Internal problem ID [9446]
Internal file name [OUTPUT/8386_Monday_June_06_2022_02_54_49_AM_71459158/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1117.
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.
\[ \boxed {x y^{\prime \prime }-\left (a +b \right ) \left (x +1\right ) y^{\prime }+a b x y=0} \]
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )-\left (a +b \right ) \left (x +1\right ) y^{\prime }+a b x y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 2 \\ {} & {} & \frac {d}{d x}y^{\prime } \\ \bullet & {} & \textrm {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\left (x +1\right ) \left (a +b \right ) y^{\prime }}{x}-a b y \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }-\frac {\left (x +1\right ) \left (a +b \right ) y^{\prime }}{x}+a b y=0 \\ \square & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & \circ & \textrm {Define functions}\hspace {3pt} \\ {} & {} & \left [P_{2}\left (x \right )=-\frac {\left (x +1\right ) \left (a +b \right )}{x}, P_{3}\left (x \right )=b a \right ] \\ {} & \circ & x \cdot P_{2}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x \cdot P_{2}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=-a -b \\ {} & \circ & x^{2}\cdot P_{3}\left (x \right )\textrm {is analytic at}\hspace {3pt} x =0 \\ {} & {} & \left (x^{2}\cdot P_{3}\left (x \right )\right )\bigg | {\mstack {}{_{x \hiderel {=}0}}}=0 \\ {} & \circ & x =0\textrm {is a regular singular point}\hspace {3pt} \\ & {} & \textrm {Check to see if}\hspace {3pt} x_{0}=0\hspace {3pt}\textrm {is a regular singular point}\hspace {3pt} \\ {} & {} & x_{0}=0 \\ \bullet & {} & \textrm {Multiply by denominators}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )-\left (a +b \right ) \left (x +1\right ) y^{\prime }+a b x y=0 \\ \bullet & {} & \textrm {Assume series solution for}\hspace {3pt} y \\ {} & {} & y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r} \\ \square & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot y\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -1 \\ {} & {} & x \cdot y=\moverset {\infty }{\munderset {k =1}{\sum }}a_{k -1} x^{k +r} \\ {} & \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 =0}{\sum }}a_{k} \left (k +r \right ) x^{k +r -1+m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1-m \\ {} & {} & x^{m}\cdot y^{\prime }=\moverset {\infty }{\munderset {k =-1+m}{\sum }}a_{k +1-m} \left (k +1-m +r \right ) x^{k +r} \\ {} & \circ & \textrm {Convert}\hspace {3pt} x \cdot \left (\frac {d}{d x}y^{\prime }\right )\hspace {3pt}\textrm {to series expansion}\hspace {3pt} \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} \left (k +r \right ) \left (k +r -1\right ) x^{k +r -1} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & x \cdot \left (\frac {d}{d x}y^{\prime }\right )=\moverset {\infty }{\munderset {k =-1}{\sum }}a_{k +1} \left (k +r +1\right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & -a_{0} r \left (1-r +a +b \right ) x^{-1+r}+\left (-a_{1} \left (1+r \right ) \left (-r +a +b \right )-a_{0} r \left (a +b \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (-a_{k +1} \left (k +r +1\right ) \left (-k -r +a +b \right )-a_{k} \left (k +r \right ) \left (a +b \right )+a b a_{k -1}\right ) x^{k +r}\right )=0 \\ \bullet & {} & a_{0}\textrm {cannot be 0 by assumption, giving the indicial equation}\hspace {3pt} \\ {} & {} & -r \left (1-r +a +b \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, a +b +1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & -a_{1} \left (1+r \right ) \left (-r +a +b \right )-a_{0} r \left (a +b \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +r +1\right ) \left (k -a -b +r \right )-a_{k} \left (k +r \right ) \left (a +b \right )+a b a_{k -1}=0 \\ \bullet & {} & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k +1 \\ {} & {} & a_{k +2} \left (k +2+r \right ) \left (k +1-a -b +r \right )-a_{k +1} \left (k +r +1\right ) \left (a +b \right )+a b a_{k}=0 \\ \bullet & {} & \textrm {Recursion relation that defines series solution to ODE}\hspace {3pt} \\ {} & {} & a_{k +2}=\frac {a b a_{k}-a k a_{k +1}-a r a_{k +1}-b k a_{k +1}-b r a_{k +1}-a a_{k +1}-b a_{k +1}}{\left (k +2+r \right ) \left (-k -1+a +b -r \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=\frac {a b a_{k}-a k a_{k +1}-b k a_{k +1}-a a_{k +1}-b a_{k +1}}{\left (k +2\right ) \left (-k -1+a +b \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =0 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k}, a_{k +2}=\frac {a b a_{k}-a k a_{k +1}-b k a_{k +1}-a a_{k +1}-b a_{k +1}}{\left (k +2\right ) \left (-k -1+a +b \right )}, -a_{1} \left (a +b \right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =a +b +1 \\ {} & {} & a_{k +2}=\frac {a b a_{k}-a k a_{k +1}-a \left (a +b +1\right ) a_{k +1}-b k a_{k +1}-b \left (a +b +1\right ) a_{k +1}-a a_{k +1}-b a_{k +1}}{\left (k +3+a +b \right ) \left (-2-k \right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =a +b +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +a +b +1}, a_{k +2}=\frac {a b a_{k}-a k a_{k +1}-a \left (a +b +1\right ) a_{k +1}-b k a_{k +1}-b \left (a +b +1\right ) a_{k +1}-a a_{k +1}-b a_{k +1}}{\left (k +3+a +b \right ) \left (-2-k \right )}, a_{1} \left (a +2+b \right )-a_{0} \left (a +b +1\right ) \left (a +b \right )=0\right ] \\ \bullet & {} & \textrm {Combine solutions and rename parameters}\hspace {3pt} \\ {} & {} & \left [y=\left (\moverset {\infty }{\munderset {k =0}{\sum }}c_{k} x^{k}\right )+\left (\moverset {\infty }{\munderset {k =0}{\sum }}d_{k} x^{k +a +b +1}\right ), c_{k +2}=\frac {a b c_{k}-a k c_{k +1}-b k c_{k +1}-a c_{k +1}-b c_{k +1}}{\left (k +2\right ) \left (-k -1+a +b \right )}, -c_{1} \left (a +b \right )=0, d_{k +2}=\frac {a b d_{k}-a k d_{k +1}-a \left (a +b +1\right ) d_{k +1}-b k d_{k +1}-b \left (a +b +1\right ) d_{k +1}-a d_{k +1}-b d_{k +1}}{\left (k +3+a +b \right ) \left (-2-k \right )}, d_{1} \left (a +2+b \right )-d_{0} \left (a +b +1\right ) \left (a +b \right )=0\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 <- hyper3 successful: received ODE is equivalent to the 1F1 ODE <- Kummer successful <- special function solution successful`
✓ Solution by Maple
Time used: 0.046 (sec). Leaf size: 82
dsolve(x*diff(diff(y(x),x),x)-(a+b)*(x+1)*diff(y(x),x)+a*b*x*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{b x} x^{b +a +1} \left (\operatorname {KummerM}\left (\frac {a^{2}+a b +a -b}{a -b}, b +2+a , x \left (a -b \right )\right ) c_{1} +\operatorname {KummerU}\left (\frac {a^{2}+a b +a -b}{a -b}, b +2+a , x \left (a -b \right )\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.093 (sec). Leaf size: 87
DSolve[a*b*x*y[x] - (a + b)*(1 + x)*y'[x] + x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{b x} x^{a+b+1} \left (c_1 \operatorname {HypergeometricU}\left (\frac {a^2+b a+a-b}{a-b},a+b+2,(a-b) x\right )+c_2 L_{-\frac {a^2+b a+a-b}{a-b}}^{a+b+1}((a-b) x)\right ) \]