Internal problem ID [10481]
Internal file name [OUTPUT/9429_Monday_June_06_2022_02_32_10_PM_31193013/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. subsection 1.2.5-1. Equations Containing
Logarithmic Functions
Problem number: 7.
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]
\[ \boxed {x^{2} y^{\prime }-x^{2} y^{2}=a \ln \left (x \right )^{2}+b \ln \left (x \right )+c} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {x^{2} y^{2}+a \ln \left (x \right )^{2}+b \ln \left (x \right )+c}{x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+\frac {a \ln \left (x \right )^{2}}{x^{2}}+\frac {b \ln \left (x \right )}{x^{2}}+\frac {c}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {a \ln \left (x \right )^{2}+b \ln \left (x \right )+c}{x^{2}}\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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 &=\frac {a \ln \left (x \right )^{2}+b \ln \left (x \right )+c}{x^{2}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\frac {\left (a \ln \left (x \right )^{2}+b \ln \left (x \right )+c \right ) u \left (x \right )}{x^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x^{-\frac {i b -\sqrt {a}}{2 \sqrt {a}}} {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}} \left (2 \ln \left (x \right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right ) c_{2} a +\operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right ) c_{2} b +c_{1} \operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {2 x^{-\frac {i b +\sqrt {a}}{2 \sqrt {a}}} \left (-\left (\frac {b \ln \left (x \right ) \left (b \ln \left (x \right )-4 c +1\right ) a^{\frac {5}{2}}}{12}+\frac {b^{2} \left (b \ln \left (x \right )-c +\frac {1}{4}\right ) a^{\frac {3}{2}}}{12}-\frac {\ln \left (x \right )^{2} \left (c -\frac {1}{4}\right ) a^{\frac {7}{2}}}{3}+\frac {b^{4} \sqrt {a}}{48}+i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {28 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )-\frac {\left (-\ln \left (x \right ) \left (c -\frac {1}{4}\right ) a^{\frac {5}{2}}+\frac {b \left (b \ln \left (x \right )-2 c +\frac {1}{2}\right ) a^{\frac {3}{2}}}{4}+\frac {b^{3} \sqrt {a}}{8}+i \left (a \ln \left (x \right )+\frac {b}{2}\right ) a^{2}\right ) c_{1} \operatorname {hypergeom}\left (\left [\frac {20 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )}{2}+\left (-\frac {a^{\frac {5}{2}} b}{4}+\left (-1-\frac {\ln \left (x \right )}{2}\right ) a^{\frac {7}{2}}+i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right ) \left (i \ln \left (x \right ) a^{3}+\frac {i a^{2} b}{2}-\frac {a^{\frac {5}{2}}}{2}\right ) c_{1}}{2}\right ) {\mathrm e}^{-\frac {i \ln \left (x \right )^{2} \sqrt {a}}{2}}}{a^{\frac {5}{2}}} \] Using the above in (1) gives the solution \[ y = \frac {2 x^{-\frac {i b +\sqrt {a}}{2 \sqrt {a}}} \left (-\left (\frac {b \ln \left (x \right ) \left (b \ln \left (x \right )-4 c +1\right ) a^{\frac {5}{2}}}{12}+\frac {b^{2} \left (b \ln \left (x \right )-c +\frac {1}{4}\right ) a^{\frac {3}{2}}}{12}-\frac {\ln \left (x \right )^{2} \left (c -\frac {1}{4}\right ) a^{\frac {7}{2}}}{3}+\frac {b^{4} \sqrt {a}}{48}+i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {28 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )-\frac {\left (-\ln \left (x \right ) \left (c -\frac {1}{4}\right ) a^{\frac {5}{2}}+\frac {b \left (b \ln \left (x \right )-2 c +\frac {1}{2}\right ) a^{\frac {3}{2}}}{4}+\frac {b^{3} \sqrt {a}}{8}+i \left (a \ln \left (x \right )+\frac {b}{2}\right ) a^{2}\right ) c_{1} \operatorname {hypergeom}\left (\left [\frac {20 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )}{2}+\left (-\frac {a^{\frac {5}{2}} b}{4}+\left (-1-\frac {\ln \left (x \right )}{2}\right ) a^{\frac {7}{2}}+i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) c_{2} \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right ) \left (i \ln \left (x \right ) a^{3}+\frac {i a^{2} b}{2}-\frac {a^{\frac {5}{2}}}{2}\right ) c_{1}}{2}\right ) x^{\frac {i b -\sqrt {a}}{2 \sqrt {a}}}}{a^{\frac {5}{2}} \left (2 \ln \left (x \right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right ) c_{2} a +\operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right ) c_{2} b +c_{1} \operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\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 (-\frac {b \ln \left (x \right ) \left (b \ln \left (x \right )-4 c +1\right ) a^{\frac {5}{2}}}{12}-\frac {b^{2} \left (b \ln \left (x \right )-c +\frac {1}{4}\right ) a^{\frac {3}{2}}}{12}+\frac {\ln \left (x \right )^{2} \left (c -\frac {1}{4}\right ) a^{\frac {7}{2}}}{3}-\frac {b^{4} \sqrt {a}}{48}-i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {28 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\left (-\frac {a^{\frac {5}{2}} b}{4}+\left (-1-\frac {\ln \left (x \right )}{2}\right ) a^{\frac {7}{2}}+i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {c_{3} \left (\left (\ln \left (x \right ) \left (c -\frac {1}{4}\right ) a^{\frac {5}{2}}-\frac {b \left (b \ln \left (x \right )-2 c +\frac {1}{2}\right ) a^{\frac {3}{2}}}{4}-\frac {b^{3} \sqrt {a}}{8}-i \left (a \ln \left (x \right )+\frac {b}{2}\right ) a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {20 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\left (i \ln \left (x \right ) a^{3}+\frac {i a^{2} b}{2}-\frac {a^{\frac {5}{2}}}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )\right )}{2}}{a^{\frac {5}{2}} \left (\left (a \ln \left (x \right )+\frac {b}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {c_{3} \operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )}{2}\right ) x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (-\frac {b \ln \left (x \right ) \left (b \ln \left (x \right )-4 c +1\right ) a^{\frac {5}{2}}}{12}-\frac {b^{2} \left (b \ln \left (x \right )-c +\frac {1}{4}\right ) a^{\frac {3}{2}}}{12}+\frac {\ln \left (x \right )^{2} \left (c -\frac {1}{4}\right ) a^{\frac {7}{2}}}{3}-\frac {b^{4} \sqrt {a}}{48}-i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {28 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\left (-\frac {a^{\frac {5}{2}} b}{4}+\left (-1-\frac {\ln \left (x \right )}{2}\right ) a^{\frac {7}{2}}+i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {c_{3} \left (\left (\ln \left (x \right ) \left (c -\frac {1}{4}\right ) a^{\frac {5}{2}}-\frac {b \left (b \ln \left (x \right )-2 c +\frac {1}{2}\right ) a^{\frac {3}{2}}}{4}-\frac {b^{3} \sqrt {a}}{8}-i \left (a \ln \left (x \right )+\frac {b}{2}\right ) a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {20 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\left (i \ln \left (x \right ) a^{3}+\frac {i a^{2} b}{2}-\frac {a^{\frac {5}{2}}}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )\right )}{2}}{a^{\frac {5}{2}} \left (\left (a \ln \left (x \right )+\frac {b}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {c_{3} \operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )}{2}\right ) x} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (-\frac {b \ln \left (x \right ) \left (b \ln \left (x \right )-4 c +1\right ) a^{\frac {5}{2}}}{12}-\frac {b^{2} \left (b \ln \left (x \right )-c +\frac {1}{4}\right ) a^{\frac {3}{2}}}{12}+\frac {\ln \left (x \right )^{2} \left (c -\frac {1}{4}\right ) a^{\frac {7}{2}}}{3}-\frac {b^{4} \sqrt {a}}{48}-i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {28 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\left (-\frac {a^{\frac {5}{2}} b}{4}+\left (-1-\frac {\ln \left (x \right )}{2}\right ) a^{\frac {7}{2}}+i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {c_{3} \left (\left (\ln \left (x \right ) \left (c -\frac {1}{4}\right ) a^{\frac {5}{2}}-\frac {b \left (b \ln \left (x \right )-2 c +\frac {1}{2}\right ) a^{\frac {3}{2}}}{4}-\frac {b^{3} \sqrt {a}}{8}-i \left (a \ln \left (x \right )+\frac {b}{2}\right ) a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {20 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\left (i \ln \left (x \right ) a^{3}+\frac {i a^{2} b}{2}-\frac {a^{\frac {5}{2}}}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )\right )}{2}}{a^{\frac {5}{2}} \left (\left (a \ln \left (x \right )+\frac {b}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {c_{3} \operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )}{2}\right ) x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-x^{2} y^{2}=a \ln \left (x \right )^{2}+b \ln \left (x \right )+c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{2} y^{2}+a \ln \left (x \right )^{2}+b \ln \left (x \right )+c}{x^{2}} \end {array} \]
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) = -(a*ln(x)^2+ln(x)*b+c)*y(x)/x^2, 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 -> 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 trying a quadrature checking if the LODE has constant coefficients checking if the LODE is of Euler type trying a symmetry of the form [xi=0, eta=F(x)] checking if the LODE is missing y -> Trying a Liouvillian solution using Kovacics algorithm <- No Liouvillian solutions exists -> 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 <- hyper3 successful: indirect Equivalence to 0F1 under \`\`^ @ Moebius\`\` is resolved <- hypergeometric successful <- special function solution successful Change of variables used: [x = exp(t)] Linear ODE actually solved: (a*t^2+b*t+c)*u(t)-diff(u(t),t)+diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 477
dsolve(x^2*diff(y(x),x)=x^2*y(x)^2+a*(ln(x))^2+b*ln(x)+c,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (-\frac {b \ln \left (x \right ) \left (b \ln \left (x \right )-4 c +1\right ) a^{\frac {5}{2}}}{12}-\frac {\left (b \ln \left (x \right )-c +\frac {1}{4}\right ) b^{2} a^{\frac {3}{2}}}{12}+\frac {\left (c -\frac {1}{4}\right ) \ln \left (x \right )^{2} a^{\frac {7}{2}}}{3}-\frac {\sqrt {a}\, b^{4}}{48}-i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) c_{1} \operatorname {hypergeom}\left (\left [\frac {28 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {5}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+c_{1} \left (-\frac {a^{\frac {5}{2}} b}{4}+\left (-1-\frac {\ln \left (x \right )}{2}\right ) a^{\frac {7}{2}}+i \left (a \ln \left (x \right )+\frac {b}{2}\right )^{2} a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {\left (\left (c -\frac {1}{4}\right ) \ln \left (x \right ) a^{\frac {5}{2}}-\frac {b \left (b \ln \left (x \right )-2 c +\frac {1}{2}\right ) a^{\frac {3}{2}}}{4}-\frac {\sqrt {a}\, b^{3}}{8}-i \left (a \ln \left (x \right )+\frac {b}{2}\right ) a^{2}\right ) \operatorname {hypergeom}\left (\left [\frac {20 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )}{2}+\frac {\left (i a^{3} \ln \left (x \right )+\frac {i a^{2} b}{2}-\frac {a^{\frac {5}{2}}}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )}{2}}{a^{\frac {5}{2}} x \left (c_{1} \left (a \ln \left (x \right )+\frac {b}{2}\right ) \operatorname {hypergeom}\left (\left [\frac {12 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )+\frac {\operatorname {hypergeom}\left (\left [\frac {4 a^{\frac {3}{2}}+i \left (4 c -1\right ) a -i b^{2}}{16 a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 a \ln \left (x \right )+b \right )^{2}}{4 a^{\frac {3}{2}}}\right )}{2}\right )} \]
✓ Solution by Mathematica
Time used: 1.151 (sec). Leaf size: 868
DSolve[x^2*y'[x]==x^2*y[x]^2+a*(Log[x])^2+b*Log[x]+c,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {i b \operatorname {ParabolicCylinderD}\left (\frac {-i b^2-4 a^{3/2}+i a (4 c-1)}{8 a^{3/2}},-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )+2 i a \log (x) \operatorname {ParabolicCylinderD}\left (\frac {-i b^2-4 a^{3/2}+i a (4 c-1)}{8 a^{3/2}},-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )-\sqrt {a} \operatorname {ParabolicCylinderD}\left (\frac {-i b^2-4 a^{3/2}+i a (4 c-1)}{8 a^{3/2}},-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )+2 (-1)^{3/4} \sqrt {2} a^{3/4} \operatorname {ParabolicCylinderD}\left (\frac {-i b^2+4 a^{3/2}+i a (4 c-1)}{8 a^{3/2}},-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )-i c_1 \left (2 a \log (x)-i \sqrt {a}+b\right ) \operatorname {ParabolicCylinderD}\left (\frac {i b^2-4 a^{3/2}+i a-4 i a c}{8 a^{3/2}},\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )+2 \sqrt [4]{-1} \sqrt {2} a^{3/4} c_1 \operatorname {ParabolicCylinderD}\left (\frac {i b^2+4 a^{3/2}-i a (4 c-1)}{8 a^{3/2}},\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )}{2 \sqrt {a} x \left (\operatorname {ParabolicCylinderD}\left (\frac {-i b^2-4 a^{3/2}+i a (4 c-1)}{8 a^{3/2}},-\frac {\left (\frac {1}{2}-\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )+c_1 \operatorname {ParabolicCylinderD}\left (\frac {i b^2-4 a^{3/2}-i a (4 c-1)}{8 a^{3/2}},\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )\right )} \\ y(x)\to -\frac {-\frac {2 \sqrt [4]{-1} \sqrt {2} \sqrt [4]{a} \operatorname {ParabolicCylinderD}\left (\frac {i b^2+4 a^{3/2}-i a (4 c-1)}{8 a^{3/2}},\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )}{\operatorname {ParabolicCylinderD}\left (\frac {i b^2-4 a^{3/2}-i a (4 c-1)}{8 a^{3/2}},\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )}+\frac {i b}{\sqrt {a}}+2 i \sqrt {a} \log (x)+1}{2 x} \\ y(x)\to -\frac {-\frac {2 \sqrt [4]{-1} \sqrt {2} \sqrt [4]{a} \operatorname {ParabolicCylinderD}\left (\frac {i b^2+4 a^{3/2}-i a (4 c-1)}{8 a^{3/2}},\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )}{\operatorname {ParabolicCylinderD}\left (\frac {i b^2-4 a^{3/2}-i a (4 c-1)}{8 a^{3/2}},\frac {\left (\frac {1}{2}+\frac {i}{2}\right ) (b+2 a \log (x))}{a^{3/4}}\right )}+\frac {i b}{\sqrt {a}}+2 i \sqrt {a} \log (x)+1}{2 x} \\ \end{align*}