Internal problem ID [8362]
Internal file name [OUTPUT/7295_Sunday_June_05_2022_05_43_11_PM_18954759/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 25.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime }+a y^{2}=b \,x^{2 \nu }+c \,x^{-1+\nu }} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -a \,y^{2}+b \,x^{2 \nu }+c \,x^{-1+\nu } \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -a \,y^{2}+b \,x^{2 \nu }+\frac {c \,x^{\nu }}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=b \,x^{2 \nu }+c \,x^{-1+\nu }\), \(f_1(x)=0\) and \(f_2(x)=-a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-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' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=a^{2} \left (b \,x^{2 \nu }+c \,x^{-1+\nu }\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -a u^{\prime \prime }\left (x \right )+a^{2} \left (b \,x^{2 \nu }+c \,x^{-1+\nu }\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{-\frac {\nu }{2}} \left (c_{1} \operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x \,x^{\nu }}{1+\nu }\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x \,x^{\nu }}{1+\nu }\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (-c_{1} \left (\sqrt {a}\, \sqrt {b}\, c -\left (\nu +2\right ) b \right ) \operatorname {WhittakerM}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )-2 b c_{2} \left (1+\nu \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )+\left (\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{1} +\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{2} \right ) \left (\sqrt {a}\, \sqrt {b}\, c +2 x^{1+\nu } \sqrt {a}\, b^{\frac {3}{2}}-\nu b \right )\right ) x^{-\frac {\nu }{2}-1}}{2 b} \] Using the above in (1) gives the solution \[ y = \frac {\left (-c_{1} \left (\sqrt {a}\, \sqrt {b}\, c -\left (\nu +2\right ) b \right ) \operatorname {WhittakerM}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )-2 b c_{2} \left (1+\nu \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )+\left (\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{1} +\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{2} \right ) \left (\sqrt {a}\, \sqrt {b}\, c +2 x^{1+\nu } \sqrt {a}\, b^{\frac {3}{2}}-\nu b \right )\right ) x^{-\frac {\nu }{2}-1} x^{\frac {\nu }{2}}}{2 b a \left (c_{1} \operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x \,x^{\nu }}{1+\nu }\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x \,x^{\nu }}{1+\nu }\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {c_{3} \left (-\sqrt {a}\, \sqrt {b}\, c +\left (\nu +2\right ) b \right ) \operatorname {WhittakerM}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )-2 b \left (1+\nu \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )+2 \left (x^{1+\nu } \sqrt {a}\, b^{\frac {3}{2}}+\frac {\sqrt {a}\, \sqrt {b}\, c}{2}-\frac {\nu b}{2}\right ) \left (\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )\right )}{2 a b x \left (\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{3} \left (-\sqrt {a}\, \sqrt {b}\, c +\left (\nu +2\right ) b \right ) \operatorname {WhittakerM}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )-2 b \left (1+\nu \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )+2 \left (x^{1+\nu } \sqrt {a}\, b^{\frac {3}{2}}+\frac {\sqrt {a}\, \sqrt {b}\, c}{2}-\frac {\nu b}{2}\right ) \left (\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )\right )}{2 a b x \left (\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{3} \left (-\sqrt {a}\, \sqrt {b}\, c +\left (\nu +2\right ) b \right ) \operatorname {WhittakerM}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )-2 b \left (1+\nu \right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )+2 \left (x^{1+\nu } \sqrt {a}\, b^{\frac {3}{2}}+\frac {\sqrt {a}\, \sqrt {b}\, c}{2}-\frac {\nu b}{2}\right ) \left (\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )\right )}{2 a b x \left (\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right ) c_{3} +\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2+2 \nu \right )}, \frac {1}{2+2 \nu }, \frac {2 \sqrt {b}\, \sqrt {a}\, x^{1+\nu }}{1+\nu }\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+a y^{2}=b \,x^{2 \nu }+c \,x^{-1+\nu } \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-a y^{2}+b \,x^{2 \nu }+c \,x^{-1+\nu } \end {array} \]
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) = a*(b*x^(2*nu)+c*x^(nu-1))*y(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 -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Whittaker successful <- special function solution successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 347
dsolve(diff(y(x),x) + a*y(x)^2 - b*x^(2*nu) - c*x^(nu-1)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\left (\frac {\nu }{2}+1\right ) \sqrt {b}-\frac {\sqrt {a}\, c}{2}\right ) \operatorname {WhittakerM}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2 \nu +2\right )}, \frac {1}{2 \nu +2}, \frac {2 \sqrt {a}\, \sqrt {b}\, x^{\nu +1}}{\nu +1}\right )-c_{1} \sqrt {b}\, \left (\nu +1\right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 \nu -2\right ) \sqrt {b}+\sqrt {a}\, c}{\sqrt {b}\, \left (2 \nu +2\right )}, \frac {1}{2 \nu +2}, \frac {2 \sqrt {a}\, \sqrt {b}\, x^{\nu +1}}{\nu +1}\right )+\left (\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2 \nu +2\right )}, \frac {1}{2 \nu +2}, \frac {2 \sqrt {a}\, \sqrt {b}\, x^{\nu +1}}{\nu +1}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2 \nu +2\right )}, \frac {1}{2 \nu +2}, \frac {2 \sqrt {a}\, \sqrt {b}\, x^{\nu +1}}{\nu +1}\right )\right ) \left (x^{\nu +1} b \sqrt {a}+\frac {\sqrt {a}\, c}{2}-\frac {\sqrt {b}\, \nu }{2}\right )}{\sqrt {b}\, \left (\operatorname {WhittakerW}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2 \nu +2\right )}, \frac {1}{2 \nu +2}, \frac {2 \sqrt {a}\, \sqrt {b}\, x^{\nu +1}}{\nu +1}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {\sqrt {a}\, c}{\sqrt {b}\, \left (2 \nu +2\right )}, \frac {1}{2 \nu +2}, \frac {2 \sqrt {a}\, \sqrt {b}\, x^{\nu +1}}{\nu +1}\right )\right ) a x} \]
✓ Solution by Mathematica
Time used: 1.092 (sec). Leaf size: 722
DSolve[y'[x] + a*y[x]^2 - b*x^(2*nu) - c*x^(nu-1)==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x^{\nu } \left (\sqrt {b} c_1 (\nu +1) \sqrt {(\nu +1)^2} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {(\nu +1)^2}}+\frac {\nu }{\nu +1}\right ),\frac {\nu }{\nu +1},\frac {2 \sqrt {a} \sqrt {b} x^{\nu +1}}{\sqrt {(\nu +1)^2}}\right )+c_1 \left (\sqrt {a} c (\nu +1)+\sqrt {b} \sqrt {(\nu +1)^2} \nu \right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {(\nu +1)^2}}+\frac {3 \nu +2}{\nu +1}\right ),\frac {\nu }{\nu +1}+1,\frac {2 \sqrt {a} \sqrt {b} x^{\nu +1}}{\sqrt {(\nu +1)^2}}\right )+\sqrt {b} (\nu +1) \sqrt {(\nu +1)^2} \left (L_{-\frac {\sqrt {a} c}{2 \sqrt {b} \sqrt {(\nu +1)^2}}-\frac {\nu }{2 (\nu +1)}}^{-\frac {1}{\nu +1}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\nu +1}}{\sqrt {(\nu +1)^2}}\right )+2 L_{-\frac {\sqrt {a} c}{2 \sqrt {b} \sqrt {(\nu +1)^2}}-\frac {3 \nu +2}{2 \nu +2}}^{\frac {\nu }{\nu +1}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\nu +1}}{\sqrt {(\nu +1)^2}}\right )\right )\right )}{\sqrt {a} (\nu +1)^2 \left (L_{-\frac {\sqrt {a} c}{2 \sqrt {b} \sqrt {(\nu +1)^2}}-\frac {\nu }{2 (\nu +1)}}^{-\frac {1}{\nu +1}}\left (\frac {2 \sqrt {a} \sqrt {b} x^{\nu +1}}{\sqrt {(\nu +1)^2}}\right )+c_1 \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {(\nu +1)^2}}+\frac {\nu }{\nu +1}\right ),\frac {\nu }{\nu +1},\frac {2 \sqrt {a} \sqrt {b} x^{\nu +1}}{\sqrt {(\nu +1)^2}}\right )\right )} \\ y(x)\to \frac {x^{\nu } \left (-\frac {\left (\sqrt {a} c (\nu +1)+\sqrt {b} \sqrt {(\nu +1)^2} \nu \right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {(\nu +1)^2}}+\frac {\nu }{\nu +1}+2\right ),\frac {\nu }{\nu +1}+1,\frac {2 \sqrt {a} \sqrt {b} x^{\nu +1}}{\sqrt {(\nu +1)^2}}\right )}{\operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {a} c}{\sqrt {b} \sqrt {(\nu +1)^2}}+\frac {\nu }{\nu +1}\right ),\frac {\nu }{\nu +1},\frac {2 \sqrt {a} \sqrt {b} x^{\nu +1}}{\sqrt {(\nu +1)^2}}\right )}-\sqrt {b} \sqrt {(\nu +1)^2} (\nu +1)\right )}{\sqrt {a} (\nu +1)^2} \\ \end{align*}