Internal problem ID [3436]
Internal file name [OUTPUT/2929_Sunday_June_05_2022_08_47_18_AM_14028383/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 7
Problem number: 180.
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]
\[ \boxed {x y^{\prime }+a \,x^{2} y^{2}+2 y=b} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {a \,x^{2} y^{2}-b +2 y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -x a \,y^{2}+\frac {b}{x}-\frac {2 y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {b}{x}\), \(f_1(x)=-\frac {2}{x}\) and \(f_2(x)=-x a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-x a 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' &=-a\\ f_1 f_2 &=2 a\\ f_2^2 f_0 &=x \,a^{2} b \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -x a u^{\prime \prime }\left (x \right )-a u^{\prime }\left (x \right )+x \,a^{2} b u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \operatorname {BesselJ}\left (0, \sqrt {-a b}\, x \right )+c_{2} \operatorname {BesselY}\left (0, \sqrt {-a b}\, x \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (-\operatorname {BesselJ}\left (1, \sqrt {-a b}\, x \right ) c_{1} -\operatorname {BesselY}\left (1, \sqrt {-a b}\, x \right ) c_{2} \right ) \sqrt {-a b} \] Using the above in (1) gives the solution \[ y = \frac {\left (-\operatorname {BesselJ}\left (1, \sqrt {-a b}\, x \right ) c_{1} -\operatorname {BesselY}\left (1, \sqrt {-a b}\, x \right ) c_{2} \right ) \sqrt {-a b}}{x a \left (c_{1} \operatorname {BesselJ}\left (0, \sqrt {-a b}\, x \right )+c_{2} \operatorname {BesselY}\left (0, \sqrt {-a b}\, x \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (-\operatorname {BesselJ}\left (1, \sqrt {-a b}\, x \right ) c_{3} -\operatorname {BesselY}\left (1, \sqrt {-a b}\, x \right )\right ) \sqrt {-a b}}{x a \left (c_{3} \operatorname {BesselJ}\left (0, \sqrt {-a b}\, x \right )+\operatorname {BesselY}\left (0, \sqrt {-a b}\, x \right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-\operatorname {BesselJ}\left (1, \sqrt {-a b}\, x \right ) c_{3} -\operatorname {BesselY}\left (1, \sqrt {-a b}\, x \right )\right ) \sqrt {-a b}}{x a \left (c_{3} \operatorname {BesselJ}\left (0, \sqrt {-a b}\, x \right )+\operatorname {BesselY}\left (0, \sqrt {-a b}\, x \right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-\operatorname {BesselJ}\left (1, \sqrt {-a b}\, x \right ) c_{3} -\operatorname {BesselY}\left (1, \sqrt {-a b}\, x \right )\right ) \sqrt {-a b}}{x a \left (c_{3} \operatorname {BesselJ}\left (0, \sqrt {-a b}\, x \right )+\operatorname {BesselY}\left (0, \sqrt {-a b}\, x \right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }+a \,x^{2} y^{2}+2 y=b \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-a \,x^{2} y^{2}-2 y+b}{x} \end {array} \]
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 to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(diff(y(x), x))/x+a*b*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: 68
dsolve(x*diff(y(x),x)+a*x^2*y(x)^2+2*y(x) = b,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-\operatorname {BesselY}\left (1, \sqrt {-a b}\, x \right ) c_{1} -\operatorname {BesselJ}\left (1, \sqrt {-a b}\, x \right )\right ) \sqrt {-a b}}{a x \left (c_{1} \operatorname {BesselY}\left (0, \sqrt {-a b}\, x \right )+\operatorname {BesselJ}\left (0, \sqrt {-a b}\, x \right )\right )} \]
✓ Solution by Mathematica
Time used: 0.255 (sec). Leaf size: 158
DSolve[x y'[x]+a x^2 y[x]^2+2 y[x]==b,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {i \sqrt {b} \left (\operatorname {BesselY}\left (1,-i \sqrt {a} \sqrt {b} x\right )-c_1 \operatorname {BesselJ}\left (1,i \sqrt {a} \sqrt {b} x\right )\right )}{\sqrt {a} x \left (\operatorname {BesselY}\left (0,-i \sqrt {a} \sqrt {b} x\right )+c_1 \operatorname {BesselJ}\left (0,i \sqrt {a} \sqrt {b} x\right )\right )} \\ y(x)\to -\frac {i \sqrt {b} \operatorname {BesselJ}\left (1,i \sqrt {a} \sqrt {b} x\right )}{\sqrt {a} x \operatorname {BesselJ}\left (0,i \sqrt {a} \sqrt {b} x\right )} \\ \end{align*}