3.114 problem 1118

3.114.1 Maple step by step solution

Internal problem ID [9447]
Internal file name [OUTPUT/8387_Monday_June_06_2022_02_55_02_AM_79382259/index.tex]

Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 2, linear second order
Problem number: 1118.
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.

\[ \boxed {x y^{\prime \prime }+\left (\left (a +b \right ) x +m +n \right ) y^{\prime }+\left (a b x +a n +b m \right ) y=0} \]

3.114.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x \left (\frac {d}{d x}y^{\prime }\right )+\left (\left (a +b \right ) x +m +n \right ) y^{\prime }+\left (\left (b x +n \right ) a +b m \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 {Isolate 2nd derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=-\frac {\left (a b x +a n +b m \right ) y}{x}-\frac {\left (a x +b x +m +n \right ) y^{\prime }}{x} \\ \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 (a x +b x +m +n \right ) y^{\prime }}{x}+\frac {\left (a b x +a n +b m \right ) y}{x}=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 {a x +b x +m +n}{x}, P_{3}\left (x \right )=\frac {a b x +a n +b m}{x}\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}}}=m +n \\ {} & \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 x +b x +m +n \right ) y^{\prime }+\left (a b x +a n +b m \right ) 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^{m}\cdot y\hspace {3pt}\textrm {to series expansion for}\hspace {3pt} m =0..1 \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k +r +m} \\ {} & \circ & \textrm {Shift index using}\hspace {3pt} k \mathrm {->}k -m \\ {} & {} & x^{m}\cdot y=\moverset {\infty }{\munderset {k =m}{\sum }}a_{k -m} 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 +1+r \right ) \left (k +r \right ) x^{k +r} \\ & {} & \textrm {Rewrite ODE with series expansions}\hspace {3pt} \\ {} & {} & a_{0} r \left (-1+r +m +n \right ) x^{-1+r}+\left (a_{1} \left (1+r \right ) \left (r +m +n \right )+a_{0} \left (a n +a r +b m +b r \right )\right ) x^{r}+\left (\moverset {\infty }{\munderset {k =1}{\sum }}\left (a_{k +1} \left (k +1+r \right ) \left (k +r +m +n \right )+a_{k} \left (a k +a n +a r +b k +b m +b r \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 +m +n \right )=0 \\ \bullet & {} & \textrm {Values of r that satisfy the indicial equation}\hspace {3pt} \\ {} & {} & r \in \left \{0, -m -n +1\right \} \\ \bullet & {} & \textrm {Each term must be 0}\hspace {3pt} \\ {} & {} & a_{1} \left (1+r \right ) \left (r +m +n \right )+a_{0} \left (a n +a r +b m +b r \right )=0 \\ \bullet & {} & \textrm {Each term in the series must be 0, giving the recursion relation}\hspace {3pt} \\ {} & {} & a_{k +1} \left (k +1+r \right ) \left (k +r +m +n \right )+a_{k} \left (a +b \right ) k +a_{k} \left (a +b \right ) r +\left (a n +b m \right ) a_{k}+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+r +m +n \right )+a_{k +1} \left (a +b \right ) \left (k +1\right )+a_{k +1} \left (a +b \right ) r +\left (a n +b m \right ) a_{k +1}+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 n a_{k +1}+a r a_{k +1}+b k a_{k +1}+b m a_{k +1}+b r a_{k +1}+a a_{k +1}+b a_{k +1}}{\left (k +2+r \right ) \left (k +1+r +m +n \right )} \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =0 \\ {} & {} & a_{k +2}=-\frac {a b a_{k}+a k a_{k +1}+a n a_{k +1}+b k a_{k +1}+b m a_{k +1}+a a_{k +1}+b a_{k +1}}{\left (k +2\right ) \left (k +1+m +n \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}+a n a_{k +1}+b k a_{k +1}+b m a_{k +1}+a a_{k +1}+b a_{k +1}}{\left (k +2\right ) \left (k +1+m +n \right )}, a_{1} \left (m +n \right )+a_{0} \left (a n +b m \right )=0\right ] \\ \bullet & {} & \textrm {Recursion relation for}\hspace {3pt} r =-m -n +1 \\ {} & {} & a_{k +2}=-\frac {a b a_{k}+a k a_{k +1}+a n a_{k +1}+a \left (-m -n +1\right ) a_{k +1}+b k a_{k +1}+b m a_{k +1}+b \left (-m -n +1\right ) a_{k +1}+a a_{k +1}+b a_{k +1}}{\left (k +3-m -n \right ) \left (k +2\right )} \\ \bullet & {} & \textrm {Solution for}\hspace {3pt} r =-m -n +1 \\ {} & {} & \left [y=\moverset {\infty }{\munderset {k =0}{\sum }}a_{k} x^{k -m -n +1}, a_{k +2}=-\frac {a b a_{k}+a k a_{k +1}+a n a_{k +1}+a \left (-m -n +1\right ) a_{k +1}+b k a_{k +1}+b m a_{k +1}+b \left (-m -n +1\right ) a_{k +1}+a a_{k +1}+b a_{k +1}}{\left (k +3-m -n \right ) \left (k +2\right )}, a_{1} \left (2-m -n \right )+a_{0} \left (a n +a \left (-m -n +1\right )+b m +b \left (-m -n +1\right )\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 -m -n +1}\right ), c_{k +2}=-\frac {a b c_{k}+a k c_{k +1}+a n c_{k +1}+b k c_{k +1}+b m c_{k +1}+a c_{k +1}+b c_{k +1}}{\left (k +2\right ) \left (k +1+m +n \right )}, c_{1} \left (m +n \right )+c_{0} \left (a n +b m \right )=0, d_{k +2}=-\frac {a b d_{k}+a k d_{k +1}+a n d_{k +1}+a \left (-m -n +1\right ) d_{k +1}+b k d_{k +1}+b m d_{k +1}+b \left (-m -n +1\right ) d_{k +1}+a d_{k +1}+b d_{k +1}}{\left (k +3-m -n \right ) \left (k +2\right )}, d_{1} \left (2-m -n \right )+d_{0} \left (a n +a \left (-m -n +1\right )+b m +b \left (-m -n +1\right )\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: 0.063 (sec). Leaf size: 39

dsolve(x*diff(diff(y(x),x),x)+((a+b)*x+m+n)*diff(y(x),x)+(a*b*x+a*n+b*m)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = {\mathrm e}^{-a x} \left (\operatorname {KummerM}\left (m , m +n , x \left (a -b \right )\right ) c_{1} +\operatorname {KummerU}\left (m , m +n , x \left (a -b \right )\right ) c_{2} \right ) \]

Solution by Mathematica

Time used: 0.079 (sec). Leaf size: 46

DSolve[(b*m + a*n + a*b*x)*y[x] + (m + n + (a + b)*x)*y'[x] + x*y''[x] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to e^{-a x} (c_1 \operatorname {HypergeometricU}(m,m+n,(a-b) x)+c_2 L_{-m}^{m+n-1}((a-b) x)) \]