Internal problem ID [7285]
Internal file name [OUTPUT/6271_Sunday_June_05_2022_04_36_40_PM_63226158/index.tex]
Book: Own collection of miscellaneous problems
Section: section 4.0
Problem number: 61.
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, _linear, _nonhomogeneous]]
\[ \boxed {\frac {x y^{\prime \prime }}{1-x}+y=\frac {1}{1-x}} \]
In normal form the ode \begin {align*} -\frac {x y^{\prime \prime }}{x -1}+y = \frac {1}{1-x} \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 )&=\frac {1-x}{x}\\ r \left (x \right )&=\frac {1}{x} \end {align*}
The Lagrange adjoint ode is given by \begin {align*} \xi ^{''}-(\xi \, p)'+\xi q &= 0\\ \xi ^{''}-\left (0\right )' + \left (\frac {\left (1-x \right ) \xi \left (x \right )}{x}\right ) &= 0\\ \xi ^{\prime \prime }\left (x \right )+\frac {\left (1-x \right ) \xi \left (x \right )}{x}&= 0 \end {align*}
Which is solved for \(\xi (x)\). Writing the ode as \begin {align*} x^{2} \xi ^{\prime \prime }\left (x \right )+\left (-x^{2}+x \right ) \xi \left (x \right ) = 0\tag {1} \end {align*}
Bessel ode has the form \begin {align*} x^{2} \xi ^{\prime \prime }\left (x \right )+\xi ^{\prime }\left (x \right ) x +\left (-n^{2}+x^{2}\right ) \xi \left (x \right ) = 0\tag {2} \end {align*}
The generalized form of Bessel ode is given by Bowman (1958) as the following \begin {align*} x^{2} \xi ^{\prime \prime }\left (x \right )+\left (1-2 \alpha \right ) x \xi ^{\prime }\left (x \right )+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) \xi \left (x \right ) = 0\tag {3} \end {align*}
With the standard solution \begin {align*} \xi \left (x \right )&=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 &= 2\\ n &= -1\\ \gamma &= {\frac {1}{2}} \end {align*}
Substituting all the above into (4) gives the solution as \begin {align*} \xi \left (x \right ) = -c_{1} \sqrt {x}\, \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )-c_{2} \sqrt {x}\, \operatorname {BesselY}\left (1, 2 \sqrt {x}\right ) \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 }-\frac {y \left (-\frac {c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )}{2 \sqrt {x}}-c_{1} \left (\operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )-\frac {\operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )}{2 \sqrt {x}}\right )-\frac {c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )}{2 \sqrt {x}}-c_{2} \left (\operatorname {BesselY}\left (0, 2 \sqrt {x}\right )-\frac {\operatorname {BesselY}\left (1, 2 \sqrt {x}\right )}{2 \sqrt {x}}\right )\right )}{-c_{1} \sqrt {x}\, \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )-c_{2} \sqrt {x}\, \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )}&=\frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{-c_{1} \sqrt {x}\, \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )-c_{2} \sqrt {x}\, \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )} \end {align*}
Which is now a first order ode. This is now solved for \(y\). In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= f( x) g(y)\\ &= \frac {\left (-1+y \right ) \left (c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )} \end {align*}
Where \(f(x)=\frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )}\) and \(g(y)=-1+y\). Integrating both sides gives \begin{align*} \frac {1}{-1+y} \,dy &= \frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )} \,d x \\ \int { \frac {1}{-1+y} \,dy} &= \int {\frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )} \,d x} \\ \ln \left (-1+y \right )&=\int \frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )}d x +c_{3} \\ \end{align*} Raising both side to exponential gives \begin {align*} -1+y &= {\mathrm e}^{\int \frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )}d x +c_{3}} \end {align*}
Which simplifies to \begin {align*} -1+y &= c_{4} {\mathrm e}^{\int \frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )}d x} \end {align*}
Hence, the solution found using Lagrange adjoint equation method is \[ y = c_{4} {\mathrm e}^{\int \frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )}d x +c_{3}}+1 \]
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{4} {\mathrm e}^{\int \frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )}d x +c_{3}}+1 \\ \end{align*}
Verification of solutions
\[ y = c_{4} {\mathrm e}^{\int \frac {c_{1} \operatorname {BesselJ}\left (0, 2 \sqrt {x}\right )+c_{2} \operatorname {BesselY}\left (0, 2 \sqrt {x}\right )}{\sqrt {x}\, \left (c_{2} \operatorname {BesselY}\left (1, 2 \sqrt {x}\right )+c_{1} \operatorname {BesselJ}\left (1, 2 \sqrt {x}\right )\right )}d x +c_{3}}+1 \] 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 -> elliptic -> Legendre <- Kummer successful <- special function solution successful <- solving first the homogeneous part of the ODE successful`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 167
dsolve(x/(1-x)*diff(y(x),x$2)+y(x)=1/(1-x),y(x), singsol=all)
\[ y \left (x \right ) = -x \left (\left (\operatorname {BesselK}\left (0, -x \right )-\operatorname {BesselK}\left (1, -x \right )\right ) \left (\int \frac {-\operatorname {BesselI}\left (0, -x \right )-\operatorname {BesselI}\left (1, -x \right )}{x \left (\operatorname {BesselI}\left (0, x\right ) \left (x +1\right ) \operatorname {BesselK}\left (1, -x \right )+1-\left (x +1\right ) \operatorname {BesselK}\left (0, -x \right ) \operatorname {BesselI}\left (1, x\right )\right )}d x \right )+\left (-\operatorname {BesselI}\left (0, -x \right )-\operatorname {BesselI}\left (1, -x \right )\right ) \left (\int \frac {-\operatorname {BesselK}\left (0, -x \right )+\operatorname {BesselK}\left (1, -x \right )}{\left (\operatorname {BesselI}\left (0, x\right ) \left (x +1\right ) \operatorname {BesselK}\left (1, -x \right )+1-\left (x +1\right ) \operatorname {BesselK}\left (0, -x \right ) \operatorname {BesselI}\left (1, x\right )\right ) x}d x \right )-\operatorname {BesselK}\left (0, -x \right ) c_{1} +\operatorname {BesselK}\left (1, -x \right ) c_{1} -\operatorname {BesselI}\left (0, -x \right ) c_{2} -\operatorname {BesselI}\left (1, -x \right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.266 (sec). Leaf size: 136
DSolve[x/(1-x)*y''[x]+y[x]==1/(1-x),y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-x} x \left (e^x (\operatorname {BesselI}(0,x)-\operatorname {BesselI}(1,x)) \int _1^x2 e^{-K[1]} \sqrt {\pi } \operatorname {HypergeometricU}\left (\frac {1}{2},2,2 K[1]\right )dK[1]-2 \sqrt {\pi } x \operatorname {HypergeometricU}\left (\frac {1}{2},2,2 x\right ) \, _1F_2\left (\frac {1}{2};1,\frac {3}{2};\frac {x^2}{4}\right )+2 \sqrt {\pi } \operatorname {HypergeometricU}\left (\frac {1}{2},2,2 x\right ) \operatorname {BesselI}(0,x)+c_1 \operatorname {HypergeometricU}\left (\frac {1}{2},2,2 x\right )+c_2 e^x \operatorname {BesselI}(0,x)-c_2 e^x \operatorname {BesselI}(1,x)\right ) \]