Problem 229

2.33.20 Problem 229

2.33.20.1 Solved as second order ode adjoint method
2.33.20.2 ✓Maple
2.33.20.3 ✓Mathematica
2.33.20.4 ✗Sympy

Internal problem ID [13890]
Book : Handbook of exact solutions for ordinary diﬀerential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 2, Second-Order Diﬀerential Equations. section 2.1.2-7
Problem number : 229
Date solved : Thursday, January 01, 2026 at 03:26:31 AM
CAS classiﬁcation : [[_2nd_order, _with_linear_symmetries]]

2.33.20.1 Solved as second order ode adjoint method

6.842 (sec)

\begin{align*} \left (a \,x^{2}+b \right )^{2} y^{\prime \prime }+\left (a \,x^{2}+b \right ) \left (c \,x^{2}+d \right ) y^{\prime }+2 \left (-a d +b c \right ) x y&=0 \\ \end{align*}

Entering second order ode lagrange adjoint equation method solverIn normal form the ode

\begin{align*} \left (a \,x^{2}+b \right )^{2} y^{\prime \prime }+\left (a \,x^{2}+b \right ) \left (c \,x^{2}+d \right ) y^{\prime }+2 \left (-a d +b c \right ) x y = 0 \tag {1} \end{align*}

Becomes

\begin{align*} y^{\prime \prime }+p \left (x \right ) y^{\prime }+q \left (x \right ) y&=r \left (x \right ) \tag {2} \end{align*}

Where

\begin{align*} p \left (x \right )&=\frac {c \,x^{2}+d}{a \,x^{2}+b}\\ q \left (x \right )&=-\frac {2 x \left (a d -b c \right )}{\left (a \,x^{2}+b \right )^{2}}\\ r \left (x \right )&=0 \end{align*}

The Lagrange adjoint ode is given by

\begin{align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (\frac {\left (c \,x^{2}+d \right ) \xi \left (x \right )}{a \,x^{2}+b}\right )' + \left (-\frac {2 x \left (a d -b c \right ) \xi \left (x \right )}{\left (a \,x^{2}+b \right )^{2}}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )-\frac {\left (c \,x^{2}+d \right ) \xi ^{\prime }\left (x \right )}{a \,x^{2}+b}&= 0 \end{align*}

Which is solved for \(\xi (x)\). Entering second order ode missing \(y\) solverThis is second order ode with missing dependent variable \(\xi \). Let

\begin{align*} u(x) &= \xi ^{\prime } \end{align*}

Then

\begin{align*} u'(x) &= \xi ^{\prime \prime } \end{align*}

Hence the ode becomes

\begin{align*} u^{\prime }\left (x \right )-\frac {\left (c \,x^{2}+d \right ) u \left (x \right )}{a \,x^{2}+b} = 0 \end{align*}

Which is now solved for \(u(x)\) as ﬁrst order ode.

Entering ﬁrst order ode linear solverIn canonical form a linear ﬁrst order is

\begin{align*} u^{\prime }\left (x \right ) + q(x)u \left (x \right ) &= p(x) \end{align*}

Comparing the above to the given ode shows that

\begin{align*} q(x) &=-\frac {c \,x^{2}+d}{a \,x^{2}+b}\\ p(x) &=0 \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int -\frac {c \,x^{2}+d}{a \,x^{2}+b}d x}\\ &= {\mathrm e}^{\frac {-c x -\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {a b}}\right )}{\sqrt {a b}}}{a}} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu u &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (u \,{\mathrm e}^{\frac {-c x -\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {a b}}\right )}{\sqrt {a b}}}{a}}\right ) &= 0 \end{align*}

Integrating gives

\begin{align*} u \,{\mathrm e}^{\frac {-c x -\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {a b}}\right )}{\sqrt {a b}}}{a}}&= \int {0 \,dx} + c_3 \\ &=c_3 \end{align*}

Dividing throughout by the integrating factor \({\mathrm e}^{\frac {-c x -\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {a b}}\right )}{\sqrt {a b}}}{a}}\) gives the ﬁnal solution

\[ u \left (x \right ) = {\mathrm e}^{-\frac {-c x -\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {a b}}\right )}{\sqrt {a b}}}{a}} c_3 \]

Simplifying the above gives

\begin{align*} u \left (x \right ) &= {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 \\ \end{align*}

In summary, these are the solution found for \(\xi \)

For solution \(u \left (x \right ) = {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3\), since \(u=\xi ^{\prime }\) then we now have a new ﬁrst order ode to solve which is

\begin{align*} \xi ^{\prime } = {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 \end{align*}

Entering ﬁrst order ode quadrature solverSince the ode has the form \(\xi ^{\prime }=f(x)\), then we only need to integrate \(f(x)\).

\begin{align*} \int {d\xi } &= \int {{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3\, dx}\\ \xi &= \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x + c_4 \end{align*}

\begin{align*} \xi &= \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c_4 \end{align*}

In summary, these are the solution found for \((\xi )\)

The original ode now reduces to ﬁrst order ode

\begin{align*} \xi \left (x \right ) y^{\prime }-y \xi ^{\prime }\left (x \right )+\xi \left (x \right ) p \left (x \right ) y&=\int \xi \left (x \right ) r \left (x \right )d x\\ y^{\prime }+y \left (p \left (x \right )-\frac {\xi ^{\prime }\left (x \right )}{\xi \left (x \right )}\right )&=\frac {\int \xi \left (x \right ) r \left (x \right )d x}{\xi \left (x \right )} \end{align*}

\begin{align*} y^{\prime }+y \left (\frac {c \,x^{2}+d}{a \,x^{2}+b}-\frac {{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3}{\int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c_4}\right )&=0 \end{align*}

Which is now a ﬁrst order ode. This is now solved for \(y\). Entering ﬁrst order ode separable solverThe ode

\begin{equation} y^{\prime } = -\frac {y \left (-{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 a \,x^{2}+c \,x^{2} \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c \,x^{2} c_4 -{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 b +d \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +d c_4 \right )}{\left (a \,x^{2}+b \right ) \left (\int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c_4 \right )} \end{equation}

is separable as it can be written as

\begin{align*} y^{\prime }&= -\frac {y \left (-{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 a \,x^{2}+c \,x^{2} \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c \,x^{2} c_4 -{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 b +d \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +d c_4 \right )}{\left (a \,x^{2}+b \right ) \left (\int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c_4 \right )}\\ &= f(x) g(y) \end{align*}

Where

\begin{align*} f(x) &= -\frac {-{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 a \,x^{2}+c \,x^{2} \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c \,x^{2} c_4 -{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 b +d \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +d c_4}{\left (a \,x^{2}+b \right ) \left (\int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c_4 \right )}\\ g(y) &= y \end{align*}

Integrating gives

\begin{align*} \int { \frac {1}{g(y)} \,dy} &= \int { f(x) \,dx} \\ \int { \frac {1}{y}\,dy} &= \int { -\frac {-{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 a \,x^{2}+c \,x^{2} \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c \,x^{2} c_4 -{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 b +d \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +d c_4}{\left (a \,x^{2}+b \right ) \left (\int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c_4 \right )} \,dx} \\ \end{align*}

\[ \ln \left (y\right )=\int -\frac {-{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 a \,x^{2}+c \,x^{2} \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c \,x^{2} c_4 -{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 b +d \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +d c_4}{\left (a \,x^{2}+b \right ) \left (\int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c_4 \right )}d x +c_5 \]

We now need to ﬁnd the singular solutions, these are found by ﬁnding for what values \(g(y)\) is zero, since we had to divide by this above. Solving \(g(y)=0\) or

\[ y=0 \]

for \(y\) gives

\begin{align*} y&=0 \end{align*}

Now we go over each such singular solution and check if it veriﬁes the ode itself and any initial conditions given. If it does not then the singular solution will not be used.

Therefore the solutions found are

\begin{align*} \ln \left (y\right ) &= \int -\frac {-{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 a \,x^{2}+c \,x^{2} \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c \,x^{2} c_4 -{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 b +d \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +d c_4}{\left (a \,x^{2}+b \right ) \left (\int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c +c x \sqrt {a b}}{a \sqrt {a b}}} c_3 d x +c_4 \right )}d x +c_5 \\ y &= 0 \\ \end{align*}

Solving for \(y\) gives

\begin{align*} y &= 0 \\ y &= {\mathrm e}^{c_3 a \int \frac {{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}} x^{2}}{\left (a \,x^{2}+b \right ) \left (c_3 \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}}d x +c_4 \right )}d x +c_3 b \int \frac {{\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}}}{\left (a \,x^{2}+b \right ) \left (c_3 \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}}d x +c_4 \right )}d x -c c_3 \int \frac {x^{2} \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}}d x}{\left (a \,x^{2}+b \right ) \left (c_3 \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}}d x +c_4 \right )}d x -d c_3 \int \frac {\int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}}d x}{\left (a \,x^{2}+b \right ) \left (c_3 \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}}d x +c_4 \right )}d x -c c_4 \int \frac {x^{2}}{\left (a \,x^{2}+b \right ) \left (c_3 \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}}d x +c_4 \right )}d x -d c_4 \int \frac {1}{\left (a \,x^{2}+b \right ) \left (c_3 \int {\mathrm e}^{\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) d}{\sqrt {a b}}} {\mathrm e}^{-\frac {\arctan \left (\frac {x a}{\sqrt {a b}}\right ) b c}{a \sqrt {a b}}} {\mathrm e}^{\frac {c x}{a}}d x +c_4 \right )}d x +c_5} \\ \end{align*}

Hence, the solution found using Lagrange adjoint equation method is

Summary of solutions found

2.33.20.2 ✓ Maple. Time used: 0.055 (sec). Leaf size: 818

ode:=(a*x^2+b)^2*diff(diff(y(x),x),x)+(a*x^2+b)*(c*x^2+d)*diff(y(x),x)+2*(-a*d+b*c)*x*y(x) = 0; 
dsolve(ode,y(x), singsol=all);

\begin{align*} \text {Solution too large to show}\end{align*}

Maple trace

Methods for second order ODEs: 
--- Trying classification methods --- 
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 
   A Liouvillian solution exists 
   Reducible group (found an exponential solution) 
   Group is reducible, not completely reducible 
   Solution has integrals. Trying a special function solution free of integrals\ 
... 
   -> 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 @ Mo\ 
ebius 
      -> Heun: Equivalence to the GHE or one of its 4 confluent cases under a \ 
power @ Moebius 
      <- Heun successful: received ODE is equivalent to the  HeunC  ODE, case  \ 
a <> 0, e <> 0, c = 0 
   <- Kovacics algorithm successful

2.33.20.3 ✓ Mathematica. Time used: 60.09 (sec). Leaf size: 104

ode=(a*x^2+b)^2*D[y[x],{x,2}]+(a*x^2+b)*(c*x^2+d)*D[y[x],x]+2*(b*c-a*d)*x*y[x]==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]

\begin{align*} y(x)&\to \exp \left (\frac {\arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right ) (b c-a d)}{a^{3/2} \sqrt {b}}-\frac {c x}{a}\right ) \left (\int _1^x\exp \left (\frac {(a d-b c) \arctan \left (\frac {\sqrt {a} K[1]}{\sqrt {b}}\right )}{a^{3/2} \sqrt {b}}+\frac {c K[1]}{a}\right ) c_1dK[1]+c_2\right ) \end{align*}

2.33.20.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
d = symbols("d") 
y = Function("y") 
ode = Eq(x*(-2*a*d + 2*b*c)*y(x) + (a*x**2 + b)**2*Derivative(y(x), (x, 2)) + (a*x**2 + b)*(c*x**2 + d)*Derivative(y(x), x),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)

False

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