34.6 problem 6

Internal problem ID [11093]
Internal file name [OUTPUT/10350_Wednesday_January_24_2024_10_18_11_PM_91888170/index.tex]

Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev. Second edition
Section: Chapter 2, Second-Order Differential Equations. section 2.1.3-1. Equations with exponential functions
Problem number: 6.
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.

\[ \boxed {y^{\prime \prime }+\left (a \,{\mathrm e}^{4 \lambda x}+b \,{\mathrm e}^{3 \lambda x}+c \,{\mathrm e}^{2 \lambda x}-\frac {\lambda ^{2}}{4}\right ) y=0} \]

Maple trace

`Methods for second order ODEs: 
--- Trying classification methods --- 
trying a symmetry of the form [xi=0, eta=F(x)] 
checking if the LODE is missing y 
-> Heun: Equivalence to the GHE or one of its 4 confluent cases under a power @ Moebius 
-> trying a solution of the form r0(x) * Y + r1(x) * Y where Y = exp(int(r(x), dx)) * 2F1([a1, a2], [b1], f) 
-> Trying changes of variables to rationalize or make the ODE simpler 
   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 
      -> Whittaker 
         -> 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 
   Change of variables used: 
      [x = ln(t)/lambda] 
   Linear ODE actually solved: 
      (4*a*t^4+4*b*t^3+4*c*t^2-lambda^2)*u(t)+4*lambda^2*t*diff(u(t),t)+4*lambda^2*t^2*diff(diff(u(t),t),t) = 0 
<- change of variables successful`
 

Solution by Maple

Time used: 0.36 (sec). Leaf size: 221

dsolve(diff(y(x),x$2)+(a*exp(4*lambda*x)+b*exp(3*lambda*x)+c*exp(2*lambda*x)-1/4*lambda^2)*y(x)=0,y(x), singsol=all)
 

\[ y \left (x \right ) = c_{1} \operatorname {hypergeom}\left (\left [\frac {4 \lambda \,a^{\frac {3}{2}}+4 i a c -i b^{2}}{16 \lambda \,a^{\frac {3}{2}}}\right ], \left [\frac {1}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{x \lambda } a +b \right )^{2}}{4 \lambda \,a^{\frac {3}{2}}}\right ) {\mathrm e}^{-\frac {i {\mathrm e}^{2 x \lambda } a +\lambda ^{2} x \sqrt {a}+i b \,{\mathrm e}^{x \lambda }}{2 \lambda \sqrt {a}}}+c_{2} \operatorname {hypergeom}\left (\left [\frac {12 \lambda \,a^{\frac {3}{2}}+4 i a c -i b^{2}}{16 \lambda \,a^{\frac {3}{2}}}\right ], \left [\frac {3}{2}\right ], \frac {i \left (2 \,{\mathrm e}^{x \lambda } a +b \right )^{2}}{4 \lambda \,a^{\frac {3}{2}}}\right ) \left (2 a \,{\mathrm e}^{-\frac {i {\mathrm e}^{2 x \lambda } a -\lambda ^{2} x \sqrt {a}+i b \,{\mathrm e}^{x \lambda }}{2 \lambda \sqrt {a}}}+b \,{\mathrm e}^{-\frac {i {\mathrm e}^{2 x \lambda } a +\lambda ^{2} x \sqrt {a}+i b \,{\mathrm e}^{x \lambda }}{2 \lambda \sqrt {a}}}\right ) \]

Solution by Mathematica

Time used: 1.895 (sec). Leaf size: 178

DSolve[y''[x]+(a*Exp[4*\[Lambda]*x]+b*Exp[3*\[Lambda]*x]+c*Exp[2*\[Lambda]*x]-1/4*\[Lambda]^2)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {e^{-\frac {i e^{\lambda x} \left (a e^{\lambda x}+b\right )}{2 \sqrt {a} \lambda }} \left (c_1 \operatorname {HermiteH}\left (\frac {i \left (b^2-4 a c+4 i a^{3/2} \lambda \right )}{8 a^{3/2} \lambda },\frac {\sqrt [4]{-1} \left (2 e^{x \lambda } a+b\right )}{2 a^{3/4} \sqrt {\lambda }}\right )+c_2 \operatorname {Hypergeometric1F1}\left (\frac {-i b^2+4 i a c+4 a^{3/2} \lambda }{16 a^{3/2} \lambda },\frac {1}{2},\frac {i \left (2 e^{x \lambda } a+b\right )^2}{4 a^{3/2} \lambda }\right )\right )}{\sqrt {e^{\lambda x}}} \]