Internal problem ID [7338]
Internal file name [OUTPUT/6319_Sunday_June_05_2022_04_39_48_PM_63185761/index.tex]
Book: First order enumerated odes
Section: section 1
Problem number: 22.
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]
\[ \boxed {y^{\prime } c -\frac {a x +b y^{2}}{x^{2} r}=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {b \,y^{2}+a x}{x^{2} r c} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {b \,y^{2}}{c \,x^{2} r}+\frac {a}{x r c} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a}{x r c}\), \(f_1(x)=0\) and \(f_2(x)=\frac {b}{c r \,x^{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {b u}{c r \,x^{2}}} \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 {2 b}{c r \,x^{3}}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {b^{2} a}{c^{3} r^{3} x^{5}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {b u^{\prime \prime }\left (x \right )}{c r \,x^{2}}+\frac {2 b u^{\prime }\left (x \right )}{c r \,x^{3}}+\frac {b^{2} a u \left (x \right )}{c^{3} r^{3} x^{5}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {c_{2} \operatorname {BesselY}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )+c_{1} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )}{\sqrt {x}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (-\operatorname {BesselY}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right ) c_{2} -\operatorname {BesselJ}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right ) c_{1} \right ) \sqrt {a b}}{x^{2} c r} \] Using the above in (1) gives the solution \[ y = -\frac {\left (-\operatorname {BesselY}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right ) c_{2} -\operatorname {BesselJ}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right ) c_{1} \right ) \sqrt {x}\, \sqrt {a b}}{b \left (c_{2} \operatorname {BesselY}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )+c_{1} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\operatorname {BesselJ}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right ) c_{3} +\operatorname {BesselY}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )\right ) \sqrt {x}\, \sqrt {a b}}{b \left (\operatorname {BesselY}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )+c_{3} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\operatorname {BesselJ}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right ) c_{3} +\operatorname {BesselY}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )\right ) \sqrt {x}\, \sqrt {a b}}{b \left (\operatorname {BesselY}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )+c_{3} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\operatorname {BesselJ}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right ) c_{3} +\operatorname {BesselY}\left (0, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )\right ) \sqrt {x}\, \sqrt {a b}}{b \left (\operatorname {BesselY}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )+c_{3} \operatorname {BesselJ}\left (1, \frac {2 \sqrt {a b}}{r c \sqrt {x}}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -y^{\prime } c \,x^{2} r +b y^{2}+a x =0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a x +b y^{2}}{x^{2} r c} \end {array} \]
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 a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists <- Abel AIR successful: ODE belongs to the 0F1 1-parameter (Bessel type) class`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 106
dsolve(c*diff(y(x),x)=(a*x+b*y(x)^2)/(r*x^2),y(x), singsol=all)
\[ y \left (x \right ) = \frac {a \left (\operatorname {BesselY}\left (0, 2 \sqrt {\frac {b a}{c^{2} r^{2} x}}\right ) c_{1} +\operatorname {BesselJ}\left (0, 2 \sqrt {\frac {b a}{c^{2} r^{2} x}}\right )\right )}{c r \sqrt {\frac {b a}{c^{2} r^{2} x}}\, \left (c_{1} \operatorname {BesselY}\left (1, 2 \sqrt {\frac {b a}{c^{2} r^{2} x}}\right )+\operatorname {BesselJ}\left (1, 2 \sqrt {\frac {b a}{c^{2} r^{2} x}}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.358 (sec). Leaf size: 492
DSolve[c*y'[x]==(a*x+b*y[x]^2)/(r*x^2),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {2 \sqrt {a} \sqrt {b} \operatorname {BesselY}\left (0,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )+\frac {2 c r \operatorname {BesselY}\left (1,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )}{\sqrt {\frac {1}{x}}}-2 \sqrt {a} \sqrt {b} \operatorname {BesselY}\left (2,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )-i \sqrt {a} \sqrt {b} c_1 \operatorname {BesselJ}\left (0,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )-\frac {i c c_1 r \operatorname {BesselJ}\left (1,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )}{\sqrt {\frac {1}{x}}}+i \sqrt {a} \sqrt {b} c_1 \operatorname {BesselJ}\left (2,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )}{2 b \sqrt {\frac {1}{x}} \left (2 \operatorname {BesselY}\left (1,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )-i c_1 \operatorname {BesselJ}\left (1,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )\right )} \\ y(x)\to \frac {x \left (\sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}} \operatorname {BesselJ}\left (0,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )+c r \operatorname {BesselJ}\left (1,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )-\sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}} \operatorname {BesselJ}\left (2,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )\right )}{2 b \operatorname {BesselJ}\left (1,\frac {2 \sqrt {a} \sqrt {b} \sqrt {\frac {1}{x}}}{c r}\right )} \\ \end{align*}