Internal problem ID [8504]
Internal file name [OUTPUT/7437_Sunday_June_05_2022_10_55_08_PM_88047147/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 168.
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]
\[ \boxed {3 \left (x^{2}-4\right ) y^{\prime }+y^{2}-y x=3} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {-x y +y^{2}-3}{3 \left (x^{2}-4\right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {y x}{3 x^{2}-12}-\frac {y^{2}}{3 \left (x^{2}-4\right )}+\frac {1}{x^{2}-4} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {1}{x^{2}-4}\), \(f_1(x)=\frac {x}{3 x^{2}-12}\) and \(f_2(x)=-\frac {1}{3 \left (x^{2}-4\right )}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {u}{3 \left (x^{2}-4\right )}} \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 x}{3 \left (x^{2}-4\right )^{2}}\\ f_1 f_2 &=-\frac {x}{9 \left (x^{2}-4\right )^{2}}\\ f_2^2 f_0 &=\frac {1}{9 \left (x^{2}-4\right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -\frac {u^{\prime \prime }\left (x \right )}{3 \left (x^{2}-4\right )}-\frac {5 x u^{\prime }\left (x \right )}{9 \left (x^{2}-4\right )^{2}}+\frac {u \left (x \right )}{9 \left (x^{2}-4\right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (x^{2}-4\right )^{{1}/{12}} \left (\operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +\operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {-2 \operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} x -2 \operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2} x +3 \operatorname {LegendreP}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +3 \operatorname {LegendreQ}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2}}{3 \left (x^{2}-4\right )^{{11}/{12}}} \] Using the above in (1) gives the solution \[ y = \frac {-2 \operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} x -2 \operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2} x +3 \operatorname {LegendreP}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +3 \operatorname {LegendreQ}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2}}{\operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +\operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2}} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-2 \operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} x -2 \operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2} x +3 \operatorname {LegendreP}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +3 \operatorname {LegendreQ}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2}}{\operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +\operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-2 \operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} x -2 \operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2} x +3 \operatorname {LegendreP}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +3 \operatorname {LegendreQ}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2}}{\operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +\operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2}} \\ \end{align*}
Verification of solutions
\[ y = \frac {-2 \operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} x -2 \operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2} x +3 \operatorname {LegendreP}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +3 \operatorname {LegendreQ}\left (\frac {5}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2}}{\operatorname {LegendreP}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{1} +\operatorname {LegendreQ}\left (-\frac {1}{6}, \frac {1}{3}, \frac {x}{2}\right ) c_{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 3 \left (x^{2}-4\right ) y^{\prime }+y^{2}-y x =3 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {-y^{2}+y x +3}{3 \left (x^{2}-4\right )} \end {array} \]
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: 167
dsolve(3*(x^2-4)*diff(y(x),x) + y(x)^2 - x*y(x) - 3=0,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\left (\left (x -2\right ) \left (-2 x -4\right )^{\frac {1}{3}} \operatorname {hypergeom}\left (\left [-\frac {1}{6}, \frac {1}{6}\right ], \left [-\frac {1}{3}\right ], -\frac {4}{x -2}\right )+24 \operatorname {hypergeom}\left (\left [\frac {5}{6}, \frac {7}{6}\right ], \left [\frac {7}{3}\right ], -\frac {4}{x -2}\right ) c_{1} \right ) \left (x +2\right )^{2}}{\left (-2 x -4\right )^{\frac {1}{3}} \left (x -2\right ) \left (x +2\right )^{2} \operatorname {hypergeom}\left (\left [-\frac {1}{6}, \frac {1}{6}\right ], \left [-\frac {1}{3}\right ], -\frac {4}{x -2}\right )+32 \left (x -\frac {5}{4}\right ) c_{1} \left (x +2\right )^{2} \operatorname {hypergeom}\left (\left [\frac {5}{6}, \frac {7}{6}\right ], \left [\frac {7}{3}\right ], -\frac {4}{x -2}\right )+4 \left (x -2\right )^{2} \left (\frac {x +2}{x -2}\right )^{\frac {1}{6}} \left (\left (-2 x -4\right )^{\frac {1}{3}} \left (x +2\right ) \operatorname {HeunCPrime}\left (0, -\frac {4}{3}, -\frac {1}{3}, 0, \frac {25}{36}, \frac {4}{x +2}\right )+24 \operatorname {HeunCPrime}\left (0, \frac {4}{3}, -\frac {1}{3}, 0, \frac {25}{36}, \frac {4}{x +2}\right ) c_{1} \right )} \]
✓ Solution by Mathematica
Time used: 0.404 (sec). Leaf size: 135
DSolve[3*(x^2-4)*y'[x] + y[x]^2 - x*y[x] - 3==0,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {-2 c_1 x P_{-\frac {1}{6}}^{\frac {1}{3}}\left (\frac {x}{2}\right )+3 c_1 P_{\frac {5}{6}}^{\frac {1}{3}}\left (\frac {x}{2}\right )-2 x Q_{-\frac {1}{6}}^{\frac {1}{3}}\left (\frac {x}{2}\right )+3 Q_{\frac {5}{6}}^{\frac {1}{3}}\left (\frac {x}{2}\right )}{Q_{-\frac {1}{6}}^{\frac {1}{3}}\left (\frac {x}{2}\right )+c_1 P_{-\frac {1}{6}}^{\frac {1}{3}}\left (\frac {x}{2}\right )} \\ y(x)\to \frac {3 P_{\frac {5}{6}}^{\frac {1}{3}}\left (\frac {x}{2}\right )}{P_{-\frac {1}{6}}^{\frac {1}{3}}\left (\frac {x}{2}\right )}-2 x \\ \end{align*}