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2.2.20 Problem 19

Solved as second order Airy ode
✓Maple
✓Mathematica
✗Sympy

Internal problem ID [10098]
Book : Own collection of miscellaneous problems
Section : section 2.0
Problem number : 19
Date solved : Thursday, November 27, 2025 at 10:18:52 AM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

Solved as second order Airy ode

Time used: 0.690 (sec)

Solve

\begin{align*} y^{\prime \prime }-y^{\prime }-x y-x^{3}+1&=0 \\ \end{align*}
This is Airy ODE. It has the general form
\[ a y^{\prime \prime } + b y^{\prime } + c x y = F(x) \]
Where in this case
\begin{align*} a &= 1\\ b &= -1\\ c &= -1\\ F &= x^{3}-1 \end{align*}

Therefore the solution to the homogeneous Airy ODE becomes

\[ y = c_1 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )+c_2 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) \]
Since this is inhomogeneous Airy ODE, then we need to find the particular solution. The particular solution \(y_p\) can be found using either the method of undetermined coefficients, or the method of variation of parameters. The method of variation of parameters will be used as it is more general and can be used when the coefficients of the ODE depend on \(x\) as well. Let
\begin{equation} \tag{1} y_p(x) = u_1 y_1 + u_2 y_2 \end{equation}
Where \(u_1,u_2\) to be determined, and \(y_1,y_2\) are the two basis solutions (the two linearly independent solutions of the homogeneous ODE) found earlier when solving the homogeneous ODE as
\begin{align*} y_1 &= {\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) \\ y_2 &= {\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) \\ \end{align*}
In the Variation of parameters \(u_1,u_2\) are found using
\begin{align*} \tag{2} u_1 &= -\int \frac {y_2 f(x)}{a W(x)} \\ \tag{3} u_2 &= \int \frac {y_1 f(x)}{a W(x)} \\ \end{align*}
Where \(W(x)\) is the Wronskian and \(a\) is the coefficient in front of \(y''\) in the given ODE. The Wronskian is given by \(W= \begin {vmatrix} y_1 & y_{2} \\ y_{1}^{\prime } & y_{2}^{\prime } \end {vmatrix} \). Hence
\[ W = \begin {vmatrix} {\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) & {\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) \\ \frac {d}{dx}\left ({\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )\right ) & \frac {d}{dx}\left ({\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )\right ) \end {vmatrix} \]
Which gives
\[ W = \begin {vmatrix} {\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) & {\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) \\ \frac {{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )}{2}-{\mathrm e}^{\frac {x}{2}} \left (-1\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, \left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) & \frac {{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )}{2}-{\mathrm e}^{\frac {x}{2}} \left (-1\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, \left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) \end {vmatrix} \]
Therefore
\[ W = \left ({\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )\right )\left (\frac {{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )}{2}-{\mathrm e}^{\frac {x}{2}} \left (-1\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, \left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )\right ) - \left ({\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )\right )\left (\frac {{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )}{2}-{\mathrm e}^{\frac {x}{2}} \left (-1\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, \left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )\right ) \]
Which simplifies to
\[ W = -{\mathrm e}^{x} \operatorname {AiryAi}\left (-\frac {\left (4 x +1\right ) \left (-1\right )^{{1}/{3}}}{4}\right ) \left (-1\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\frac {\left (4 x +1\right ) \left (-1\right )^{{1}/{3}}}{4}\right )+{\mathrm e}^{x} \operatorname {AiryBi}\left (-\frac {\left (4 x +1\right ) \left (-1\right )^{{1}/{3}}}{4}\right ) \left (-1\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\frac {\left (4 x +1\right ) \left (-1\right )^{{1}/{3}}}{4}\right ) \]
Which simplifies to
\[ W = -\frac {{\mathrm e}^{x} \left (1+i \sqrt {3}\right )}{2 \pi } \]
Therefore Eq. (2) becomes
\[ u_1 = -\int \frac {{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) \left (x^{3}-1\right )}{-\frac {{\mathrm e}^{x} \left (1+i \sqrt {3}\right )}{2 \pi }}\,dx \]
Which simplifies to
\[ u_1 = - \int \frac {\left (-2 x^{3}+2\right ) \pi \,{\mathrm e}^{-\frac {x}{2}} \operatorname {AiryBi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )}{1+i \sqrt {3}}d x \]
Hence
\[ u_1 = -\int _{0}^{x}\frac {\left (-2 \alpha ^{3}+2\right ) \pi \,{\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryBi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )}{1+i \sqrt {3}}d \alpha \]
And Eq. (3) becomes
\[ u_2 = \int \frac {{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) \left (x^{3}-1\right )}{-\frac {{\mathrm e}^{x} \left (1+i \sqrt {3}\right )}{2 \pi }}\,dx \]
Which simplifies to
\[ u_2 = \int \frac {\left (-2 x^{3}+2\right ) \pi \,{\mathrm e}^{-\frac {x}{2}} \operatorname {AiryAi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )}{1+i \sqrt {3}}d x \]
Hence
\[ u_2 = \int _{0}^{x}\frac {\left (-2 \alpha ^{3}+2\right ) \pi \,{\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryAi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )}{1+i \sqrt {3}}d \alpha \]
Therefore the particular solution, from equation (1) is
\[ y_p(x) = -\int _{0}^{x}\frac {\left (-2 \alpha ^{3}+2\right ) \pi \,{\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryBi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )}{1+i \sqrt {3}}d \alpha {\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )+{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right ) \int _{0}^{x}\frac {\left (-2 \alpha ^{3}+2\right ) \pi \,{\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryAi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )}{1+i \sqrt {3}}d \alpha \]
Which simplifies to
\[ y_p(x) = -\frac {2 \,{\mathrm e}^{\frac {x}{2}} \pi \left (\operatorname {AiryBi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right ) \int _{0}^{x}\left (\alpha ^{3}-1\right ) {\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryAi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )d \alpha -\operatorname {AiryAi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right ) \int _{0}^{x}\left (\alpha ^{3}-1\right ) {\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryBi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )d \alpha \right )}{1+i \sqrt {3}} \]
Therefore the general solution is
\begin{align*} y &= y_h + y_p \\ &= \left (c_1 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )+c_2 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )\right ) + \left (-\frac {2 \,{\mathrm e}^{\frac {x}{2}} \pi \left (\operatorname {AiryBi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right ) \int _{0}^{x}\left (\alpha ^{3}-1\right ) {\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryAi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )d \alpha -\operatorname {AiryAi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right ) \int _{0}^{x}\left (\alpha ^{3}-1\right ) {\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryBi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )d \alpha \right )}{1+i \sqrt {3}}\right ) \\ &= c_1 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )+c_2 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )-\frac {2 \,{\mathrm e}^{\frac {x}{2}} \pi \left (\operatorname {AiryBi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right ) \int _{0}^{x}\left (\alpha ^{3}-1\right ) {\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryAi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )d \alpha -\operatorname {AiryAi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right ) \int _{0}^{x}\left (\alpha ^{3}-1\right ) {\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryBi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )d \alpha \right )}{1+i \sqrt {3}} \\ \end{align*}

Summary of solutions found

\begin{align*} y &= c_1 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )+c_2 \,{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\left (-x -\frac {1}{4}\right ) \left (-1\right )^{{1}/{3}}\right )-\frac {2 \,{\mathrm e}^{\frac {x}{2}} \pi \left (\operatorname {AiryBi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right ) \int _{0}^{x}\left (\alpha ^{3}-1\right ) {\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryAi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )d \alpha -\operatorname {AiryAi}\left (-\frac {\left (x +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right ) \int _{0}^{x}\left (\alpha ^{3}-1\right ) {\mathrm e}^{-\frac {\alpha }{2}} \operatorname {AiryBi}\left (-\frac {\left (\alpha +\frac {1}{4}\right ) \left (1+i \sqrt {3}\right )}{2}\right )d \alpha \right )}{1+i \sqrt {3}} \\ \end{align*}
✓ Maple. Time used: 0.005 (sec). Leaf size: 67
ode:=diff(diff(y(x),x),x)-diff(y(x),x)-y(x)*x-x^3+1 = 0; 
dsolve(ode,y(x), singsol=all);
\[ y = {\mathrm e}^{\frac {x}{2}} \left (\pi \int {\mathrm e}^{-\frac {x}{2}} \operatorname {AiryAi}\left (\frac {1}{4}+x \right ) \left (x^{3}-1\right )d x \operatorname {AiryBi}\left (\frac {1}{4}+x \right )-\pi \int {\mathrm e}^{-\frac {x}{2}} \operatorname {AiryBi}\left (\frac {1}{4}+x \right ) \left (x^{3}-1\right )d x \operatorname {AiryAi}\left (\frac {1}{4}+x \right )+\operatorname {AiryBi}\left (\frac {1}{4}+x \right ) c_1 +\operatorname {AiryAi}\left (\frac {1}{4}+x \right ) c_2 \right ) \]

Maple trace 

Methods for second order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
trying high order exact linear fully integrable 
trying differential order: 2; linear nonhomogeneous with symmetry [0,1] 
trying a double symmetry of the form [xi=0, eta=F(x)] 
-> Try solving first the homogeneous part of the ODE 
   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 
      <- Bessel successful 
   <- special function solution successful 
<- solving first the homogeneous part of the ODE successful
✓ Mathematica. Time used: 0.096 (sec). Leaf size: 107
ode=D[y[x],{x,2}]-D[y[x],x]-x*y[x]-x^3+1==0; 
ic={}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
\begin{align*} y(x)&\to e^{x/2} \left (\operatorname {AiryAi}\left (x+\frac {1}{4}\right ) \int _1^x-e^{-\frac {K[1]}{2}} \pi \operatorname {AiryBi}\left (K[1]+\frac {1}{4}\right ) \left (K[1]^3-1\right )dK[1]+\operatorname {AiryBi}\left (x+\frac {1}{4}\right ) \int _1^xe^{-\frac {K[2]}{2}} \pi \operatorname {AiryAi}\left (K[2]+\frac {1}{4}\right ) \left (K[2]^3-1\right )dK[2]+c_1 \operatorname {AiryAi}\left (x+\frac {1}{4}\right )+c_2 \operatorname {AiryBi}\left (x+\frac {1}{4}\right )\right ) \end{align*}
✗ Sympy
from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(-x**3 - x*y(x) - Derivative(y(x), x) + Derivative(y(x), (x, 2)) + 1,0) 
ics = {} 
dsolve(ode,func=y(x),ics=ics)
                                                                                         
                                                                                         
 

NotImplementedError : The given ODE x**3 + x*y(x) + Derivative(y(x), x) - Derivative(y(x), (x, 2)) - 1 cannot be solved by the factorable group method

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