Internal problem ID [11053]
Internal file name [OUTPUT/10310_Wednesday_January_24_2024_10_07_02_PM_22005237/index.tex]
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 Equation of form
\((a_4 x^4+a_3 x^3+a_2 x^2 x+a_1 x+a_0) y''+f(x)y'+g(x)y=0\)
Problem number: 229.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "unknown"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {\left (a \,x^{2}+b \right )^{2} y^{\prime \prime }+\left (a \,x^{2}+b \right ) \left (c \,x^{2}+d \right ) y^{\prime }+2 \left (-a d +b c \right ) x y=0} \]
In 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 }+\left (-2 a d x +2 b c x \right ) 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}+\left (-\frac {2 c x}{a \,x^{2}+b}+\frac {2 \left (c \,x^{2}+d \right ) a x}{\left (a \,x^{2}+b \right )^{2}}-\frac {2 x \left (a d -b c \right )}{\left (a \,x^{2}+b \right )^{2}}\right ) \xi \left (x \right )&= 0 \end {align*}
Which is solved for \(\xi (x)\). This is second order ode with missing dependent variable \(\xi \left (x \right )\). Let \begin {align*} p(x) &= \xi ^{\prime }\left (x \right ) \end {align*}
Then \begin {align*} p'(x) &= \xi ^{\prime \prime }\left (x \right ) \end {align*}
Hence the ode becomes \begin {align*} p^{\prime }\left (x \right ) \left (a \,x^{2}+b \right )+\left (-c \,x^{2}-d \right ) p \left (x \right ) = 0 \end {align*}
Which is now solve for \(p(x)\) as first order ode. In canonical form the ODE is \begin {align*} p' &= F(x,p)\\ &= f( x) g(p)\\ &= \frac {p \left (c \,x^{2}+d \right )}{a \,x^{2}+b} \end {align*}
Where \(f(x)=\frac {c \,x^{2}+d}{a \,x^{2}+b}\) and \(g(p)=p\). Integrating both sides gives \begin {align*} \frac {1}{p} \,dp &= \frac {c \,x^{2}+d}{a \,x^{2}+b} \,d x\\ \int { \frac {1}{p} \,dp} &= \int {\frac {c \,x^{2}+d}{a \,x^{2}+b} \,d x}\\ \ln \left (p \right )&=\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}+c_{1}\\ p&={\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}+c_{1}}\\ &=c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} \end {align*}
Since \(p=\xi ^{\prime }\left (x \right )\) then the new first order ode to solve is \begin {align*} \xi ^{\prime }\left (x \right ) = c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} \end {align*}
Integrating both sides gives \begin {align*} \xi \left (x \right ) = \int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x +c_{2} \end {align*}
The original ode (2) 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 )}\\ y^{\prime }+y \left (\frac {c \,x^{2}+d}{a \,x^{2}+b}-\frac {c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}}{\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x +c_{2}}\right )&=0 \end {align*}
Which is now a first order ode. This is now solved for \(y\).
Entering Linear first order ODE solver. In canonical form a linear first order is \begin {align*} y^{\prime } + p(x)y &= q(x) \end {align*}
Where here \begin {align*} p(x) &=-\frac {{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} a \,x^{2}-c \,x^{2} \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-c \,x^{2} c_{2} +{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} b -d \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-d c_{2}}{\left (a \,x^{2}+b \right ) \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x +c_{2} \right )}\\ q(x) &=0 \end {align*}
Hence the ode is \begin {align*} y^{\prime }-\frac {y \left ({\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} a \,x^{2}-c \,x^{2} \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-c \,x^{2} c_{2} +{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} b -d \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-d c_{2} \right )}{\left (a \,x^{2}+b \right ) \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x +c_{2} \right )} = 0 \end {align*}
The integrating factor \(\mu \) is \[ \mu = {\mathrm e}^{\int -\frac {{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} a \,x^{2}-c \,x^{2} \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-c \,x^{2} c_{2} +{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} b -d \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-d c_{2}}{\left (a \,x^{2}+b \right ) \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x +c_{2} \right )}d x} \] The ode becomes \begin {align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \mu y &= 0 \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left ({\mathrm e}^{\int -\frac {{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} a \,x^{2}-c \,x^{2} \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-c \,x^{2} c_{2} +{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} b -d \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-d c_{2}}{\left (a \,x^{2}+b \right ) \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x +c_{2} \right )}d x} y\right ) &= 0 \end {align*}
Integrating gives \begin {align*} {\mathrm e}^{\int -\frac {{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} a \,x^{2}-c \,x^{2} \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-c \,x^{2} c_{2} +{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} b -d \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-d c_{2}}{\left (a \,x^{2}+b \right ) \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x +c_{2} \right )}d x} y &= c_{3} \end {align*}
Dividing both sides by the integrating factor \(\mu ={\mathrm e}^{\int -\frac {{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} a \,x^{2}-c \,x^{2} \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-c \,x^{2} c_{2} +{\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}} c_{1} b -d \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x \right )-d c_{2}}{\left (a \,x^{2}+b \right ) \left (\int c_{1} {\mathrm e}^{\frac {c x}{a}+\frac {\left (a d -b c \right ) \arctan \left (\frac {x a}{\sqrt {b a}}\right )}{a \sqrt {b a}}}d x +c_{2} \right )}d x}\) results in \begin {align*} y &= c_{3} {\mathrm e}^{\int \frac {-c_{1} \left (c \,x^{2}+d \right ) \left (\int {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}d x \right )+c_{1} \left (a \,x^{2}+b \right ) {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}-c_{2} \left (c \,x^{2}+d \right )}{\left (a \,x^{2}+b \right ) \left (c_{1} \left (\int {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}d x \right )+c_{2} \right )}d x} \end {align*}
Hence, the solution found using Lagrange adjoint equation method is \[ y = c_{3} {\mathrm e}^{\int \frac {-c_{1} \left (c \,x^{2}+d \right ) \left (\int {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}d x \right )+c_{1} \left (a \,x^{2}+b \right ) {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}-c_{2} \left (c \,x^{2}+d \right )}{\left (a \,x^{2}+b \right ) \left (c_{1} \left (\int {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}d x \right )+c_{2} \right )}d x} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{3} {\mathrm e}^{\int \frac {-c_{1} \left (c \,x^{2}+d \right ) \left (\int {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}d x \right )+c_{1} \left (a \,x^{2}+b \right ) {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}-c_{2} \left (c \,x^{2}+d \right )}{\left (a \,x^{2}+b \right ) \left (c_{1} \left (\int {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}d x \right )+c_{2} \right )}d x} \\ \end{align*}
Verification of solutions
\[ y = c_{3} {\mathrm e}^{\int \frac {-c_{1} \left (c \,x^{2}+d \right ) \left (\int {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}d x \right )+c_{1} \left (a \,x^{2}+b \right ) {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}-c_{2} \left (c \,x^{2}+d \right )}{\left (a \,x^{2}+b \right ) \left (c_{1} \left (\int {\mathrm e}^{\frac {c x \sqrt {b a}+\arctan \left (\frac {x a}{\sqrt {b a}}\right ) a d -\arctan \left (\frac {x a}{\sqrt {b a}}\right ) b c}{a \sqrt {b a}}}d x \right )+c_{2} \right )}d x} \] Verified OK.
Maple trace Kovacic algorithm successful
`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 @ Moebius -> 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`
✓ Solution by Maple
Time used: 0.312 (sec). Leaf size: 866
dsolve((a*x^2+b)^2*diff(y(x),x$2)+(a*x^2+b)*(c*x^2+d)*diff(y(x),x)+2*(b*c-a*d)*x*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (-a x +\sqrt {-a b}\right )^{\frac {2 a^{2} b +\sqrt {-a b \left (4 \sqrt {-a b}\, a^{2} d -4 \sqrt {-a b}\, a b c -4 a^{3} b +a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}}{4 a^{2} b}} \left (c_{1} \left (a x +\sqrt {-a b}\right )^{\frac {2 a^{2} b +\sqrt {4 a^{2} b \left (a d -b c \right ) \sqrt {-a b}+4 a^{4} b^{2}-a^{3} b \,d^{2}+2 d \,b^{2} c \,a^{2}-b^{3} c^{2} a}}{4 a^{2} b}} {\mathrm e}^{\frac {\sqrt {-a b}\, c}{2 a^{2}}-\frac {\arctan \left (\frac {\sqrt {a}\, x}{\sqrt {b}}\right ) d}{2 \sqrt {a}\, \sqrt {b}}+\frac {\sqrt {b}\, \arctan \left (\frac {\sqrt {a}\, x}{\sqrt {b}}\right ) c}{2 a^{\frac {3}{2}}}} \operatorname {HeunC}\left (\frac {2 \sqrt {-\frac {b}{a}}\, c}{a}, \frac {\sqrt {4 a^{2} b \left (a d -b c \right ) \sqrt {-a b}+4 a^{4} b^{2}-a^{3} b \,d^{2}+2 d \,b^{2} c \,a^{2}-b^{3} c^{2} a}}{2 a^{2} b}, \frac {\sqrt {-a b \left (4 \sqrt {-a b}\, a^{2} d -4 \sqrt {-a b}\, a b c -4 a^{3} b +a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}}{2 a^{2} b}, 0, \frac {1}{2}-\frac {d^{2}}{8 a b}-\frac {c d}{4 a^{2}}+\frac {3 b \,c^{2}}{8 a^{3}}, \frac {a x}{2 \sqrt {-a b}}+\frac {1}{2}\right )+c_{2} \left (a x +\sqrt {-a b}\right )^{-\frac {-2 a^{2} b +\sqrt {4 a^{2} b \left (a d -b c \right ) \sqrt {-a b}+4 a^{4} b^{2}-a^{3} b \,d^{2}+2 d \,b^{2} c \,a^{2}-b^{3} c^{2} a}}{4 a^{2} b}} \operatorname {HeunC}\left (\frac {2 \sqrt {-\frac {b}{a}}\, c}{a}, -\frac {\sqrt {4 a^{2} b \left (a d -b c \right ) \sqrt {-a b}+4 a^{4} b^{2}-a^{3} b \,d^{2}+2 d \,b^{2} c \,a^{2}-b^{3} c^{2} a}}{2 a^{2} b}, \frac {\sqrt {-a b \left (4 \sqrt {-a b}\, a^{2} d -4 \sqrt {-a b}\, a b c -4 a^{3} b +a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}}{2 a^{2} b}, 0, \frac {1}{2}-\frac {d^{2}}{8 a b}-\frac {c d}{4 a^{2}}+\frac {3 b \,c^{2}}{8 a^{3}}, \frac {a x}{2 \sqrt {-a b}}+\frac {1}{2}\right ) {\mathrm e}^{\frac {i \pi \sqrt {4 a^{2} b \left (a d -b c \right ) \sqrt {-a b}+4 a^{4} b^{2}-a^{3} b \,d^{2}+2 d \,b^{2} c \,a^{2}-b^{3} c^{2} a}-i \pi \sqrt {-a b \left (4 \sqrt {-a b}\, a^{2} d -4 \sqrt {-a b}\, a b c -4 a^{3} b +a^{2} d^{2}-2 a b c d +b^{2} c^{2}\right )}-4 \left (a^{2} \left (\frac {d}{\sqrt {b}\, \sqrt {a}}-\frac {\sqrt {b}\, c}{a^{\frac {3}{2}}}\right ) \arctan \left (\frac {\sqrt {a}\, x}{\sqrt {b}}\right )-\sqrt {-a b}\, c \right ) b}{8 a^{2} b}}\right ) \]
✓ Solution by Mathematica
Time used: 0.165 (sec). Leaf size: 104
DSolve[(a*x^2+b)^2*y''[x]+(a*x^2+b)*(c*x^2+d)*y'[x]+2*(b*c-a*d)*x*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \exp \left (\frac {\arctan \left (\frac {\sqrt {a} x}{\sqrt {b}}\right ) (b c-a d)}{a^{3/2} \sqrt {b}}-\frac {c x}{a}\right ) \left (\int _1^x\exp \left (\frac {(a d-b c) \arctan \left (\frac {\sqrt {a} K[1]}{\sqrt {b}}\right )}{a^{3/2} \sqrt {b}}+\frac {c K[1]}{a}\right ) c_1dK[1]+c_2\right ) \]