2.2.31 problem 30

Internal problem ID [8511]
Book : Own collection of miscellaneous problems
Section : section 2.0
Problem number : 30
Date solved : Thursday, December 12, 2024 at 09:23:36 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

Solve

\begin{align*} y^{\prime \prime }-x y-x&=0 \end{align*}

Solved as second order Airy ode

Time used: 0.052 (sec)

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 &= 0\\ c &= -1\\ F &= x \end{align*}

Therefore the solution to the homogeneous Airy ODE becomes

\[ y = c_1 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_2 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right ) \]

Since this is inhomogeneous Airy ODE, then we need to find the particular solution. The particular solution is now found using the method of undetermined coefficients. Looking at the RHS of the ode, which is

\[ x \]

Shows that the corresponding undetermined set of the basis functions (UC_set) for the trial solution is

\[ [\{1, x\}] \]

While the set of the basis functions for the homogeneous solution found earlier is

\[ \left \{\operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right ), \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right )\right \} \]

Since there is no duplication between the basis function in the UC_set and the basis functions of the homogeneous solution, the trial solution is a linear combination of all the basis in the UC_set.

\[ y_p = A_{2} x +A_{1} \]

The unknowns \(\{A_{1}, A_{2}\}\) are found by substituting the above trial solution \(y_p\) into the ODE and comparing coefficients. Substituting the trial solution into the ODE and simplifying gives

\[ -x \left (A_{2} x +A_{1}\right )-x = 0 \]

Solving for the unknowns by comparing coefficients results in

\[ [A_{1} = -1, A_{2} = 0] \]

Substituting the above back in the above trial solution \(y_p\), gives the particular solution

\[ y_p = -1 \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_1 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_2 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right )\right ) + \left (-1\right ) \\ &= c_1 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_2 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right )-1 \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_2 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right )-1 \\ \end{align*}

Solved as second order Bessel ode

Time used: 0.611 (sec)

Writing the ode as

\begin{align*} x^{2} y^{\prime \prime }-x^{3} y = x^{3}\tag {1} \end{align*}

Let the solution be

\begin{align*} y &= y_h + y_p \end{align*}

Where \(y_h\) is the solution to the homogeneous ODE and \(y_p\) is a particular solution to the non-homogeneous ODE. Bessel ode has the form

\begin{align*} x^{2} y^{\prime \prime }+x y^{\prime }+\left (-n^{2}+x^{2}\right ) y = 0\tag {2} \end{align*}

The generalized form of Bessel ode is given by Bowman (1958) as the following

\begin{align*} x^{2} y^{\prime \prime }+\left (1-2 \alpha \right ) x y^{\prime }+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) y = 0\tag {3} \end{align*}

With the standard solution

\begin{align*} y&=x^{\alpha } \left (c_1 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_2 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}

Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives

\begin{align*} \alpha &= {\frac {1}{2}}\\ \beta &= \frac {2 i}{3}\\ n &= {\frac {1}{3}}\\ \gamma &= {\frac {3}{2}} \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} y = c_1 \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )+c_2 \sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \end{align*}

Therefore the homogeneous solution \(y_h\) is

\[ y_h = c_1 \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )+c_2 \sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \]

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 &= \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \\ y_2 &= \sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{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

\[ W = \begin {vmatrix} \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) & \sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \\ \frac {d}{dx}\left (\sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )\right ) & \frac {d}{dx}\left (\sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )\right ) \end {vmatrix} \]

Which gives

\[ W = \begin {vmatrix} \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) & \sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \\ \frac {\operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{2 \sqrt {x}}+i x \left (-\operatorname {BesselJ}\left (\frac {4}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )-\frac {i \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{2 x^{{3}/{2}}}\right ) & \frac {\operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{2 \sqrt {x}}+i x \left (-\operatorname {BesselY}\left (\frac {4}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )-\frac {i \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{2 x^{{3}/{2}}}\right ) \end {vmatrix} \]

Therefore

\[ W = \left (\sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )\right )\left (\frac {\operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{2 \sqrt {x}}+i x \left (-\operatorname {BesselY}\left (\frac {4}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )-\frac {i \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{2 x^{{3}/{2}}}\right )\right ) - \left (\sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )\right )\left (\frac {\operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{2 \sqrt {x}}+i x \left (-\operatorname {BesselJ}\left (\frac {4}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )-\frac {i \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{2 x^{{3}/{2}}}\right )\right ) \]

Which simplifies to

\[ W = -i x^{{3}/{2}} \left (\operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \operatorname {BesselY}\left (\frac {4}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )-\operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \operatorname {BesselJ}\left (\frac {4}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )\right ) \]

Which simplifies to

\[ W = \frac {3}{\pi } \]

Therefore Eq. (2) becomes

\[ u_1 = -\int \frac {x^{{7}/{2}} \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{\frac {3 x^{2}}{\pi }}\,dx \]

Which simplifies to

\[ u_1 = - \int \frac {x^{{3}/{2}} \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \pi }{3}d x \]

Hence

\[ u_1 = -\int _{0}^{x}\frac {\alpha ^{{3}/{2}} \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right ) \pi }{3}d \alpha \]

And Eq. (3) becomes

\[ u_2 = \int \frac {x^{{7}/{2}} \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )}{\frac {3 x^{2}}{\pi }}\,dx \]

Which simplifies to

\[ u_2 = \int \frac {x^{{3}/{2}} \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \pi }{3}d x \]

Hence

\[ u_2 = \int _{0}^{x}\frac {\alpha ^{{3}/{2}} \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right ) \pi }{3}d \alpha \]

Therefore the particular solution, from equation (1) is

\[ y_p(x) = -\int _{0}^{x}\frac {\alpha ^{{3}/{2}} \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right ) \pi }{3}d \alpha \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )+\sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \int _{0}^{x}\frac {\alpha ^{{3}/{2}} \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right ) \pi }{3}d \alpha \]

Which simplifies to

\[ y_p(x) = -\frac {\pi \sqrt {x}\, \left (\int _{0}^{x}\alpha ^{{3}/{2}} \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right )d \alpha \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )-\operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \int _{0}^{x}\alpha ^{{3}/{2}} \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right )d \alpha \right )}{3} \]

Therefore the general solution is

\begin{align*} y &= y_h + y_p \\ &= \left (c_1 \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )+c_2 \sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )\right ) + \left (-\frac {\pi \sqrt {x}\, \left (\int _{0}^{x}\alpha ^{{3}/{2}} \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right )d \alpha \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )-\operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \int _{0}^{x}\alpha ^{{3}/{2}} \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right )d \alpha \right )}{3}\right ) \\ \end{align*}

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_1 \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )+c_2 \sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )-\frac {\pi \sqrt {x}\, \left (\int _{0}^{x}\alpha ^{{3}/{2}} \operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right )d \alpha \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right )-\operatorname {BesselY}\left (\frac {1}{3}, \frac {2 i x^{{3}/{2}}}{3}\right ) \int _{0}^{x}\alpha ^{{3}/{2}} \operatorname {BesselJ}\left (\frac {1}{3}, \frac {2 i \alpha ^{{3}/{2}}}{3}\right )d \alpha \right )}{3} \\ \end{align*}

Solved as second order ode adjoint method

Time used: 0.782 (sec)

In normal form the ode

\begin{align*} y^{\prime \prime }-x y-x = 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 )&=0\\ q \left (x \right )&=-x\\ r \left (x \right )&=x \end{align*}

The Lagrange adjoint ode is given by

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

Which is solved for \(\xi (x)\). This is Airy ODE. It has the general form

\[ a \xi ^{\prime \prime } + b \xi ^{\prime } + c x \xi = F(x) \]

Where in this case

\begin{align*} a &= 1\\ b &= 0\\ c &= -1\\ F &= 0 \end{align*}

Therefore the solution to the homogeneous Airy ODE becomes

\[ \xi = c_3 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_4 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right ) \]

Will add steps showing solving for IC soon.

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*}

Or

\begin{align*} y^{\prime }-\frac {y \left (-c_3 \left (-1\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )-c_4 \left (-1\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )\right )}{c_3 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_4 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right )}&=\frac {-\frac {c_3 \operatorname {AiryAi}\left (1, -x \left (-1\right )^{{1}/{3}}\right ) \sqrt {-3}}{2}-\frac {c_3 \operatorname {AiryAi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )}{2}-\frac {c_4 \operatorname {AiryBi}\left (1, -x \left (-1\right )^{{1}/{3}}\right ) \sqrt {-3}}{2}-\frac {c_4 \operatorname {AiryBi}\left (1, -x \left (-1\right )^{{1}/{3}}\right )}{2}}{c_3 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_4 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right )} \end{align*}

Which is now a first order ode. This is now solved for \(y\). In canonical form a linear first order is

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

Comparing the above to the given ode shows that

\begin{align*} q(x) &=\frac {\left (1+i \sqrt {3}\right ) \left (c_3 \operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{2 c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+2 c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )}\\ p(x) &=-\frac {\left (1+i \sqrt {3}\right ) \left (c_3 \operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{2 c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+2 c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )} \end{align*}

The integrating factor \(\mu \) is

\begin{align*} \mu &= e^{\int {q\,dx}}\\ &= {\mathrm e}^{\int \frac {\left (1+i \sqrt {3}\right ) \left (c_3 \operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{2 c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+2 c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )}d x}\\ &= \frac {1}{c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )} \end{align*}

The ode becomes

\begin{align*} \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \mu p \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}}\left ( \mu y\right ) &= \left (\mu \right ) \left (-\frac {\left (1+i \sqrt {3}\right ) \left (c_3 \operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{2 c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+2 c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )}\right ) \\ \frac {\mathop {\mathrm {d}}}{ \mathop {\mathrm {d}x}} \left (\frac {y}{c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )}\right ) &= \left (\frac {1}{c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )}\right ) \left (-\frac {\left (1+i \sqrt {3}\right ) \left (c_3 \operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{2 c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+2 c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )}\right ) \\ \mathrm {d} \left (\frac {y}{c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )}\right ) &= \left (-\frac {\left (1+i \sqrt {3}\right ) \left (c_3 \operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{\left (2 c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+2 c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right ) \left (c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}\right )\, \mathrm {d} x \\ \end{align*}

Integrating gives

\begin{align*} \frac {y}{c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )}&= \int {-\frac {\left (1+i \sqrt {3}\right ) \left (c_3 \operatorname {AiryAi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (1, -\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )}{\left (2 c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+2 c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right ) \left (c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )\right )} \,dx} \\ &=\frac {1+i \sqrt {3}}{2 \left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) \left (c_3 \operatorname {AiryAi}\left (\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x \right )+c_4 \operatorname {AiryBi}\left (\left (-\frac {1}{2}-\frac {i \sqrt {3}}{2}\right ) x \right )\right )} + c_5 \end{align*}

Dividing throughout by the integrating factor \(\frac {1}{c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )}\) gives the final solution

\[ y = \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right ) c_4 c_5 +\operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right ) c_3 c_5 -1 \]

Hence, the solution found using Lagrange adjoint equation method is

\begin{align*} y &= \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right ) c_4 c_5 +\operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right ) c_3 c_5 -1 \\ \end{align*}

The constants can be merged to give

\[ y = c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )-1 \]

Will add steps showing solving for IC soon.

Summary of solutions found

\begin{align*} y &= c_4 \operatorname {AiryBi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )+c_3 \operatorname {AiryAi}\left (-\frac {x \left (1+i \sqrt {3}\right )}{2}\right )-1 \\ \end{align*}

Maple step by step solution

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`
 

Maple dsolve solution

Solving time : 0.007 (sec)
Leaf size : 14

dsolve(diff(diff(y(x),x),x)-x*y(x)-x = 0, 
       y(x),singsol=all)
 

\[ y = \operatorname {AiryAi}\left (x \right ) c_{2} +\operatorname {AiryBi}\left (x \right ) c_{1} -1 \]

Mathematica DSolve solution

Solving time : 0.036 (sec)
Leaf size : 28

DSolve[{D[y[x],{x,2}]-x*y[x]-x==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to \pi \operatorname {AiryAiPrime}(x) \operatorname {AiryBi}(x)+c_2 \operatorname {AiryBi}(x)+\operatorname {AiryAi}(x) (-\pi \operatorname {AiryBiPrime}(x)+c_1) \]