Internal problem ID [8475]
Internal file name [OUTPUT/7408_Sunday_June_05_2022_10_54_37_PM_12682561/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 139.
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]
\[ \boxed {x^{2} \left (y^{\prime }+y^{2}\right )=-a \,x^{k}+b \left (b -1\right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {y^{2} x^{2}+a \,x^{k}-b^{2}+b}{x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -y^{2}-\frac {a \,x^{k}}{x^{2}}+\frac {b^{2}}{x^{2}}-\frac {b}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {a \,x^{k}-b^{2}+b}{x^{2}}\), \(f_1(x)=0\) and \(f_2(x)=-1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-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' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=-\frac {a \,x^{k}-b^{2}+b}{x^{2}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (x \right )-\frac {\left (a \,x^{k}-b^{2}+b \right ) u \left (x \right )}{x^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \sqrt {x}\, \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{1} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {-\sqrt {a}\, x^{\frac {k}{2}} \operatorname {BesselJ}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{1} -\sqrt {a}\, x^{\frac {k}{2}} \operatorname {BesselY}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{2} +b \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{1} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{2} \right )}{\sqrt {x}} \] Using the above in (1) gives the solution \[ y = \frac {-\sqrt {a}\, x^{\frac {k}{2}} \operatorname {BesselJ}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{1} -\sqrt {a}\, x^{\frac {k}{2}} \operatorname {BesselY}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{2} +b \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{1} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{2} \right )}{x \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{1} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\left (\operatorname {BesselJ}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{3} +\operatorname {BesselY}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right ) x^{\frac {k}{2}} \sqrt {a}+b \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{3} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right )}{x \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{3} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\left (\operatorname {BesselJ}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{3} +\operatorname {BesselY}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right ) x^{\frac {k}{2}} \sqrt {a}+b \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{3} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right )}{x \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{3} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\left (\operatorname {BesselJ}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{3} +\operatorname {BesselY}\left (\frac {2 b +k -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right ) x^{\frac {k}{2}} \sqrt {a}+b \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{3} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right )}{x \left (\operatorname {BesselJ}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{3} +\operatorname {BesselY}\left (\frac {2 b -1}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \left (y^{\prime }+y^{2}\right )=-a \,x^{k}+b \left (b -1\right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {y^{2} x^{2}+a \,x^{k}-b^{2}+b}{x^{2}} \end {array} \]
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 Special trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -(x^(-2+k)*a*x^2-b^2+b)*y(x)/x^2, 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: 229
dsolve(x^2*(diff(y(x),x)+y(x)^2) + a*x^k - b*(b-1)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-x^{\frac {k}{2}} \left (\operatorname {BesselY}\left (\frac {\operatorname {csgn}\left (2 b -1\right ) \left (2 b -1\right )+k}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\operatorname {csgn}\left (2 b -1\right ) \left (2 b -1\right )+k}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right ) \sqrt {a}+\left (\frac {1}{2}+\operatorname {csgn}\left (2 b -1\right ) b -\frac {\operatorname {csgn}\left (2 b -1\right )}{2}\right ) \left (\operatorname {BesselY}\left (\frac {\operatorname {csgn}\left (2 b -1\right ) \left (2 b -1\right )}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\operatorname {csgn}\left (2 b -1\right ) \left (2 b -1\right )}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right )}{x \left (\operatorname {BesselY}\left (\frac {\operatorname {csgn}\left (2 b -1\right ) \left (2 b -1\right )}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\operatorname {csgn}\left (2 b -1\right ) \left (2 b -1\right )}{k}, \frac {2 \sqrt {a}\, x^{\frac {k}{2}}}{k}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.496 (sec). Leaf size: 627
DSolve[x^2*(y'[x]+y[x]^2) + a*x^k - b*(b-1)==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sqrt {a} x^k \operatorname {Gamma}\left (\frac {2 b+k-1}{k}\right ) \operatorname {BesselJ}\left (-\frac {-2 b+k+1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )-\sqrt {a} x^k \operatorname {Gamma}\left (\frac {2 b+k-1}{k}\right ) \operatorname {BesselJ}\left (\frac {2 b+k-1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )+\sqrt {x^k} \operatorname {Gamma}\left (\frac {2 b+k-1}{k}\right ) \operatorname {BesselJ}\left (\frac {2 b-1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )-\sqrt {a} c_1 x^k \operatorname {Gamma}\left (\frac {-2 b+k+1}{k}\right ) \operatorname {BesselJ}\left (\frac {-2 b+k+1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )+\sqrt {a} c_1 x^k \operatorname {Gamma}\left (\frac {-2 b+k+1}{k}\right ) \operatorname {BesselJ}\left (-\frac {2 b+k-1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )+c_1 \sqrt {x^k} \operatorname {Gamma}\left (\frac {-2 b+k+1}{k}\right ) \operatorname {BesselJ}\left (\frac {1-2 b}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )}{2 x \sqrt {x^k} \left (\operatorname {Gamma}\left (\frac {2 b+k-1}{k}\right ) \operatorname {BesselJ}\left (\frac {2 b-1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )+c_1 \operatorname {Gamma}\left (\frac {-2 b+k+1}{k}\right ) \operatorname {BesselJ}\left (\frac {1-2 b}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )\right )} \\ y(x)\to \frac {\frac {\sqrt {a} \sqrt {x^k} \left (\operatorname {BesselJ}\left (-\frac {2 b+k-1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )-\operatorname {BesselJ}\left (\frac {-2 b+k+1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )\right )}{\operatorname {BesselJ}\left (\frac {1-2 b}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )}+1}{2 x} \\ y(x)\to \frac {\frac {\sqrt {a} \sqrt {x^k} \left (\operatorname {BesselJ}\left (-\frac {2 b+k-1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )-\operatorname {BesselJ}\left (\frac {-2 b+k+1}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )\right )}{\operatorname {BesselJ}\left (\frac {1-2 b}{k},\frac {2 \sqrt {a} \sqrt {x^k}}{k}\right )}+1}{2 x} \\ \end{align*}