Internal problem ID [3321]
Internal file name [OUTPUT/2813_Sunday_June_05_2022_08_40_56_AM_59766821/index.tex]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 3
Problem number: 57.
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]
\[ \boxed {y^{\prime }-c y^{2}=a \,x^{n -1}+b \,x^{2 n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,x^{n -1}+b \,x^{2 n}+c \,y^{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,x^{n}}{x}+b \,x^{2 n}+c \,y^{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \,x^{n -1}+b \,x^{2 n}\), \(f_1(x)=0\) and \(f_2(x)=c\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{c 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 &=c^{2} \left (a \,x^{n -1}+b \,x^{2 n}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} c u^{\prime \prime }\left (x \right )+c^{2} \left (a \,x^{n -1}+b \,x^{2 n}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{-\frac {n}{2}} \left (c_{1} \operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (-\left (i \sqrt {b}\, \sqrt {c}\, a -b \left (n +2\right )\right ) c_{1} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n +1}}{n +1}\right )-2 c_{2} b \left (n +1\right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n +1}}{n +1}\right )+\left (i \sqrt {b}\, \sqrt {c}\, a +2 i b^{\frac {3}{2}} \sqrt {c}\, x^{n +1}-b n \right ) \left (\operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n +1}}{n +1}\right ) c_{1} +\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n +1}}{n +1}\right ) c_{2} \right )\right ) x^{-\frac {n}{2}-1}}{2 b} \] Using the above in (1) gives the solution \[ y = -\frac {\left (-\left (i \sqrt {b}\, \sqrt {c}\, a -b \left (n +2\right )\right ) c_{1} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n +1}}{n +1}\right )-2 c_{2} b \left (n +1\right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n +1}}{n +1}\right )+\left (i \sqrt {b}\, \sqrt {c}\, a +2 i b^{\frac {3}{2}} \sqrt {c}\, x^{n +1}-b n \right ) \left (\operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n +1}}{n +1}\right ) c_{1} +\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n +1}}{n +1}\right ) c_{2} \right )\right ) x^{-\frac {n}{2}-1} x^{\frac {n}{2}}}{2 b c \left (c_{1} \operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+c_{2} \operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (i \sqrt {b}\, \sqrt {c}\, a -b \left (n +2\right )\right ) c_{3} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+2 b \left (n +1\right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )-\left (i \sqrt {b}\, \sqrt {c}\, a +2 i b^{\frac {3}{2}} \sqrt {c}\, x^{n} x -b n \right ) \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right )}{2 b c x \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (i \sqrt {b}\, \sqrt {c}\, a -b \left (n +2\right )\right ) c_{3} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+2 b \left (n +1\right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )-\left (i \sqrt {b}\, \sqrt {c}\, a +2 i b^{\frac {3}{2}} \sqrt {c}\, x^{n} x -b n \right ) \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right )}{2 b c x \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (i \sqrt {b}\, \sqrt {c}\, a -b \left (n +2\right )\right ) c_{3} \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+2 b \left (n +1\right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )-\left (i \sqrt {b}\, \sqrt {c}\, a +2 i b^{\frac {3}{2}} \sqrt {c}\, x^{n} x -b n \right ) \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right )}{2 b c x \left (c_{3} \operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2+2 n \right )}, \frac {1}{2+2 n}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-c y^{2}=a \,x^{n -1}+b \,x^{2 n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,x^{n -1}+b \,x^{2 n}+c y^{2} \end {array} \]
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) = -c*(a*x^(n-1)+b*x^(2*n))*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: 354
dsolve(diff(y(x),x) = a*x^(n-1)+b*x^(2*n)+c*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\left (\left (2+n \right ) \sqrt {b}-i \sqrt {c}\, a \right ) \operatorname {WhittakerM}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )-2 c_{1} \sqrt {b}\, \left (n +1\right ) \operatorname {WhittakerW}\left (-\frac {\left (-2 n -2\right ) \sqrt {b}+i \sqrt {c}\, a}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )+\left (-\sqrt {b}\, n +i \left (2 x^{n} b x +a \right ) \sqrt {c}\right ) \left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right )}{2 \sqrt {b}\, \left (\operatorname {WhittakerW}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right ) c_{1} +\operatorname {WhittakerM}\left (-\frac {i \sqrt {c}\, a}{\sqrt {b}\, \left (2 n +2\right )}, \frac {1}{2 n +2}, \frac {2 i \sqrt {b}\, \sqrt {c}\, x^{n} x}{n +1}\right )\right ) c x} \]
✓ Solution by Mathematica
Time used: 1.115 (sec). Leaf size: 982
DSolve[y'[x]==a x^(n-1)+b x^(2 n)+c y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x^n \left (\sqrt {b} c_1 (n+1) \sqrt {-(n+1)^2} \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} a}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+c_1 \left (a \sqrt {c} (n+1)+\sqrt {b} \sqrt {-(n+1)^2} n\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} a}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {3 n+2}{n+1}\right ),\frac {n}{n+1}+1,\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+\sqrt {b} (n+1) \sqrt {-(n+1)^2} \left (L_{-\frac {\sqrt {c} a}{2 \sqrt {b} \sqrt {-(n+1)^2}}-\frac {n}{2 (n+1)}}^{-\frac {1}{n+1}}\left (\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+2 L_{-\frac {\sqrt {c} a}{2 \sqrt {b} \sqrt {-(n+1)^2}}-\frac {3 n+2}{2 n+2}}^{\frac {n}{n+1}}\left (\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )\right )\right )}{\sqrt {c} (n+1)^2 \left (L_{-\frac {\sqrt {c} a}{2 \sqrt {b} \sqrt {-(n+1)^2}}-\frac {n}{2 (n+1)}}^{-\frac {1}{n+1}}\left (\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )+c_1 \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} a}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )\right )} \\ y(x)\to \frac {x^n \left (-\frac {\left (a \sqrt {c} (n+1)+\sqrt {b} \sqrt {-(n+1)^2} n\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} a}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}+2\right ),\frac {n}{n+1}+1,\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}{\operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} a}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}-\sqrt {b} \sqrt {-(n+1)^2} (n+1)\right )}{\sqrt {c} (n+1)^2} \\ y(x)\to \frac {x^n \left (-\frac {\left (a \sqrt {c} (n+1)+\sqrt {b} \sqrt {-(n+1)^2} n\right ) \operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} a}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}+2\right ),\frac {n}{n+1}+1,\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}{\operatorname {HypergeometricU}\left (\frac {1}{2} \left (\frac {\sqrt {c} a}{\sqrt {b} \sqrt {-(n+1)^2}}+\frac {n}{n+1}\right ),\frac {n}{n+1},\frac {2 \sqrt {b} \sqrt {c} x^{n+1}}{\sqrt {-(n+1)^2}}\right )}-\sqrt {b} \sqrt {-(n+1)^2} (n+1)\right )}{\sqrt {c} (n+1)^2} \\ \end{align*}