Internal problem ID [7150]
Internal file name [OUTPUT/6136_Sunday_June_05_2022_04_24_34_PM_92301162/index.tex]
Book: Own collection of miscellaneous problems
Section: section 2.0
Problem number: 14.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_airy"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }-y^{\prime }-x y=x} \]
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 \end {align*}
Therefore the solution to the homogeneous Airy ODE becomes \[ y = {\mathrm e}^{-\frac {b x}{2 a}} \left (c_{1} \operatorname {AiryAi}\left (\frac {\left (-\frac {c}{a}\right )^{\frac {1}{3}} \left (4 c x a +b^{2}\right )}{4 c a}\right )+c_{2} \operatorname {AiryBi}\left (\frac {\left (-\frac {c}{a}\right )^{\frac {1}{3}} \left (4 c x a +b^{2}\right )}{4 c a}\right )\right ) \] Substituting the values for \(a,b,c\) gives \[ y = {\mathrm e}^{\frac {x}{2}} \left (c_{1} \operatorname {AiryAi}\left (x +\frac {1}{4}\right )+c_{2} \operatorname {AiryBi}\left (x +\frac {1}{4}\right )\right ) \] Since this is inhomogeneous Airy ODE, then we need to find the particular solution and add that to the homogeneous above. The particular solution is found using variation of parameters. 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 &= \operatorname {AiryAi}\left (x +\frac {1}{4}\right ) \\ y_2 &= \operatorname {AiryBi}\left (x +\frac {1}{4}\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 \[ W = \begin {vmatrix} \operatorname {AiryAi}\left (x +\frac {1}{4}\right ) & \operatorname {AiryBi}\left (x +\frac {1}{4}\right ) \\ \frac {d}{dx}\left (\operatorname {AiryAi}\left (x +\frac {1}{4}\right )\right ) & \frac {d}{dx}\left (\operatorname {AiryBi}\left (x +\frac {1}{4}\right )\right ) \end {vmatrix} \] Which gives \[ W = \begin {vmatrix} \operatorname {AiryAi}\left (x +\frac {1}{4}\right ) & \operatorname {AiryBi}\left (x +\frac {1}{4}\right ) \\ \operatorname {AiryAi}\left (1, x +\frac {1}{4}\right ) & \operatorname {AiryBi}\left (1, x +\frac {1}{4}\right ) \end {vmatrix} \] Therefore \[ W = \left (\operatorname {AiryAi}\left (x +\frac {1}{4}\right )\right )\left (\operatorname {AiryBi}\left (1, x +\frac {1}{4}\right )\right ) - \left (\operatorname {AiryBi}\left (x +\frac {1}{4}\right )\right )\left (\operatorname {AiryAi}\left (1, x +\frac {1}{4}\right )\right ) \] Which simplifies to \[ W = \operatorname {AiryAi}\left (x +\frac {1}{4}\right ) \operatorname {AiryBi}\left (1, x +\frac {1}{4}\right )-\operatorname {AiryBi}\left (x +\frac {1}{4}\right ) \operatorname {AiryAi}\left (1, x +\frac {1}{4}\right ) \] Which simplifies to \[ W = \frac {1}{\pi } \] Therefore Eq. (2) becomes \[ u_1 = -\int \frac {\operatorname {AiryBi}\left (x +\frac {1}{4}\right ) x}{\frac {1}{\pi }}\,dx \] Which simplifies to \[ u_1 = - \int \operatorname {AiryBi}\left (x +\frac {1}{4}\right ) x \pi d x \] Hence \[ u_1 = -\left (\int _{0}^{x}\operatorname {AiryBi}\left (\alpha +\frac {1}{4}\right ) \alpha \pi d \alpha \right ) \] And Eq. (3) becomes \[ u_2 = \int \frac {\operatorname {AiryAi}\left (x +\frac {1}{4}\right ) x}{\frac {1}{\pi }}\,dx \] Which simplifies to \[ u_2 = \int \operatorname {AiryAi}\left (x +\frac {1}{4}\right ) x \pi d x \] Hence \[ u_2 = \int _{0}^{x}\operatorname {AiryAi}\left (\alpha +\frac {1}{4}\right ) \alpha \pi d \alpha \] Which simplifies to \begin{align*} u_1 &= -\pi \left (\int _{0}^{x}\operatorname {AiryBi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \\ u_2 &= \pi \left (\int _{0}^{x}\operatorname {AiryAi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \\ \end{align*} Therefore the particular solution, from equation (1) is \[ y_p(x) = -\pi \left (\int _{0}^{x}\operatorname {AiryBi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryAi}\left (x +\frac {1}{4}\right )+\pi \left (\int _{0}^{x}\operatorname {AiryAi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryBi}\left (x +\frac {1}{4}\right ) \] Which simplifies to \[ y_p(x) = \pi \left (\left (\int _{0}^{x}\operatorname {AiryAi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryBi}\left (x +\frac {1}{4}\right )-\left (\int _{0}^{x}\operatorname {AiryBi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryAi}\left (x +\frac {1}{4}\right )\right ) \] Therefore the general solution is \begin{align*} y &= y_h + y_p \\ &= \left ({\mathrm e}^{\frac {x}{2}} \left (c_{1} \operatorname {AiryAi}\left (x +\frac {1}{4}\right )+c_{2} \operatorname {AiryBi}\left (x +\frac {1}{4}\right )\right )\right ) + \left (\pi \left (\left (\int _{0}^{x}\operatorname {AiryAi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryBi}\left (x +\frac {1}{4}\right )-\left (\int _{0}^{x}\operatorname {AiryBi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryAi}\left (x +\frac {1}{4}\right )\right )\right ) \\ &= \pi \left (\left (\int _{0}^{x}\operatorname {AiryAi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryBi}\left (x +\frac {1}{4}\right )-\left (\int _{0}^{x}\operatorname {AiryBi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryAi}\left (x +\frac {1}{4}\right )\right )+{\mathrm e}^{\frac {x}{2}} \left (c_{1} \operatorname {AiryAi}\left (x +\frac {1}{4}\right )+c_{2} \operatorname {AiryBi}\left (x +\frac {1}{4}\right )\right ) \\ \end{align*}
The solution(s) found are the following \begin{align*} \tag{1} y &= \pi \left (\left (\int _{0}^{x}\operatorname {AiryAi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryBi}\left (x +\frac {1}{4}\right )-\left (\int _{0}^{x}\operatorname {AiryBi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryAi}\left (x +\frac {1}{4}\right )\right )+{\mathrm e}^{\frac {x}{2}} \left (c_{1} \operatorname {AiryAi}\left (x +\frac {1}{4}\right )+c_{2} \operatorname {AiryBi}\left (x +\frac {1}{4}\right )\right ) \\ \end{align*}
Verification of solutions
\[ y = \pi \left (\left (\int _{0}^{x}\operatorname {AiryAi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryBi}\left (x +\frac {1}{4}\right )-\left (\int _{0}^{x}\operatorname {AiryBi}\left (\alpha +\frac {1}{4}\right ) \alpha d \alpha \right ) \operatorname {AiryAi}\left (x +\frac {1}{4}\right )\right )+{\mathrm e}^{\frac {x}{2}} \left (c_{1} \operatorname {AiryAi}\left (x +\frac {1}{4}\right )+c_{2} \operatorname {AiryBi}\left (x +\frac {1}{4}\right )\right ) \] Verified OK.
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`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 26
dsolve(diff(y(x),x$2)-diff(y(x),x)-x*y(x)-x=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{\frac {x}{2}} \operatorname {AiryAi}\left (\frac {1}{4}+x \right ) c_{2} +{\mathrm e}^{\frac {x}{2}} \operatorname {AiryBi}\left (\frac {1}{4}+x \right ) c_{1} -1 \]
✓ Solution by Mathematica
Time used: 13.6 (sec). Leaf size: 99
DSolve[y''[x]-y'[x]-x*y[x]-x==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{x/2} \left (\operatorname {AiryAi}\left (x+\frac {1}{4}\right ) \int _1^x-e^{-\frac {K[1]}{2}} \pi \operatorname {AiryBi}\left (K[1]+\frac {1}{4}\right ) K[1]dK[1]+\operatorname {AiryBi}\left (x+\frac {1}{4}\right ) \int _1^xe^{-\frac {K[2]}{2}} \pi \operatorname {AiryAi}\left (K[2]+\frac {1}{4}\right ) K[2]dK[2]+c_1 \operatorname {AiryAi}\left (x+\frac {1}{4}\right )+c_2 \operatorname {AiryBi}\left (x+\frac {1}{4}\right )\right ) \]