Internal problem ID [3638]
Internal file name [OUTPUT/3131_Sunday_June_05_2022_08_53_07_AM_31378124/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 14
Problem number: 382.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {x^{n} y^{\prime }+y^{2}=x^{2 n -1}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \left (x^{2 n -1}-y^{2}\right ) x^{-n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {x^{n}}{x}-x^{-n} y^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x^{2 n -1} x^{-n}\), \(f_1(x)=0\) and \(f_2(x)=-x^{-n}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-x^{-n} u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=\frac {x^{-n} n}{x}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=x^{-3 n} x^{2 n -1} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -x^{-n} u^{\prime \prime }\left (x \right )-\frac {x^{-n} n u^{\prime }\left (x \right )}{x}+x^{-3 n} x^{2 n -1} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{-\frac {n}{2}} \left (\sqrt {x}\, \operatorname {BesselI}\left (n +1, 2 \sqrt {x}\right ) c_{1} +\sqrt {x}\, \operatorname {BesselK}\left (n +1, 2 \sqrt {x}\right ) c_{2} +n \left (c_{1} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-c_{2} \operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = x^{-\frac {n}{2}} \left (c_{1} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-c_{2} \operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right ) \] Using the above in (1) gives the solution \[ y = \frac {\left (c_{1} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-c_{2} \operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right ) x^{n}}{\sqrt {x}\, \operatorname {BesselI}\left (n +1, 2 \sqrt {x}\right ) c_{1} +\sqrt {x}\, \operatorname {BesselK}\left (n +1, 2 \sqrt {x}\right ) c_{2} +n \left (c_{1} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-c_{2} \operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (c_{3} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-\operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right ) x^{n}}{\sqrt {x}\, \operatorname {BesselI}\left (n +1, 2 \sqrt {x}\right ) c_{3} +\operatorname {BesselK}\left (n +1, 2 \sqrt {x}\right ) \sqrt {x}+n \left (c_{3} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-\operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-\operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right ) x^{n}}{\sqrt {x}\, \operatorname {BesselI}\left (n +1, 2 \sqrt {x}\right ) c_{3} +\operatorname {BesselK}\left (n +1, 2 \sqrt {x}\right ) \sqrt {x}+n \left (c_{3} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-\operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-\operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right ) x^{n}}{\sqrt {x}\, \operatorname {BesselI}\left (n +1, 2 \sqrt {x}\right ) c_{3} +\operatorname {BesselK}\left (n +1, 2 \sqrt {x}\right ) \sqrt {x}+n \left (c_{3} \operatorname {BesselI}\left (n , 2 \sqrt {x}\right )-\operatorname {BesselK}\left (n , 2 \sqrt {x}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{n} y^{\prime }+y^{2}=x^{2 n -1} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{2 n -1}-y^{2}}{x^{n}} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -n*(diff(y(x), x))/x+x^(-n)*x^(n-1)*y(x), y(x)` *** Sublevel 2 ** Methods for second order ODEs: --- Trying classification methods --- trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel <- Bessel successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 80
dsolve(x^n*diff(y(x),x) = x^(2*n-1)-y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\operatorname {BesselK}\left (n , 2 \sqrt {x}\right ) c_{1} -\operatorname {BesselI}\left (n , 2 \sqrt {x}\right )\right ) x^{n}}{-\operatorname {BesselI}\left (n +1, 2 \sqrt {x}\right ) \sqrt {x}-\sqrt {x}\, \operatorname {BesselK}\left (n +1, 2 \sqrt {x}\right ) c_{1} +n \left (\operatorname {BesselK}\left (n , 2 \sqrt {x}\right ) c_{1} -\operatorname {BesselI}\left (n , 2 \sqrt {x}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.366 (sec). Leaf size: 293
DSolve[x^n y'[x]==x^(2 n -1)-y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^{n-1} \left (-\left ((n-1) \operatorname {Gamma}(2-n) \operatorname {BesselI}\left (1-n,2 \sqrt {x}\right )\right )+\sqrt {x} \operatorname {Gamma}(2-n) \operatorname {BesselI}\left (2-n,2 \sqrt {x}\right )+\sqrt {x} \operatorname {Gamma}(2-n) \operatorname {BesselI}\left (-n,2 \sqrt {x}\right )-c_1 (-1)^n \sqrt {x} \operatorname {Gamma}(n) \operatorname {BesselI}\left (n-2,2 \sqrt {x}\right )-c_1 (-1)^n \operatorname {Gamma}(n) \operatorname {BesselI}\left (n-1,2 \sqrt {x}\right )+c_1 (-1)^n n \operatorname {Gamma}(n) \operatorname {BesselI}\left (n-1,2 \sqrt {x}\right )-c_1 (-1)^n \sqrt {x} \operatorname {Gamma}(n) \operatorname {BesselI}\left (n,2 \sqrt {x}\right )\right )}{2 \left (\operatorname {Gamma}(2-n) \operatorname {BesselI}\left (1-n,2 \sqrt {x}\right )-c_1 (-1)^n \operatorname {Gamma}(n) \operatorname {BesselI}\left (n-1,2 \sqrt {x}\right )\right )} \\ y(x)\to \frac {x^{n-1} \left (\sqrt {x} \operatorname {BesselI}\left (n-2,2 \sqrt {x}\right )-(n-1) \operatorname {BesselI}\left (n-1,2 \sqrt {x}\right )+\sqrt {x} \operatorname {BesselI}\left (n,2 \sqrt {x}\right )\right )}{2 \operatorname {BesselI}\left (n-1,2 \sqrt {x}\right )} \\ \end{align*}