1.107 problem 107

1.107.1 Solving as riccati ode
1.107.2 Maple step by step solution

Internal problem ID [8444]
Internal file name [OUTPUT/7377_Sunday_June_05_2022_10_53_55_PM_74397981/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 107.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program : "riccati"

Maple gives the following as the ode type

[_rational, _Riccati]

\[ \boxed {x y^{\prime }+a \,x^{\alpha } y^{2}+y b=c \,x^{\beta }} \]

1.107.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {a \,x^{\alpha } y^{2}-c \,x^{\beta }+b y}{x} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {a \,x^{\alpha } y^{2}}{x}+\frac {c \,x^{\beta }}{x}-\frac {b y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {c \,x^{\beta }}{x}\), \(f_1(x)=-\frac {b}{x}\) and \(f_2(x)=-\frac {a \,x^{\alpha }}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {a \,x^{\alpha } u}{x}} \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 {a \,x^{\alpha }}{x^{2}}-\frac {a \,x^{\alpha } \alpha }{x^{2}}\\ f_1 f_2 &=\frac {b a \,x^{\alpha }}{x^{2}}\\ f_2^2 f_0 &=\frac {a^{2} x^{2 \alpha } c \,x^{\beta }}{x^{3}} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} -\frac {a \,x^{\alpha } u^{\prime \prime }\left (x \right )}{x}-\left (\frac {a \,x^{\alpha }}{x^{2}}-\frac {a \,x^{\alpha } \alpha }{x^{2}}+\frac {b a \,x^{\alpha }}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {a^{2} x^{2 \alpha } c \,x^{\beta } u \left (x \right )}{x^{3}} &=0 \end {align*}

Solving the above ODE (this ode solved using Maple, not this program), gives

\[ u \left (x \right ) = -\frac {\left (x^{\frac {\alpha }{2}+\frac {\beta }{2}} \left (\operatorname {BesselJ}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{1} +\operatorname {BesselY}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{2} \right ) \sqrt {-a c}-\left (b +\beta \right ) \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{1} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{2} \right )\right ) x^{-\frac {b}{2}-\frac {\beta }{2}}}{\sqrt {-a c}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {a c \,x^{\frac {\beta }{2}-1+\alpha -\frac {b}{2}} \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{1} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{2} \right )}{\sqrt {-a c}} \] Using the above in (1) gives the solution \[ y = -\frac {c \,x^{\frac {\beta }{2}-1+\alpha -\frac {b}{2}} \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{1} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{2} \right ) x \,x^{-\alpha } x^{\frac {b}{2}+\frac {\beta }{2}}}{x^{\frac {\alpha }{2}+\frac {\beta }{2}} \left (\operatorname {BesselJ}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{1} +\operatorname {BesselY}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{2} \right ) \sqrt {-a c}-\left (b +\beta \right ) \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{1} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = -\frac {c \,x^{\beta } \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right )}{x^{\frac {\alpha }{2}+\frac {\beta }{2}} \left (\operatorname {BesselJ}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right ) \sqrt {-a c}-\left (b +\beta \right ) \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {c \,x^{\beta } \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right )}{x^{\frac {\alpha }{2}+\frac {\beta }{2}} \left (\operatorname {BesselJ}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right ) \sqrt {-a c}-\left (b +\beta \right ) \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right )} \\ \end{align*}

Verification of solutions

\[ y = -\frac {c \,x^{\beta } \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right )}{x^{\frac {\alpha }{2}+\frac {\beta }{2}} \left (\operatorname {BesselJ}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right ) \sqrt {-a c}-\left (b +\beta \right ) \left (\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right )} \] Verified OK.

1.107.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }+a \,x^{\alpha } y^{2}+y b =c \,x^{\beta } \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-a \,x^{\alpha } y^{2}-y b +c \,x^{\beta }}{x} \end {array} \]

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) = (-b-1+alpha)*(diff(y(x), x))/x+x^(-1+alpha)*a*x^(-1+beta)*c*y(x), y(x) 
      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: 240

dsolve(x*diff(y(x),x) + a*x^alpha*y(x)^2 + b*y(x) - c*x^beta=0,y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {c \,x^{\beta } \left (\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{1} +\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right )}{-\left (\operatorname {BesselY}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{1} +\operatorname {BesselJ}\left (\frac {b +2 \beta +\alpha }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right ) x^{\frac {\alpha }{2}+\frac {\beta }{2}} \sqrt {-a c}+\left (b +\beta \right ) \left (\operatorname {BesselY}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right ) c_{1} +\operatorname {BesselJ}\left (\frac {b +\beta }{\alpha +\beta }, \frac {2 \sqrt {-a c}\, x^{\frac {\alpha }{2}+\frac {\beta }{2}}}{\alpha +\beta }\right )\right )} \]

Solution by Mathematica

Time used: 0.977 (sec). Leaf size: 1474

DSolve[x*y'[x] + a*x^\[Alpha]*y[x]^2 + b*y[x] - c*x^\[Beta]==0,y[x],x,IncludeSingularSolutions -> True]
 

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