Internal problem ID [4386]
Internal file name [OUTPUT/3879_Sunday_June_05_2022_11_34_20_AM_19298257/index.tex]
Book: Differential Equations, By George Boole F.R.S. 1865
Section: Chapter 3
Problem number: 10.
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]
\[ \boxed {\frac {x^{n} y^{\prime }}{b y^{2}-c \,x^{2 a}}-\frac {a y x^{a -1}}{b y^{2}-c \,x^{2 a}}=-x^{a -1}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -x^{a -1} \left (b \,y^{2}-c \,x^{2 a}-y a \right ) x^{-n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {x^{a} x^{-n} b \,y^{2}}{x}+\frac {x^{3 a} x^{-n} c}{x}+\frac {x^{a} x^{-n} y a}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=x^{-n} x^{2 a} x^{a -1} c\), \(f_1(x)=a \,x^{a -1} x^{-n}\) and \(f_2(x)=-b \,x^{a -1} x^{-n}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-b \,x^{a -1} x^{-n} 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' &=-\frac {b \,x^{a -1} \left (a -1\right ) x^{-n}}{x}+\frac {b \,x^{a -1} x^{-n} n}{x}\\ f_1 f_2 &=-a \,x^{-2+2 a} x^{-2 n} b\\ f_2^2 f_0 &=b^{2} x^{3 a -3} x^{-3 n} x^{2 a} c \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -b \,x^{a -1} x^{-n} u^{\prime \prime }\left (x \right )-\left (-\frac {b \,x^{a -1} \left (a -1\right ) x^{-n}}{x}+\frac {b \,x^{a -1} x^{-n} n}{x}-a \,x^{-2+2 a} x^{-2 n} b \right ) u^{\prime }\left (x \right )+b^{2} x^{3 a -3} x^{-3 n} x^{2 a} c u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (1-a +n -a \,x^{a -n}\right )}{x}-b c \,x^{4 a -2 n -2} \textit {\_Y} \left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {d}{d x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (1-a +n -a \,x^{a -n}\right )}{x}-b c \,x^{4 a -2 n -2} \textit {\_Y} \left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = \frac {\left (\frac {d}{d x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (1-a +n -a \,x^{a -n}\right )}{x}-b c \,x^{4 a -2 n -2} \textit {\_Y} \left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) x^{-a +1} x^{n}}{b \operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (1-a +n -a \,x^{a -n}\right )}{x}-b c \,x^{4 a -2 n -2} \textit {\_Y} \left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {x^{-a +n +1} \left (\frac {d}{d x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (1-a +n -a \,x^{a -n}\right )}{x}-b c \,x^{4 a -2 n -2} \textit {\_Y} \left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right )}{b \operatorname {DESol}\left (\left \{\frac {-b c \textit {\_Y} \left (x \right ) x^{4 a -2 n}-a \,x^{a -n +1} \textit {\_Y}^{\prime }\left (x \right )-\left (-\textit {\_Y}^{\prime \prime }\left (x \right ) x +\textit {\_Y}^{\prime }\left (x \right ) \left (a -1-n \right )\right ) x}{x^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {x^{-a +n +1} \left (\frac {d}{d x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (1-a +n -a \,x^{a -n}\right )}{x}-b c \,x^{4 a -2 n -2} \textit {\_Y} \left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right )}{b \operatorname {DESol}\left (\left \{\frac {-b c \textit {\_Y} \left (x \right ) x^{4 a -2 n}-a \,x^{a -n +1} \textit {\_Y}^{\prime }\left (x \right )-\left (-\textit {\_Y}^{\prime \prime }\left (x \right ) x +\textit {\_Y}^{\prime }\left (x \right ) \left (a -1-n \right )\right ) x}{x^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {x^{-a +n +1} \left (\frac {d}{d x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )+\frac {\textit {\_Y}^{\prime }\left (x \right ) \left (1-a +n -a \,x^{a -n}\right )}{x}-b c \,x^{4 a -2 n -2} \textit {\_Y} \left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right )}{b \operatorname {DESol}\left (\left \{\frac {-b c \textit {\_Y} \left (x \right ) x^{4 a -2 n}-a \,x^{a -n +1} \textit {\_Y}^{\prime }\left (x \right )-\left (-\textit {\_Y}^{\prime \prime }\left (x \right ) x +\textit {\_Y}^{\prime }\left (x \right ) \left (a -1-n \right )\right ) x}{x^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{2} x^{a -1} b -a y x^{a -1}-x^{2 a} x^{a -1} c +x^{n} y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-y^{2} x^{a -1} b +x^{2 a} x^{a -1} c +a y x^{a -1}}{x^{n}} \end {array} \]
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) = (x^(a-1-n)*a*x+a-n-1)*(diff(y(x), x))/x+x^(a-1-n)*b*x^(3*a-1-n)*c*y(x) 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 solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Kummer -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) 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 solution in terms of special functions: -> Bessel -> elliptic -> Legendre -> Whittaker -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius -> Mathieu -> Equivalence to the rational form of Mathieu ODE under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius -> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) -> Trying changes of variables to rationalize or make the ODE simpler <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> Trying a change of variables to reduce to Bernoulli -> Calling odsolve with the ODE`, diff(y(x), x)-(-x^(a-1-n)*b*y(x)^2+y(x)+x^(a-1-n)*a*y(x)*x+x^2*x^(3*a-1-n)*c)/x, y(x), explicit 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 inverse_Riccati trying 1st order ODE linearizable_by_differentiation -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 2`
✗ Solution by Maple
dsolve( x^n/(b*y(x)^2-c*x^(2*a))*diff(y(x),x) - a*y(x)*x^(a-1)/(b*y(x)^2-c*x^(2*a)) + x^(a-1)=0,y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[x^n/(b*y[x]^2-c*x^(2*a))*y'[x] - a*y[x]*x^(a-1)/(b*y[x]^2-c*x^(2*a)) + x^(a-1)==0,y[x],x,IncludeSingularSolutions -> True]
Not solved