2.4.19 Problem 40

Internal problem ID [13319]
Book : Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section : Chapter 1, section 1.2. Riccati Equation. subsection 1.2.3-2. Equations with power and exponential functions
Problem number : 40
Date solved : Sunday, January 18, 2026 at 07:15:20 PM
CAS classification : [_Riccati]

2.4.19.1 Solved using first_order_ode_riccati

1.186 (sec)

Entering first order ode riccati solver

\begin{align*} x^{4} \left (y^{\prime }-y^{2}\right )&=a +b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}} \\ \end{align*}

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {y^{2} x^{4}+b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}}+a}{x^{4}} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = y^{2}+\frac {b \,{\mathrm e}^{\frac {k}{x}}}{x^{4}}+\frac {c \,{\mathrm e}^{\frac {2 k}{x}}}{x^{4}}+\frac {a}{x^{4}} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=\frac {a}{x^{4}}+\frac {b \,{\mathrm e}^{\frac {k}{x}}}{x^{4}}+\frac {c \,{\mathrm e}^{\frac {2 k}{x}}}{x^{4}}\), \(f_1(x)=0\) 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}{x^{4}}+\frac {b \,{\mathrm e}^{\frac {k}{x}}}{x^{4}}+\frac {c \,{\mathrm e}^{\frac {2 k}{x}}}{x^{4}} \end{align*}

Substituting the above terms back in equation (2) gives

\[ u^{\prime \prime }\left (x \right )+\left (\frac {a}{x^{4}}+\frac {b \,{\mathrm e}^{\frac {k}{x}}}{x^{4}}+\frac {c \,{\mathrm e}^{\frac {2 k}{x}}}{x^{4}}\right ) u \left (x \right ) = 0 \]

Unable to solve. Will ask Maple to solve this ode now.

The solution for \(u \left (x \right )\) is

\begin{equation} \tag{3} u \left (x \right ) = c_1 \,{\mathrm e}^{-\frac {k}{2 x}} x \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )+c_2 \,{\mathrm e}^{-\frac {k}{2 x}} x \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) \end{equation}

Taking derivative gives

\begin{equation} \tag{4} u^{\prime }\left (x \right ) = \frac {c_1 k \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 x}+c_1 \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\frac {2 i c_1 \,{\mathrm e}^{-\frac {k}{2 x}} \left (\left (\frac {1}{2}+\frac {b \,{\mathrm e}^{-\frac {k}{x}}}{4 c}\right ) \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\frac {i \left (\frac {1}{2}+\frac {i \sqrt {a}}{k}-\frac {i b}{2 k \sqrt {c}}\right ) k \,{\mathrm e}^{-\frac {k}{x}} \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 \sqrt {c}}\right ) \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{x}+\frac {c_2 k \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 x}+c_2 \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\frac {2 i c_2 \,{\mathrm e}^{-\frac {k}{2 x}} \left (\left (\frac {1}{2}+\frac {b \,{\mathrm e}^{-\frac {k}{x}}}{4 c}\right ) \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )+\frac {i k \,{\mathrm e}^{-\frac {k}{x}} \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 \sqrt {c}}\right ) \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{x} \end{equation}

Substituting equations (3,4) into (1) results in

\begin{align*} y &= \frac {-u'}{f_2 u} \\ y &= \frac {-u'}{u} \\ y &= -\frac {\frac {c_1 k \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 x}+c_1 \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\frac {2 i c_1 \,{\mathrm e}^{-\frac {k}{2 x}} \left (\left (\frac {1}{2}+\frac {b \,{\mathrm e}^{-\frac {k}{x}}}{4 c}\right ) \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\frac {i \left (\frac {1}{2}+\frac {i \sqrt {a}}{k}-\frac {i b}{2 k \sqrt {c}}\right ) k \,{\mathrm e}^{-\frac {k}{x}} \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 \sqrt {c}}\right ) \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{x}+\frac {c_2 k \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 x}+c_2 \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\frac {2 i c_2 \,{\mathrm e}^{-\frac {k}{2 x}} \left (\left (\frac {1}{2}+\frac {b \,{\mathrm e}^{-\frac {k}{x}}}{4 c}\right ) \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )+\frac {i k \,{\mathrm e}^{-\frac {k}{x}} \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 \sqrt {c}}\right ) \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{x}}{c_1 \,{\mathrm e}^{-\frac {k}{2 x}} x \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )+c_2 \,{\mathrm e}^{-\frac {k}{2 x}} x \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )} \\ \end{align*}

Doing change of constants, the above solution becomes

\[ y = -\frac {\frac {k \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 x}+{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\frac {2 i {\mathrm e}^{-\frac {k}{2 x}} \left (\left (\frac {1}{2}+\frac {b \,{\mathrm e}^{-\frac {k}{x}}}{4 c}\right ) \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\frac {i \left (\frac {1}{2}+\frac {i \sqrt {a}}{k}-\frac {i b}{2 k \sqrt {c}}\right ) k \,{\mathrm e}^{-\frac {k}{x}} \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 \sqrt {c}}\right ) \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{x}+\frac {c_3 k \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 x}+c_3 \,{\mathrm e}^{-\frac {k}{2 x}} \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\frac {2 i c_3 \,{\mathrm e}^{-\frac {k}{2 x}} \left (\left (\frac {1}{2}+\frac {b \,{\mathrm e}^{-\frac {k}{x}}}{4 c}\right ) \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )+\frac {i k \,{\mathrm e}^{-\frac {k}{x}} \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )}{2 \sqrt {c}}\right ) \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{x}}{{\mathrm e}^{-\frac {k}{2 x}} x \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )+c_3 \,{\mathrm e}^{-\frac {k}{2 x}} x \operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )} \]

Simplifying the above gives

\begin{align*} y &= \frac {{\mathrm e}^{\frac {k}{x}} {\mathrm e}^{-\frac {k}{x}} \left (\left (i \sqrt {a}+\frac {k}{2}\right ) c^{{3}/{2}}-\frac {i b c}{2}\right ) \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c^{{3}/{2}} {\mathrm e}^{\frac {k}{x}} {\mathrm e}^{-\frac {k}{x}} c_3 k +\frac {\left (i {\mathrm e}^{\frac {k}{x}} {\mathrm e}^{-\frac {k}{x}} b c +2 i {\mathrm e}^{\frac {k}{x}} c^{2}-\left (k +2 x \right ) c^{{3}/{2}}\right ) \left (\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_3 +\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )}{2}}{c^{{3}/{2}} x^{2} \left (\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_3 +\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= \frac {{\mathrm e}^{\frac {k}{x}} {\mathrm e}^{-\frac {k}{x}} \left (\left (i \sqrt {a}+\frac {k}{2}\right ) c^{{3}/{2}}-\frac {i b c}{2}\right ) \operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}+1, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c^{{3}/{2}} {\mathrm e}^{\frac {k}{x}} {\mathrm e}^{-\frac {k}{x}} c_3 k +\frac {\left (i {\mathrm e}^{\frac {k}{x}} {\mathrm e}^{-\frac {k}{x}} b c +2 i {\mathrm e}^{\frac {k}{x}} c^{2}-\left (k +2 x \right ) c^{{3}/{2}}\right ) \left (\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_3 +\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )}{2}}{c^{{3}/{2}} x^{2} \left (\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_3 +\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )} \\ \end{align*}

2.4.19.2 ✓ Maple. Time used: 0.004 (sec). Leaf size: 302

ode:=x^4*(diff(y(x),x)-y(x)^2) = a+b*exp(k/x)+c*exp(2*k/x); 
dsolve(ode,y(x), singsol=all);
 

\[ y = \frac {\left (\left (i \sqrt {a}+\frac {k}{2}\right ) \sqrt {c}-\frac {i b}{2}\right ) \operatorname {WhittakerM}\left (-\frac {i b -2 k \sqrt {c}}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )-c_1 k \operatorname {WhittakerW}\left (-\frac {i b -2 k \sqrt {c}}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) \sqrt {c}+\left (i {\mathrm e}^{\frac {k}{x}} c +\left (-\frac {k}{2}-x \right ) \sqrt {c}+\frac {i b}{2}\right ) \left (\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_1 +\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )}{\sqrt {c}\, x^{2} \left (\operatorname {WhittakerW}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right ) c_1 +\operatorname {WhittakerM}\left (-\frac {i b}{2 k \sqrt {c}}, \frac {i \sqrt {a}}{k}, \frac {2 i \sqrt {c}\, {\mathrm e}^{\frac {k}{x}}}{k}\right )\right )} \]

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+b*exp(k/x)+c*exp 
(2*k/x))/x^4*y(x), 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 symmetry of the form [xi=0, eta=F(x)] 
         checking if the LODE is missing y 
         -> Trying an equivalence, under non-integer power transformations, 
            to LODEs admitting Liouvillian solutions. 
            -> Trying a Liouvillian solution using Kovacics algorithm 
            <- No Liouvillian solutions exists 
         -> Trying a solution in terms of special functions: 
            -> Bessel 
            -> elliptic 
            -> Legendre 
            -> Whittaker 
               -> hyper3: Equivalence to 1F1 under a power @ Moebius 
               <- hyper3 successful: received ODE is equivalent to the 1F1 ODE 
            <- Whittaker successful 
         <- special function solution successful 
         Change of variables used: 
            [x = 1/ln(t)] 
         Linear ODE actually solved: 
            (ln(t)*a+ln(t)*b*t^k+ln(t)*c*(t^k)^2)*u(t)+(t*ln(t)+2*t)*diff(u(t),\ 
t)+ln(t)*t^2*diff(diff(u(t),t),t) = 0 
      <- change of variables successful 
   <- Riccati to 2nd Order successful
 

Maple step by step

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{4} \left (\frac {d}{d x}y \left (x \right )-y \left (x \right )^{2}\right )=a +b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {x^{4} y \left (x \right )^{2}+b \,{\mathrm e}^{\frac {k}{x}}+c \,{\mathrm e}^{\frac {2 k}{x}}+a}{x^{4}} \end {array} \]

2.4.19.3 ✓ Mathematica. Time used: 1.375 (sec). Leaf size: 940

ode=x^4*(D[y[x],x]-y[x]^2)==a+b*Exp[k/x]+c*Exp[2*k/x]; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

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

2.4.19.4 ✗ Sympy

from sympy import * 
x = symbols("x") 
a = symbols("a") 
b = symbols("b") 
c = symbols("c") 
k = symbols("k") 
y = Function("y") 
ode = Eq(-a - b*exp(k/x) - c*exp(2*k/x) + x**4*(-y(x)**2 + Derivative(y(x), x)),0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
 

NotImplementedError : The given ODE Derivative(y(x), x) - (a + b*exp(k/x) + c*exp(2*k/x) + x**4*y(x)
 

Python version: 3.12.3 (main, Aug 14 2025, 17:47:21) [GCC 13.3.0] 
Sympy version 1.14.0
 

classify_ode(ode,func=y(x)) 
 
('factorable', 'lie_group')