2.30   ODE No. 30

1. Problem in Latex
2. Mathematica input
3. Maple input

$$x^{-a-1}y(x{)}^2-x^a+y^{\prime }(x)=0$$ ✓ Mathematica : cpu = 0.289385 (sec), leaf count = 197

$$\left\{\{y(x)\to \frac{x^a((-1{)}^a{c}_1\sqrt{x}\mathrm{\Gamma }(a+1)I_{a-1}\left(2\sqrt{x}\right)+(-1{)}^{a+1}a{c}_1\mathrm{\Gamma }(a+1)I_a\left(2\sqrt{x}\right)+(-1{)}^a{c}_1\sqrt{x}\mathrm{\Gamma }(a+1)I_{a+1}\left(2\sqrt{x}\right)+\sqrt{x}\mathrm{\Gamma }(1-a)I_{-a-1}\left(2\sqrt{x}\right)+\sqrt{x}\mathrm{\Gamma }(1-a)I_{1-a}\left(2\sqrt{x}\right)-a\mathrm{\Gamma }(1-a)I_{-a}\left(2\sqrt{x}\right))}{2((-1{)}^a{c}_1\mathrm{\Gamma }(a+1)I_a\left(2\sqrt{x}\right)+\mathrm{\Gamma }(1-a)I_{-a}\left(2\sqrt{x}\right))}\}\right\}$$

✓ Maple : cpu = 0.097 (sec), leaf count = 54

$$\{y\left(x\right)=x^{a+1}(-{\mathit{K}}_{a+1}\left(2\sqrt{x}\right)\mathit{\_}\mathit{C}\mathit{1}+{\mathit{I}}_{a+1}\left(2\sqrt{x}\right))\frac{1}{\sqrt{x}}{({\mathit{K}}_a\left(2\sqrt{x}\right)\mathit{\_}\mathit{C}\mathit{1}+{\mathit{I}}_a\left(2\sqrt{x}\right))}^{-1}\}$$

Hand solution

$$\begin{array}{rl}{y}^{\prime }+x^{-a-1}{y}^2-x^a& =0\\ y^{\prime }& =x^a-x^{-a-1}{y}^2\\ \text{(1)}& & =P\left(x\right)+Q\left(x\right)y+R\left(x\right)y^2\end{array}$$

This is Ricatti first order non-linear ODE. Using standard transformation$$y=-\frac{u^{\prime }}{uR\left(x\right)}=x^{a+1}\frac{u^{\prime }}{u}$$

Hence

$$y^{\prime }=(a+1)x^a\frac{u^{\prime }}{u}+x^{a+1}\frac{u^{\prime \prime }}{u}-x^{a+1}\frac{{\left(u^{\prime }\right)}^2}{u^2}$$

Comparing to (1) gives

$$\begin{array}{rl}{x}^a-x^{-a-1}{y}^2& =(a+1)x^a\frac{u^{\prime }}{u}+x^{a+1}\frac{u^{\prime \prime }}{u}-x^{a+1}\frac{{\left(u^{\prime }\right)}^2}{u^2}\\ x^a-x^{-a-1}{\left(x^{a+1}\frac{u^{\prime }}{u}\right)}^2& =(a+1)x^a\frac{u^{\prime }}{u}+x^{a+1}\frac{u^{\prime \prime }}{u}-x^{a+1}\frac{{\left(u^{\prime }\right)}^2}{u^2}\\ 1-\frac{x^{-a-1}}{x^a}{x}^{2a+2}\frac{{\left(u^{\prime }\right)}^2}{u^2}& =(a+1)\frac{u^{\prime }}{u}+x\frac{u^{\prime \prime }}{u}-x\frac{{\left(u^{\prime }\right)}^2}{u^2}\\ 1-x\frac{{\left(u^{\prime }\right)}^2}{u^2}& =(a+1)\frac{u^{\prime }}{u}+x\frac{u^{\prime \prime }}{u}-x\frac{{\left(u^{\prime }\right)}^2}{u^2}\\ 1& =(a+1)\frac{u^{\prime }}{u}+x\frac{u^{\prime \prime }}{u}\\ \text{(2)}& x{u}^{\prime \prime }+(1+a)u^{\prime }-u& =0\end{array}$$

In standard form $u^{\prime \prime }+\frac{1}{x}(1+a)u^{\prime }-\frac{1}{x}u=0$ or $u^{\prime \prime }+p\left(x\right)(1+a)u^{\prime }+q\left(x\right)u=0$. We see that $p\left(x\right)$ 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 $u=\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n$ is used when the expansion point is not singular point. (i.e. $p\left(x\right)$ and $q\left(x\right)$ are analytic there). Forbenius series $u=x^r\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n$ is used when there is a removable singular point (called also regular singular point), as in this case. Starting with$$u=x^r\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n=\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^{n+r}$$ Hence $$\begin{array}{rl}{u}^{\prime }& =\sum _{n=0}^{\mathrm{\infty }}(n+r)c_n{x}^{n+r-1}\\ u^{\prime \prime }& =\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)c_n{x}^{n+r-2}\end{array}$$

Substituting in (2) gives$$\begin{array}{rl}x\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)c_n{x}^{n+r-2}+(1+a)\sum _{n=0}^{\mathrm{\infty }}(n+r)c_n{x}^{n+r-1}-\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^{n+r}& =0\\ \sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)c_n{x}^{n+r-1}+(1+a)\sum _{n=0}^{\mathrm{\infty }}(n+r)c_n{x}^{n+r-1}-\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^{n+r}& =0\end{array}$$

Dividing out $x^r$$$\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)c_n{x}^{n-1}+(1+a)\sum _{n=0}^{\mathrm{\infty }}(n+r)c_n{x}^{n-1}-\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n=0$$ Each term should have $x^{n-1}$ in it. So we adjust the last term$$\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)c_n{x}^{n-1}+(1+a)\sum _{n=0}^{\mathrm{\infty }}(n+r)c_n{x}^{n-1}-\sum _{n=1}^{\mathrm{\infty }}{c}_{n-1}{x}^{n-1}=0$$ Expanding the second term$$\sum _{n=0}^{\mathrm{\infty }}(n+r)(n+r-1)c_n{x}^{n-1}+\sum _{n=0}^{\mathrm{\infty }}(n+r)c_n{x}^{n-1}+\sum _{n=0}^{\mathrm{\infty }}a(n+r)c_n{x}^{n-1}-\sum _{n=1}^{\mathrm{\infty }}{c}_{n-1}{x}^{n-1}=0$$ Hence for $n=0$$$\begin{array}{rl}(n+r)(n+r-1)c_n{x}^{n-1}+(n+r)c_n{x}^{n-1}+a(n+r)c_n{x}^{n-1}& =0\\ r(r-1)c_0+r{c}_0+ar{c}_0& =0\end{array}$$

Since $c_0\ne 0$ then$$r(r-1)+r+ar=0$$ Hence $r=$ $-a$ or $r=0$. Now for $n\ge 1$$$\begin{array}{rl}(n+r)(n+r-1)c_n{x}^{n-1}+(n+r)c_n{x}^{n-1}+a(n+r)c_n{x}^{n-1}-c_{n-1}{x}^{n-1}& =0\\ (n+r)(n+r-1)c_n+(n+r)c_n+a(n+r)c_n-c_{n-1}& =0\\ ((n+r)(n+r-1)+(n+r)+a(n+r))c_n& =c_{n-1}\\ c_n& =\frac{c_{n-1}}{(n+r)(n+r-1)+(n+r)+a(n+r)}\end{array}$$

For $r=0$, we obtain$$\begin{array}{}\text{(3)}& c_n=\frac{c_{n-1}}{n(n-1)+n+an}\end{array}$$ For $r=-a$$$\begin{array}{}\text{(4)}& c_n=\frac{c_{n-1}}{(n-a)(n-a-1)+(n-a)+a(n-a)}\end{array}$$ There are two solutions. Looking at (3) for now, for $n=1$$$c_1=\frac{c_0}{1+a}$$ For $n=2$$$c_2=\frac{c_1}{4+2a}=\frac{c_0}{1+a}\frac{1}{2(2+a)}$$ For $n=3$$$c_3=\frac{c_2}{3\left(2\right)+3+3a}=\frac{c_2}{3(3+a)}=\frac{c_0}{1+a}\frac{1}{2(2+a)}\frac{1}{3(3+a)}$$ And so on. Since the solution is assumed to be $x^r\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n$ and we are looking at case $r=0$ then$$\begin{array}{rl}{u}_{r=0}\left(x\right)& =\sum _{n=1}^{\mathrm{\infty }}{c}_n{x}^n\\ & =c_0+c_{1}x+c_2{x}^2+\cdots \\ & =c_0{x}^0+\frac{c_0}{1+a}x+\frac{c_0}{1+a}\frac{1}{2(2+a)}{x}^2+\frac{c_0}{1+a}\frac{1}{2(2+a)}\frac{1}{3(3+a)}{x}^3+\cdots \\ \text{(5)}& & =c_0(x^0+\frac{1}{1+a}x+\frac{1}{(1+a)}\frac{1}{2(2+a)}{x}^2+\frac{1}{(1+a)}\frac{1}{2(2+a)}\frac{1}{3(3+a)}{x}^3+\cdots )\end{array}$$

Since $$\mathrm{\Gamma }\left(n\right)=(n-1)!$$ and $$a(1+a)(2+a)\cdots (n+a)=\frac{\mathrm{\Gamma }(a+n+1)}{\mathrm{\Gamma }\left(a\right)}$$ Then$$(1+a)(2+a)\cdots (n+a)=\frac{\mathrm{\Gamma }(a+n+1)}{a\mathrm{\Gamma }\left(a\right)}$$ And (5) can now be written as$$\begin{array}{}\text{(6)}& y_{r=0}\left(x\right)=c_0\sum _{n=1}^{\mathrm{\infty }}\frac{1}{n!}\frac{a\mathrm{\Gamma }\left(a\right)}{\mathrm{\Gamma }(a+n+1)}{x}^n\end{array}$$ But modified Bessel function of first kind is $$\mathrm{BesselI}(a,z)=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\frac{1}{\mathrm{\Gamma }(a+n+1)}{\left(\frac{z}{2}\right)}^{2n+a}$$ So if we let $z=2\sqrt{x}$ we obtain$$\begin{array}{rl}\mathrm{BesselI}(a,2\sqrt{x})& =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\frac{1}{\mathrm{\Gamma }(a+n+1)}{\left(\frac{2\sqrt{x}}{2}\right)}^{2n+a}\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\frac{1}{\mathrm{\Gamma }(a+n+1)}{\left(\sqrt{x}\right)}^{2n}{\left(\sqrt{x}\right)}^a\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\frac{1}{\mathrm{\Gamma }(a+n+1)}{x}^n{\left(\sqrt{x}\right)}^a\end{array}$$

Hence$$\begin{array}{}\text{(7)}& \frac{1}{\sqrt{x^a}}\mathrm{BesselI}(a,2\sqrt{x})=\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\frac{1}{\mathrm{\Gamma }(a+n+1)}{x}^n\end{array}$$ If we now compare (6) and (7), we see that if we set $c_0$, which is arbitrary, to be $c_0=\frac{1}{a\mathrm{\Gamma }\left(a\right)}$, then we obtain$$\begin{array}{rl}{u}_{r=0}\left(x\right)& =\frac{1}{a\mathrm{\Gamma }\left(a\right)}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\frac{a\mathrm{\Gamma }\left(a\right)}{\mathrm{\Gamma }(a+n+1)}{x}^n\\ & =\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\frac{1}{\mathrm{\Gamma }(a+n+1)}{x}^n\end{array}$$

But this is (7). Hence we found the first solution, which is $$\begin{array}{}\text{(8)}& u_{r=0}\left(x\right)=\frac{1}{\sqrt{x^a}}\mathrm{BesselI}(a,2\sqrt{x})\end{array}$$

The above was for  $r=0$. Now we find the second solution for $r=-a$. From (4)

$$c_n=\frac{c_{n-1}}{(n-a)(n-a-1)+(n-a)+a(n-a)}$$

For $n=1$

$$c_1=\frac{c_0}{-a(1-a)+(1-a)+a(1-a)}=\frac{c_0}{(1-a)}$$

For $n=2$

$$c_2=\frac{c_1}{(2-a)(1-a)+(2-a)+a(2-a)}=\frac{c_1}{4-2a}=\frac{c_0}{(1-a)}\frac{1}{2(2-a)}$$

For $n=3$

$$c_3=\frac{c_2}{(3-a)(2-a)+(3-a)+a(3-a)}=\frac{c_2}{3(3-a)}=\frac{c_0}{(1-a)}\frac{1}{2(2-a)}\frac{1}{3(3-a)}$$

And so on. Since the solution is assumed to be $x^r\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n$ then

$$\begin{array}{rl}{u}_{r=-a}& =x^{-a}\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^n\\ & =\sum _{n=0}^{\mathrm{\infty }}{c}_n{x}^{n-a}\\ & =c_0{x}^{-a}\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}(\frac{1}{(1-a)}\frac{1}{(2-a)}\frac{1}{(3-a)}\cdots \frac{1}{(n-a)})x^{n-a}\end{array}$$

But as we found above, we obtain that $(1-a)(2-a)\cdots (n-a)=\frac{\mathrm{\Gamma }(-a+n+1)}{-a\mathrm{\Gamma }(-a)}$, therefore

$$u_{r=-a}=c_0\sum _{n=0}^{\mathrm{\infty }}\frac{1}{n!}\frac{-a\mathrm{\Gamma }(-a)}{\mathrm{\Gamma }(-a+n+1)}{x}^{n-a}$$

Modified Bessel function of second kind is $\mathrm{BesselK}(a,z)=\frac{\pi }{2}\frac{1}{\sin \left(a\pi \right)}(\mathrm{BesselI}(-a,z)-\mathrm{BesselI}(a,z))$. The above should result in $\frac{1}{\sqrt{x^a}}\mathrm{BesselK}(a,2\sqrt{x})$ 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 $u_{r=-a}\left(x\right)=\frac{1}{\sqrt{x^a}}\mathrm{BesselK}(a,2\sqrt{x})$. Therefore, the solution is$$u=C_1\frac{1}{\sqrt{x^a}}\mathrm{BesselI}(a,2\sqrt{x})+C_2\frac{1}{\sqrt{x^a}}\mathrm{BesselK}(a,2\sqrt{x})$$ But $$\begin{array}{rl}\frac{d}{dx}\frac{1}{\sqrt{x^a}}\mathrm{BesselI}(a,2\sqrt{x})& =\frac{1}{\sqrt{x^{1+a}}}\mathrm{BesselI}(1+a,2\sqrt{x})\\ \frac{d}{dx}\frac{1}{\sqrt{x^a}}\mathrm{BesselI}(a,2\sqrt{x})& =-\frac{1}{\sqrt{x^{1+a}}}\mathrm{BesselK}(1+a,2\sqrt{x})\end{array}$$

Hence$$u^{\prime }=C_1\frac{1}{\sqrt{x^{1+a}}}\mathrm{BesselI}(1+a,2\sqrt{x})-C_2\frac{1}{\sqrt{x^{1+a}}}\mathrm{BesselK}(1+a,2\sqrt{x})$$ And from $y=x^{a+1}\frac{u^{\prime }}{u}$$$y=x^{1+a}\frac{C_1\frac{1}{\sqrt{x^{1+a}}}\mathrm{BesselI}(1+a,2\sqrt{x})-C_2\frac{1}{\sqrt{x^{1+a}}}\mathrm{BesselK}(1+a,2\sqrt{x})}{C_1\frac{1}{\sqrt{x^a}}\mathrm{BesselI}(a,2\sqrt{x})+C_2\frac{1}{\sqrt{x^a}}\mathrm{BesselK}(a,2\sqrt{x})}$$ Let $C=\frac{C_2}{C_1}$ hence$$y=x^{1+a}\frac{\frac{1}{\sqrt{x^{1+a}}}\mathrm{BesselI}(1+a,2\sqrt{x})-C\frac{1}{\sqrt{x^{1+a}}}\mathrm{BesselK}(1+a,2\sqrt{x})}{\frac{1}{\sqrt{x^a}}\mathrm{BesselI}(a,2\sqrt{x})+C\frac{1}{\sqrt{x^a}}\mathrm{BesselK}(a,2\sqrt{x})}$$ Or$$\begin{array}{rl}y& =x^{1+a}\frac{x^{-\frac{1}{2}}\mathrm{BesselI}(1+a,2\sqrt{x})-C{x}^{-\frac{1}{2}}\mathrm{BesselK}(1+a,2\sqrt{x})}{\mathrm{BesselI}(a,2\sqrt{x})+C\mathrm{BesselK}(a,2\sqrt{x})}\\ & =\frac{x^{\frac{1}{2}+a}\mathrm{BesselI}(1+a,2\sqrt{x})-C{x}^{\frac{1}{2}+a}\mathrm{BesselK}(1+a,2\sqrt{x})}{\mathrm{BesselI}(a,2\sqrt{x})+C\mathrm{BesselK}(a,2\sqrt{x})}\end{array}$$

Verification

eq:=diff(y(x),x)+x^(-a-1)*y(x)^2-x^a = 0; 
num:=x^(1/2+a)*BesselI(1+a,2*sqrt(x))-_C1*x^(1/2+a)*BesselK(1+a,2*sqrt(x)); 
den:=BesselI(a,2*sqrt(x))+_C1*BesselK(a,2*sqrt(x)); 
my_sol:=num/den; 
odetest(y(x)=my_sol,eq); 
0