Internal problem ID [8493]
Internal file name [OUTPUT/7426_Sunday_June_05_2022_10_54_57_PM_40545454/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 157.
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]
\[ \boxed {\left (x^{2}-1\right ) y^{\prime }+a \left (y^{2}-2 y x +1\right )=0} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {a \left (-2 x y +y^{2}+1\right )}{x^{2}-1} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {2 a x y}{x^{2}-1}-\frac {a \,y^{2}}{x^{2}-1}-\frac {a}{x^{2}-1} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {a}{x^{2}-1}\), \(f_1(x)=\frac {2 a x}{x^{2}-1}\) and \(f_2(x)=-\frac {a}{x^{2}-1}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {a u}{x^{2}-1}} \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 {2 a x}{\left (x^{2}-1\right )^{2}}\\ f_1 f_2 &=-\frac {2 a^{2} x}{\left (x^{2}-1\right )^{2}}\\ f_2^2 f_0 &=-\frac {a^{3}}{\left (x^{2}-1\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {a u^{\prime \prime }\left (x \right )}{x^{2}-1}-\left (-\frac {2 a^{2} x}{\left (x^{2}-1\right )^{2}}+\frac {2 a x}{\left (x^{2}-1\right )^{2}}\right ) u^{\prime }\left (x \right )-\frac {a^{3} u \left (x \right )}{\left (x^{2}-1\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (\operatorname {LegendreQ}\left (-1+a , x\right ) c_{2} +\operatorname {LegendreP}\left (-1+a , x\right ) c_{1} \right ) \left (x^{2}-1\right )^{\frac {a}{2}} \] The above shows that \[ u^{\prime }\left (x \right ) = a \left (x^{2}-1\right )^{-1+\frac {a}{2}} \left (\operatorname {LegendreQ}\left (a , x\right ) c_{2} +\operatorname {LegendreP}\left (a , x\right ) c_{1} \right ) \] Using the above in (1) gives the solution \[ y = \frac {\left (x^{2}-1\right )^{-1+\frac {a}{2}} \left (\operatorname {LegendreQ}\left (a , x\right ) c_{2} +\operatorname {LegendreP}\left (a , x\right ) c_{1} \right ) \left (x^{2}-1\right ) \left (x^{2}-1\right )^{-\frac {a}{2}}}{\operatorname {LegendreQ}\left (-1+a , x\right ) c_{2} +\operatorname {LegendreP}\left (-1+a , x\right ) c_{1}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\operatorname {LegendreQ}\left (a , x\right ) c_{2} +\operatorname {LegendreP}\left (a , x\right ) c_{1}}{\operatorname {LegendreQ}\left (-1+a , x\right ) c_{2} +\operatorname {LegendreP}\left (-1+a , x\right ) c_{1}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\operatorname {LegendreQ}\left (a , x\right ) c_{2} +\operatorname {LegendreP}\left (a , x\right ) c_{1}}{\operatorname {LegendreQ}\left (-1+a , x\right ) c_{2} +\operatorname {LegendreP}\left (-1+a , x\right ) c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \frac {\operatorname {LegendreQ}\left (a , x\right ) c_{2} +\operatorname {LegendreP}\left (a , x\right ) c_{1}}{\operatorname {LegendreQ}\left (-1+a , x\right ) c_{2} +\operatorname {LegendreP}\left (-1+a , x\right ) c_{1}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (x^{2}-1\right ) y^{\prime }+a \left (y^{2}-2 y x +1\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-\frac {a \left (y^{2}-2 y x +1\right )}{x^{2}-1} \end {array} \]
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: <- Abel AIR successful: ODE belongs to the 2F1 3-parameter class`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 280
dsolve((x^2-1)*diff(y(x),x) + a*(y(x)^2-2*x*y(x)+1)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 \left (-\frac {x}{2}-\frac {1}{2}\right )^{2 a} \left (-\frac {\left (-\frac {x}{2}-\frac {1}{2}\right )^{-2 a} \left (x +1\right ) \left (x -1\right )^{2} \operatorname {HeunCPrime}\left (0, 2 a -1, 0, 0, a^{2}-a +\frac {1}{2}, \frac {2}{x +1}\right )}{8}-\left (x -1\right )^{2} \operatorname {HeunCPrime}\left (0, -2 a +1, 0, 0, a^{2}-a +\frac {1}{2}, \frac {2}{x +1}\right ) c_{1} +\left (c_{1} \left (\frac {x +1}{x -1}\right )^{-a} \left (\left (a -\frac {1}{2}\right ) x -\frac {a}{2}+\frac {1}{2}\right ) \operatorname {hypergeom}\left (\left [-a +1, -a +1\right ], \left [-2 a +2\right ], -\frac {2}{x -1}\right )+\frac {\operatorname {hypergeom}\left (\left [a , a\right ], \left [2 a \right ], -\frac {2}{x -1}\right ) \left (\frac {x +1}{x -1}\right )^{a} a \left (-\frac {x}{2}-\frac {1}{2}\right )^{-2 a} \left (x -1\right )}{16}\right ) \left (x +1\right )^{2}\right ) \left (\frac {x +1}{x -1}\right )^{a}}{a \left (x +1\right )^{2} \left (c_{1} \operatorname {hypergeom}\left (\left [-a +1, -a +1\right ], \left [-2 a +2\right ], -\frac {2}{x -1}\right ) \left (-\frac {x}{2}-\frac {1}{2}\right )^{2 a}+\frac {\operatorname {hypergeom}\left (\left [a , a\right ], \left [2 a \right ], -\frac {2}{x -1}\right ) \left (\frac {x +1}{x -1}\right )^{2 a} \left (x -1\right )}{8}\right )} \]
✓ Solution by Mathematica
Time used: 0.355 (sec). Leaf size: 47
DSolve[(x^2-1)*y'[x] + a*(y[x]^2-2*x*y[x]+1)==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\operatorname {LegendreQ}(a,x)+c_1 \operatorname {LegendreP}(a,x)}{\operatorname {LegendreQ}(a-1,x)+c_1 \operatorname {LegendreP}(a-1,x)} \\ y(x)\to \frac {\operatorname {LegendreP}(a,x)}{\operatorname {LegendreP}(a-1,x)} \\ \end{align*}