Internal problem ID [3333]
Internal file name [OUTPUT/2825_Sunday_June_05_2022_08_41_12_AM_18465556/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 3
Problem number: 69.
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]
\[ \boxed {y^{\prime }-x^{n} y^{2} b=a \,x^{m}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,x^{m}+x^{n} b \,y^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,x^{m}+x^{n} b \,y^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \,x^{m}\), \(f_1(x)=0\) and \(f_2(x)=b \,x^{n}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{b \,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 {b \,x^{n} n}{x}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=b^{2} x^{2 n} a \,x^{m} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} b \,x^{n} u^{\prime \prime }\left (x \right )-\frac {b \,x^{n} n u^{\prime }\left (x \right )}{x}+b^{2} x^{2 n} a \,x^{m} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {BesselY}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{1} \right ) x^{\frac {n}{2}+\frac {1}{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = x^{\frac {1}{2}+n +\frac {m}{2}} \sqrt {a b}\, \left (-\operatorname {BesselY}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{2} -\operatorname {BesselJ}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{1} \right ) \] Using the above in (1) gives the solution \[ y = -\frac {x^{\frac {1}{2}+n +\frac {m}{2}} \sqrt {a b}\, \left (-\operatorname {BesselY}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{2} -\operatorname {BesselJ}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{1} \right ) x^{-n} x^{-\frac {n}{2}-\frac {1}{2}}}{b \left (\operatorname {BesselY}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{2} +\operatorname {BesselJ}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{1} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x^{\frac {m}{2}-\frac {n}{2}} \sqrt {a b}\, \left (\operatorname {BesselJ}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{3} +\operatorname {BesselY}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right )\right )}{b \left (\operatorname {BesselY}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right )+\operatorname {BesselJ}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{\frac {m}{2}-\frac {n}{2}} \sqrt {a b}\, \left (\operatorname {BesselJ}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{3} +\operatorname {BesselY}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right )\right )}{b \left (\operatorname {BesselY}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right )+\operatorname {BesselJ}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{\frac {m}{2}-\frac {n}{2}} \sqrt {a b}\, \left (\operatorname {BesselJ}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{3} +\operatorname {BesselY}\left (\frac {1+m}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right )\right )}{b \left (\operatorname {BesselY}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right )+\operatorname {BesselJ}\left (\frac {-n -1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-x^{n} y^{2} b =a \,x^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,x^{m}+x^{n} y^{2} b \end {array} \]
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-b*x^n*a*x^m*y(x), y(x)` *** Sublevel 2 *** 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 <- Bessel successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 170
dsolve(diff(y(x),x) = a*x^m+b*x^n*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {x^{-\frac {n}{2}+\frac {m}{2}} \sqrt {a b}\, \left (\operatorname {BesselY}\left (\frac {m +1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {m +1}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right )\right )}{b \left (\operatorname {BesselY}\left (\frac {-1-n}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {-1-n}{m +n +2}, \frac {2 \sqrt {a b}\, x^{\frac {m}{2}+\frac {n}{2}+1}}{m +n +2}\right )\right )} \]
✓ Solution by Mathematica
Time used: 1.822 (sec). Leaf size: 1805
DSolve[y'[x]==a x^m+ b x^n y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {a^{-\frac {n+1}{2 (m+n+2)}} b^{-\frac {2 m+3 n+5}{2 (m+n+2)}} (m+n+1)^{\frac {n+1}{m+n+2}} \left ((m+n+1)^2\right )^{\frac {n+1}{m+n+2}-\frac {1}{2}} x^{-n-1} \left (x^{m+n+1}\right )^{-\frac {n+1}{2 (m+n+1)}} \left (a^{\frac {n+1}{2 (m+n+2)}} b^{\frac {n+1}{2 (m+n+2)}} (m+n+1)^{-\frac {n+1}{m+n+2}} (m+n+2) \left (-\sqrt {a} \sqrt {b} (m+n+1) \operatorname {BesselJ}\left (\frac {m+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right ) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}+\sqrt {a} \sqrt {b} (m+n+1) \operatorname {BesselJ}\left (-\frac {m+2 n+3}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right ) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}+(n+1) \sqrt {(m+n+1)^2} \operatorname {BesselJ}\left (-\frac {n+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )\right ) c_1 \operatorname {Gamma}\left (\frac {m+1}{m+n+2}\right ) \left (x^{m+n+1}\right )^{\frac {n+1}{2 (m+n+1)}}+a^{\frac {n+1}{2 (m+n+2)}} b^{\frac {n+1}{2 (m+n+2)}} (n+1)^2 (m+n+1)^{\frac {n+1}{m+n+2}} \left ((m+n+1)^2\right )^{\frac {m-n}{2 (m+n+2)}} \operatorname {BesselJ}\left (\frac {n+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right ) \operatorname {Gamma}\left (\frac {n+1}{m+n+2}\right ) \left (x^{m+n+1}\right )^{\frac {n+1}{2 (m+n+1)}}+a^{\frac {m+2 n+3}{2 (m+n+2)}} b^{\frac {m+2 n+3}{2 (m+n+2)}} (n+1) (m+n+1)^{\frac {m+2 n+3}{m+n+2}} \left ((m+n+1)^2\right )^{-\frac {n+1}{m+n+2}} \left (\operatorname {BesselJ}\left (-\frac {m+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )-\operatorname {BesselJ}\left (\frac {n+1}{m+n+2}+1,\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )\right ) \operatorname {Gamma}\left (\frac {n+1}{m+n+2}\right ) \left (x^{m+n+1}\right )^{\frac {m+2 n+3}{2 (m+n+1)}}\right )}{2 \left ((m+n+2) \operatorname {BesselJ}\left (-\frac {n+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right ) c_1 \operatorname {Gamma}\left (\frac {m+1}{m+n+2}\right ) \left ((m+n+1)^2\right )^{\frac {n+1}{m+n+2}}+(n+1) (m+n+1)^{\frac {2 (n+1)}{m+n+2}} \operatorname {BesselJ}\left (\frac {n+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right ) \operatorname {Gamma}\left (\frac {n+1}{m+n+2}\right )\right )} \\ y(x)\to \frac {x^{-n-1} \left (\sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (\frac {1}{m+n+1}+1\right )} \operatorname {BesselJ}\left (\frac {m+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )-\sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (\frac {1}{m+n+1}+1\right )} \operatorname {BesselJ}\left (-\frac {m+2 n+3}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )-(n+1) \sqrt {(m+n+1)^2} \operatorname {BesselJ}\left (-\frac {n+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )\right )}{2 b \sqrt {(m+n+1)^2} \operatorname {BesselJ}\left (-\frac {n+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )} \\ y(x)\to \frac {x^{-n-1} \left (\sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (\frac {1}{m+n+1}+1\right )} \operatorname {BesselJ}\left (\frac {m+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )-\sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (\frac {1}{m+n+1}+1\right )} \operatorname {BesselJ}\left (-\frac {m+2 n+3}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )-(n+1) \sqrt {(m+n+1)^2} \operatorname {BesselJ}\left (-\frac {n+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )\right )}{2 b \sqrt {(m+n+1)^2} \operatorname {BesselJ}\left (-\frac {n+1}{m+n+2},\frac {2 \sqrt {a} \sqrt {b} (m+n+1) \left (x^{m+n+1}\right )^{\frac {1}{2} \left (1+\frac {1}{m+n+1}\right )}}{\sqrt {(m+n+1)^2} (m+n+2)}\right )} \\ \end{align*}