Internal problem ID [3425]
Internal file name [OUTPUT/2918_Sunday_June_05_2022_08_47_03_AM_14127343/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 6
Problem number: 169.
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]
\[ \boxed {x y^{\prime }-y b -c y^{2}=a \,x^{n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,x^{n}+b y +c \,y^{2}}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {c \,y^{2}}{x}+\frac {a \,x^{n}}{x}+\frac {b y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a \,x^{n}}{x}\), \(f_1(x)=\frac {b}{x}\) and \(f_2(x)=\frac {c}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {c 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 {c}{x^{2}}\\ f_1 f_2 &=\frac {b c}{x^{2}}\\ f_2^2 f_0 &=\frac {c^{2} a \,x^{n}}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {c u^{\prime \prime }\left (x \right )}{x}-\left (\frac {b c}{x^{2}}-\frac {c}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {c^{2} a \,x^{n} u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right ) x^{\frac {b}{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = -x^{-1+\frac {b}{2}} \left (\sqrt {a c}\, x^{\frac {n}{2}} \operatorname {BesselJ}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\sqrt {a c}\, x^{\frac {n}{2}} \operatorname {BesselY}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{2} -b \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right )\right ) \] Using the above in (1) gives the solution \[ y = \frac {x^{-1+\frac {b}{2}} \left (\sqrt {a c}\, x^{\frac {n}{2}} \operatorname {BesselJ}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\sqrt {a c}\, x^{\frac {n}{2}} \operatorname {BesselY}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{2} -b \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right )\right ) x \,x^{-\frac {b}{2}}}{c \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\sqrt {a c}\, \left (\operatorname {BesselJ}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right ) x^{\frac {n}{2}}-b \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right )}{c \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\sqrt {a c}\, \left (\operatorname {BesselJ}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right ) x^{\frac {n}{2}}-b \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right )}{c \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\sqrt {a c}\, \left (\operatorname {BesselJ}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right ) x^{\frac {n}{2}}-b \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right )}{c \left (\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{3} +\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x y^{\prime }-y b -c y^{2}=a \,x^{n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,x^{n}+y b +c y^{2}}{x} \end {array} \]
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)*(diff(y(x), x))/x-c*a*x^(n-1)*y(x)/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: 164
dsolve(x*diff(y(x),x) = a*x^n+b*y(x)+c*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\operatorname {BesselY}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {b +n}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right ) \sqrt {a c}\, x^{\frac {n}{2}}-b \left (\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right )}{c \left (\operatorname {BesselY}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {b}{n}, \frac {2 \sqrt {a c}\, x^{\frac {n}{2}}}{n}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.307 (sec). Leaf size: 402
DSolve[x y'[x]==a x^n+b y[x]+c y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sqrt {a} \sqrt {c} x^{n/2} \left (-2 \operatorname {BesselJ}\left (\frac {b}{n}-1,\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )+c_1 \left (\operatorname {BesselJ}\left (1-\frac {b}{n},\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )-\operatorname {BesselJ}\left (-\frac {b+n}{n},\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )\right )\right )-b c_1 \operatorname {BesselJ}\left (-\frac {b}{n},\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )}{2 c \left (\operatorname {BesselJ}\left (\frac {b}{n},\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )+c_1 \operatorname {BesselJ}\left (-\frac {b}{n},\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )\right )} \\ y(x)\to -\frac {-\sqrt {a} \sqrt {c} x^{n/2} \operatorname {BesselJ}\left (1-\frac {b}{n},\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )+\sqrt {a} \sqrt {c} x^{n/2} \operatorname {BesselJ}\left (-\frac {b+n}{n},\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )+b \operatorname {BesselJ}\left (-\frac {b}{n},\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )}{2 c \operatorname {BesselJ}\left (-\frac {b}{n},\frac {2 \sqrt {a} \sqrt {c} x^{n/2}}{n}\right )} \\ \end{align*}