}{\left (2-a\right ) \left (1-a\right ) \ +\ \left ( 2-a\right ) \ +\ a\left (2-a\right ) }=\frac {c_{1}}{4-2a}=\frac {c_{0}}{\ \left (1-a\right ) \ }\frac {1}{2\left (2-a\right ) }\]
For \(n=3\)
\[ c_{3}\ \ =\frac {c_{2}}{\left (3-a\right ) \left (2-a\right ) \ +\ \left ( 3-a\right ) \ +\ a\left (3-a\right ) }=\frac {c_{2}}{3\left (3-a\right ) }=\frac {c_{0}}{\ \left (1-a\right ) \ }\frac {1}{2\left (2-a\right ) }\frac {1}{3\left (3-a\right ) }\]
And so on. Since the solution is assumed to be \(x^{r}\sum _{n=0}^{\infty }c_{n}x^{n}\) then
\begin {align*} u_{r=-a} & =x^{-a}\sum _{n=0}^{\infty }c_{n}x^{n}\\ & =\sum _{n=0}^{\infty }c_{n}x^{n-a}\\ & =c_{0}x^{-a}\sum _{n=0}^{\infty }\frac {1}{\ n!}\left (\frac {1}{\left ( 1-a\right ) }\frac {1}{\left (2-a\right ) }\frac {1}{\left (3-a\right ) }\cdots \frac {1}{\left (n-a\right ) }\right ) x^{n-a} \end {align*}
But as we found above, we obtain that \(\left (1-a\right ) \left (2-a\right ) \cdots \left (n-a\right ) =\frac {\Gamma \left (-a+n+1\right ) }{-a\Gamma \left (-a\right ) }\), therefore
\[ u_{r=-a}=c_{0}\sum _{n=0}^{\infty }\frac {1}{\ n!}\frac {-a\Gamma \left ( -a\right ) }{\Gamma \left (-a+n+1\right ) }x^{n-a}\]
Modified Bessel function of second kind is \(\operatorname {BesselK}\left ( a,z\right ) =\frac {\pi }{2}\frac {1}{\sin \left (a\pi \right ) }\left ( \operatorname {BesselI}\left (-a,z\right ) -\operatorname {BesselI}\left ( a,z\right ) \right ) \). The above should result in \(\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left (a,2\sqrt {x}\right ) \) for \(z=2\sqrt {x}\) by setting \(c_{0}\) to appropriate arbitrary value. I need to work out this final manipulation later. Hence we find \(u_{r=-a}\relax (x) =\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left (a,2\sqrt {x}\right ) \). Therefore, the solution is\[ u=C_{1}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left (a,2\sqrt {x}\right ) +C_{2}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left (a,2\sqrt {x}\right ) \] But \begin {align*} \frac {d}{dx}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left (a,2\sqrt {x}\right ) & =\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) \\ \frac {d}{dx}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left (a,2\sqrt {x}\right ) & =-\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselK}\left ( 1+a,2\sqrt {x}\right ) \end {align*}
Hence\[ u^{\prime }=C_{1}\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) -C_{2}\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselK}\left (1+a,2\sqrt {x}\right ) \] And from \(y=x^{a+1}\frac {u^{\prime }}{u}\)\[ y=x^{1+a}\frac {C_{1}\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) -C_{2}\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselK}\left (1+a,2\sqrt {x}\right ) }{C_{1}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left (a,2\sqrt {x}\right ) +C_{2}\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left (a,2\sqrt {x}\right ) }\] Let \(C=\frac {C_{2}}{C_{1}}\) hence\[ y=x^{1+a}\frac {\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselI}\left ( 1+a,2\sqrt {x}\right ) -C\frac {1}{\sqrt {x^{1+a}}}\operatorname {BesselK}\left ( 1+a,2\sqrt {x}\right ) }{\frac {1}{\sqrt {x^{a}}}\operatorname {BesselI}\left ( a,2\sqrt {x}\right ) +C\frac {1}{\sqrt {x^{a}}}\operatorname {BesselK}\left ( a,2\sqrt {x}\right ) }\] Or\begin {align*} y & =x^{1+a}\frac {x^{-\frac {1}{2}}\operatorname {BesselI}\left (1+a,2\sqrt {x}\right ) -Cx^{-\frac {1}{2}}\operatorname {BesselK}\left (1+a,2\sqrt {x}\right ) }{\operatorname {BesselI}\left (a,2\sqrt {x}\right ) +C\operatorname {BesselK}\left (a,2\sqrt {x}\right ) }\\ & =\frac {x^{\frac {1}{2}+a}\operatorname {BesselI}\left (1+a,2\sqrt {x}\right ) -Cx^{\frac {1}{2}+a}\operatorname {BesselK}\left (1+a,2\sqrt {x}\right ) }{\operatorname {BesselI}\left (a,2\sqrt {x}\right ) +C\operatorname {BesselK}\left (a,2\sqrt {x}\right ) } \end {align*}
Verification