Internal problem ID [9387]
Internal file name [OUTPUT/8324_Monday_June_06_2022_02_43_38_AM_45216467/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1055.
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, _with_linear_symmetries]]
Unable to solve or complete the solution.
\[ \boxed {y^{\prime \prime }+\left (a x +b \right ) y^{\prime }+\left (\operatorname {a1} \,x^{2}+\operatorname {b1} x +\operatorname {c1} \right ) y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }+\left (a x +b \right ) y^{\prime }+\left (\mathit {a1} \,x^{2}+\mathit {b1} x +\mathit {c1} \right ) y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..2 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )}{\sum }}a_{k} x^{k +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =\max \left (0, -m \right )+m}{\sum }}a_{k -m} x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x^{m}\cdot y^{\prime }\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =\max \left (0, 1-m \right )}{\sum }}a_{k} k \,x^{k -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =\max \left (0, 1-m \right )+m -1}{\sum }}a_{k +1-m} \left (k +1-m \right ) x^{k} \\ {} & \circ & \textrm {Convert}\hspace {3pt} \frac {d}{d x}y^{\prime }\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =2}{\sum }}a_{k} k \left (k -1\right ) x^{k -2} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \frac {d}{d x}y^{\prime }=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k +2} \left (k +2\right ) \left (k +1\right ) x^{k} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{1} b +a_{0} \mathit {c1} +2 a_{2}+\left (6 a_{3}+2 a_{2} b +a_{1} \left (a +\mathit {c1} \right )+a_{0} \mathit {b1} \right ) x +\left (\moverset {\infty }{\munderset {k =2}{\sum }}\left (a_{k +2} \left (k +2\right ) \left (k +1\right )+a_{k +1} \left (k +1\right ) b +a_{k} \left (a k +\mathit {c1} \right )+a_{k -1} \mathit {b1} +a_{k -2} \mathit {a1} \right ) x^{k}\right )=0 \\ \bullet & {} & \textrm {The coefficients of each power of}\hspace {3pt} x \hspace {3pt}\textrm {must be 0}\hspace {3pt} \\ {} & {} & \left [2 a_{2}+a_{1} b +a_{0} \mathit {c1} =0, 6 a_{3}+2 a_{2} b +a_{1} \left (a +\mathit {c1} \right )+a_{0} \mathit {b1} =0\right ] \\ \bullet & {} & \textrm {Solve for the dependent coefficient(s)}\hspace {3pt} \\ {} & {} & \left \{a_{2}=-\frac {a_{1} b}{2}-\frac {a_{0} \mathit {c1}}{2}, a_{3}=\frac {1}{6} a_{1} b^{2}+\frac {1}{6} a_{0} b \mathit {c1} -\frac {1}{6} a_{1} a -\frac {1}{6} a_{0} \mathit {b1} -\frac {1}{6} a_{1} \mathit {c1} \right \} \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & k^{2} a_{k +2}+\left (a a_{k}+a_{k +1} b +3 a_{k +2}\right ) k +a_{k +1} b +a_{k -2} \mathit {a1} +a_{k -1} \mathit {b1} +a_{k} \mathit {c1} +2 a_{k +2}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +2 \\ {} & {} & \left (k +2\right )^{2} a_{k +4}+\left (a a_{k +2}+a_{k +3} b +3 a_{k +4}\right ) \left (k +2\right )+a_{k +3} b +a_{k} \mathit {a1} +a_{k +1} \mathit {b1} +a_{k +2} \mathit {c1} +2 a_{k +4}=0 \\ \bullet & {} & \textrm {Recursion relation that defines the series solution to the ODE}\hspace {3pt} \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +4}=-\frac {a k a_{k +2}+b k a_{k +3}+2 a a_{k +2}+a_{k} \mathit {a1} +3 a_{k +3} b +a_{k +1} \mathit {b1} +a_{k +2} \mathit {c1}}{k^{2}+7 k +12}, a_{2}=-\frac {a_{1} b}{2}-\frac {a_{0} \mathit {c1}}{2}, a_{3}=\frac {1}{6} a_{1} b^{2}+\frac {1}{6} a_{0} b \mathit {c1} -\frac {1}{6} a_{1} a -\frac {1}{6} a_{0} \mathit {b1} -\frac {1}{6} a_{1} \mathit {c1} \right ] \end {array} \]
Maple trace
`Methods for second order ODEs: --- Trying classification methods --- trying a quadrature 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 -> hyper3: Equivalence to 1F1 under a power @ Moebius -> hypergeometric -> heuristic approach -> hyper3: Equivalence to 2F1, 1F1 or 0F1 under a power @ Moebius <- hyper3 successful: indirect Equivalence to 0F1 under \`\`^ @ Moebius\`\` is resolved <- hypergeometric successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.063 (sec). Leaf size: 254
dsolve(diff(diff(y(x),x),x)+(a*x+b)*diff(y(x),x)+(a1*x^2+b1*x+c1)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \left (c_{2} \left (a^{2} x +a b -4 \operatorname {a1} x -2 \operatorname {b1} \right ) \operatorname {hypergeom}\left (\left [\frac {3 \left (a^{2}-4 \operatorname {a1} \right )^{\frac {3}{2}}+a^{3}-2 a^{2} \operatorname {c1} +2 \left (\operatorname {b1} b -2 \operatorname {a1} \right ) a +2 \left (-b^{2}+4 \operatorname {c1} \right ) \operatorname {a1} -2 \operatorname {b1}^{2}}{4 \left (a^{2}-4 \operatorname {a1} \right )^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {\left (a^{2} x +a b -4 \operatorname {a1} x -2 \operatorname {b1} \right )^{2}}{2 \left (a^{2}-4 \operatorname {a1} \right )^{\frac {3}{2}}}\right )+\operatorname {hypergeom}\left (\left [\frac {\left (a^{2}-4 \operatorname {a1} \right )^{\frac {3}{2}}+a^{3}-2 a^{2} \operatorname {c1} +\left (2 \operatorname {b1} b -4 \operatorname {a1} \right ) a +\left (-2 b^{2}+8 \operatorname {c1} \right ) \operatorname {a1} -2 \operatorname {b1}^{2}}{4 \left (a^{2}-4 \operatorname {a1} \right )^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {\left (a^{2} x +a b -4 \operatorname {a1} x -2 \operatorname {b1} \right )^{2}}{2 \left (a^{2}-4 \operatorname {a1} \right )^{\frac {3}{2}}}\right ) c_{1} \right ) {\mathrm e}^{-\frac {x \left (\left (a x +2 b \right ) \sqrt {a^{2}-4 \operatorname {a1}}+x \left (a^{2}-4 \operatorname {a1} \right )+2 a b -4 \operatorname {b1} \right )}{4 \sqrt {a^{2}-4 \operatorname {a1}}}} \]
✓ Solution by Mathematica
Time used: 0.278 (sec). Leaf size: 305
DSolve[(c1 + b1*x + a1*x^2)*y[x] + (b + a*x)*y'[x] + y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \exp \left (-\frac {x \left (a \left (x \sqrt {a^2-4 \text {a1}}+2 b\right )+2 b \sqrt {a^2-4 \text {a1}}+a^2 x-4 (\text {a1} x+\text {b1})\right )}{4 \sqrt {a^2-4 \text {a1}}}\right ) \left (c_1 \operatorname {HermiteH}\left (\frac {-a^3-\left (\sqrt {a^2-4 \text {a1}}-2 \text {c1}\right ) a^2+(4 \text {a1}-2 b \text {b1}) a+2 \left (\text {b1}^2+\text {a1} \left (b^2-4 \text {c1}+2 \sqrt {a^2-4 \text {a1}}\right )\right )}{2 \left (a^2-4 \text {a1}\right )^{3/2}},\frac {x a^2+b a-2 (\text {b1}+2 \text {a1} x)}{\sqrt {2} \left (a^2-4 \text {a1}\right )^{3/4}}\right )+c_2 \operatorname {Hypergeometric1F1}\left (\frac {a^3+\left (\sqrt {a^2-4 \text {a1}}-2 \text {c1}\right ) a^2+(2 b \text {b1}-4 \text {a1}) a-2 \left (\text {b1}^2+\text {a1} \left (b^2-4 \text {c1}+2 \sqrt {a^2-4 \text {a1}}\right )\right )}{4 \left (a^2-4 \text {a1}\right )^{3/2}},\frac {1}{2},\frac {\left (x a^2+b a-2 (\text {b1}+2 \text {a1} x)\right )^2}{2 \left (a^2-4 \text {a1}\right )^{3/2}}\right )\right ) \]