Internal problem ID [3639]
Internal file name [OUTPUT/3132_Sunday_June_05_2022_08_53_09_AM_87430216/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 14
Problem number: 384.
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]
\[ \boxed {x^{n} y^{\prime }+y^{2}=-x^{-2+2 n}-\left (-n +1\right ) x^{n -1}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \left (n \,x^{n -1}-y^{2}-x^{-2+2 n}-x^{n -1}\right ) x^{-n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {n}{x}-x^{-n} y^{2}-\frac {x^{n}}{x^{2}}-\frac {1}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\left (n \,x^{n -1}-x^{n -1}-x^{-2+2 n}\right ) x^{-n}\), \(f_1(x)=0\) 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} \left (n \,x^{n -1}-x^{n -1}-x^{-2+2 n}\right ) \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} \left (n \,x^{n -1}-x^{n -1}-x^{-2+2 n}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{-\frac {n}{2}} \sqrt {x}\, \left (c_{1} \operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}} \sqrt {x}}{\sqrt {-n +1}}\right )+c_{2} \operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}} \sqrt {x}}{\sqrt {-n +1}}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (\left (n -1\right ) \left (\operatorname {BesselJ}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{1} +\operatorname {BesselY}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{2} \right ) x^{-\frac {n}{2}+\frac {1}{2}}+\frac {\left (\operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{2} +\operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{1} \right ) \left (\sqrt {n^{2}-2 n -3}-n +1\right ) \sqrt {-n +1}}{2}\right ) x^{-\frac {n}{2}-\frac {1}{2}}}{\sqrt {-n +1}} \] Using the above in (1) gives the solution \[ y = \frac {\left (\left (n -1\right ) \left (\operatorname {BesselJ}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{1} +\operatorname {BesselY}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{2} \right ) x^{-\frac {n}{2}+\frac {1}{2}}+\frac {\left (\operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{2} +\operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{1} \right ) \left (\sqrt {n^{2}-2 n -3}-n +1\right ) \sqrt {-n +1}}{2}\right ) x^{-\frac {n}{2}-\frac {1}{2}} x^{n} x^{\frac {n}{2}}}{\sqrt {-n +1}\, \sqrt {x}\, \left (c_{1} \operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}} \sqrt {x}}{\sqrt {-n +1}}\right )+c_{2} \operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}} \sqrt {x}}{\sqrt {-n +1}}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x^{n -1} \left (2 \left (n -1\right ) \left (\operatorname {BesselJ}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{3} +\operatorname {BesselY}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right )\right ) x^{-\frac {n}{2}+\frac {1}{2}}+\left (\operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right )+\operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{3} \right ) \left (\sqrt {n^{2}-2 n -3}-n +1\right ) \sqrt {-n +1}\right )}{\sqrt {-n +1}\, \left (2 \operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{3} +2 \operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{n -1} \left (2 \left (n -1\right ) \left (\operatorname {BesselJ}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{3} +\operatorname {BesselY}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right )\right ) x^{-\frac {n}{2}+\frac {1}{2}}+\left (\operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right )+\operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{3} \right ) \left (\sqrt {n^{2}-2 n -3}-n +1\right ) \sqrt {-n +1}\right )}{\sqrt {-n +1}\, \left (2 \operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{3} +2 \operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{n -1} \left (2 \left (n -1\right ) \left (\operatorname {BesselJ}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{3} +\operatorname {BesselY}\left (\frac {-\sqrt {n^{2}-2 n -3}+n -1}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right )\right ) x^{-\frac {n}{2}+\frac {1}{2}}+\left (\operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right )+\operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{3} \right ) \left (\sqrt {n^{2}-2 n -3}-n +1\right ) \sqrt {-n +1}\right )}{\sqrt {-n +1}\, \left (2 \operatorname {BesselJ}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right ) c_{3} +2 \operatorname {BesselY}\left (-\frac {\sqrt {n^{2}-2 n -3}}{n -1}, \frac {2 x^{-\frac {n}{2}+\frac {1}{2}}}{\sqrt {-n +1}}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{n} y^{\prime }+y^{2}=-x^{-2+2 n}-\left (-n +1\right ) x^{n -1} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {x^{-2+2 n}+y^{2}+\left (-n +1\right ) x^{n -1}}{x^{n}} \end {array} \]
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-2)*x-n+1)*y(x)/x, y(x)` *** Sub Methods for second order ODEs: --- Trying classification methods --- trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying an equivalence, under non-integer power transformations, to LODEs admitting Liouvillian solutions. -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> Trying a solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 0F1 ODE <- hypergeometric successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 1225
dsolve(x^n*diff(y(x),x)+x^(2*n-2)+y(x)^2+(1-n)*x^(n-1) = 0,y(x), singsol=all)
\[ \text {Expression too large to display} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[x^n*y'[x]+x^(2*n-2)+y[x]^2+(1-n)*x^(n-1)==0,y[x],x,IncludeSingularSolutions -> True]
Not solved