Internal problem ID [11097]
Internal file name [OUTPUT/10354_Wednesday_January_24_2024_10_18_13_PM_80628882/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: 10.
ODE order: 2.
ODE degree: 1.
The type(s) of ODE detected by this program : "second_order_bessel_ode_form_A"
Maple gives the following as the ode type
[[_2nd_order, _with_linear_symmetries]]
\[ \boxed {y^{\prime \prime }+a y^{\prime }+\left (b \,{\mathrm e}^{\lambda x}+c \right ) y=0} \]
Writing the ode as \begin {align*} y^{\prime \prime }+a y^{\prime }+\left (b \,{\mathrm e}^{\lambda x}+c \right ) y = 0\tag {1} \end {align*}
An ode of the form\begin {equation} ay^{\prime \prime }+by^{\prime }+(ce^{rx}+m)y=0\tag {1} \end {equation} can be transformed to Bessel ode using the transformation\begin {align*} x & =\ln \left ( t\right ) \\ e^{x} & =t \end {align*}
Where \(a,b,c,m\) are not functions of \(x\) and where \(b\) and \(m\) are allowed to be be zero. Using this transformation gives\begin {align} \frac {dy}{dx} & =\frac {dy}{dt}\frac {dt}{dx}\nonumber \\ & =\frac {dy}{dt}e^{x}\nonumber \\ & =t\frac {dy}{dt}\tag {2} \end {align}
And\begin {align} \frac {d^{2}y}{dx^{2}} & =\frac {d}{dx}\left ( \frac {dy}{dx}\right ) \nonumber \\ & =\frac {d}{dx}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =\frac {d}{dt}\frac {dt}{dx}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =\frac {dt}{dx}\frac {d}{dt}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =t\frac {d}{dt}\left ( t\frac {dy}{dt}\right ) \nonumber \\ & =t\left ( \frac {dy}{dt}+t\frac {d^{2}y}{dt^{2}}\right ) \tag {3} \end {align}
Substituting (2,3) into (1) gives\begin {align} at\left ( \frac {dy}{dt}+t\frac {d^{2}y}{dt^{2}}\right ) +bt\frac {dy}{dt}+(ce^{rx}+m)y & =0\nonumber \\ \left ( aty^{\prime }+at^{2}y^{\prime \prime }\right ) +bty^{\prime }+(ct^{r}+m)y & =0\nonumber \\ at^{2}y^{\prime \prime }+\left ( b+a\right ) ty^{\prime }+(ct^{r}+m)y & =0\nonumber \\ t^{2}y^{\prime \prime }+\frac {b+a}{a}ty^{\prime }+\left ( \frac {c}{a}t^{r}+\frac {m}{a}\right ) y & =0\tag {4} \end {align}
Which is Bessel ODE. Comparing the above to the general known Bowman form of Bessel ode which is\begin {equation} t^{2}y^{\prime \prime }+\left ( 1-2\alpha \right ) ty^{\prime }+\left ( \beta ^{2}\gamma ^{2}t^{2\gamma }-\left ( n^{2}\gamma ^{2}-\alpha ^{2}\right ) \right ) y=0\tag {C} \end {equation} And now comparing (4) and (C) shows that\begin {align} \left ( 1-2\alpha \right ) & =\frac {b+a}{a}\tag {5}\\ \beta ^{2}\gamma ^{2} & =\frac {c}{a}\tag {6}\\ 2\gamma & =r\tag {7}\\ \left ( n^{2}\gamma ^{2}-\alpha ^{2}\right ) & =-\frac {m}{a}\tag {8} \end {align}
(5) gives \(\alpha =\frac {1}{2}-\frac {b+a}{2a}\). (7) gives \(\gamma =\frac {r}{2}\). (8) now becomes \(\left ( n^{2}\left ( \frac {r}{2}\right ) ^{2}-\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}\right ) =-\frac {m}{a}\) or \(n^{2}=\frac {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}{\left ( \frac {r}{2}\right ) ^{2}}\). Hence \(n=\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}\ \)by taking the positive root. And finally (6) gives \(\beta ^{2}=\frac {c}{a\gamma ^{2}}\) or \(\beta =\sqrt {\frac {c}{a}}\frac {1}{\gamma }=\sqrt {\frac {c}{a}}\frac {2}{r}\) (also taking the positive root). Hence\begin {align*} \alpha & =\frac {1}{2}-\frac {b+a}{2a}\\ n & =\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}\\ \beta & =\sqrt {\frac {c}{a}}\frac {2}{r}\\ \gamma & =\frac {r}{2} \end {align*}
But the solution to (C) which is general form of Bessel ode is known and given by \[ y\left ( t\right ) =t^{\alpha }\left ( c_{1}J_{n}\left ( \beta t^{\gamma }\right ) +c_{2}Y_{n}\left ( \beta t^{\gamma }\right ) \right ) \] Substituting the above values found into this solution gives\[ y\left ( t\right ) =t^{\frac {1}{2}-\frac {b+a}{2a}}\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}t^{\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}t^{\frac {r}{2}}\right ) \right ) \] Since \(e^{x}=t\) then the above becomes\begin {align} y\left ( x\right ) & =e^{x\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {1}{2}-\frac {b+a}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {-b}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\left ( \frac {-b}{2a}\right ) ^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {m}{a}+\frac {b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {m}{a}+\frac {b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {2}{r}\sqrt {-\frac {4ma+b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {2}{r}\sqrt {-\frac {4ma+b^{2}}{4a^{2}}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \nonumber \\ & =e^{x\left ( \frac {-b}{2a}\right ) }\left ( c_{1}J_{\frac {1}{ra}\sqrt {-4ma+b^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) +c_{2}Y_{\frac {1}{ra}\sqrt {-4ma+b^{2}}}\left ( \sqrt {\frac {c}{a}}\frac {2}{r}e^{x\frac {r}{2}}\right ) \right ) \tag {9} \end {align}
Equation (9) above is the solution to \(ay^{\prime \prime }+by^{\prime }+(ce^{rx}+m)y=0\). Therefore we just need now to compare this form to the ode given and use (9) to obtain the final solution.
Comparing form (1) to the ode we are solving shows that \begin {align*} a &= 1\\ b &= a\\ c &= b\\ r &= \lambda \\ m &= c \end {align*}
Substituting these in (9) gives the solution as \begin {align*} y&=c_{1} {\mathrm e}^{-\frac {a x}{2}} \operatorname {BesselJ}\left (\frac {\sqrt {a^{2}-4 c}}{\lambda }, \frac {2 \sqrt {b}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right )+c_{2} {\mathrm e}^{-\frac {a x}{2}} \operatorname {BesselY}\left (\frac {\sqrt {a^{2}-4 c}}{\lambda }, \frac {2 \sqrt {b}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right ) \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} {\mathrm e}^{-\frac {a x}{2}} \operatorname {BesselJ}\left (\frac {\sqrt {a^{2}-4 c}}{\lambda }, \frac {2 \sqrt {b}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right )+c_{2} {\mathrm e}^{-\frac {a x}{2}} \operatorname {BesselY}\left (\frac {\sqrt {a^{2}-4 c}}{\lambda }, \frac {2 \sqrt {b}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right ) \\ \end{align*}
Verification of solutions
\[ y = c_{1} {\mathrm e}^{-\frac {a x}{2}} \operatorname {BesselJ}\left (\frac {\sqrt {a^{2}-4 c}}{\lambda }, \frac {2 \sqrt {b}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right )+c_{2} {\mathrm e}^{-\frac {a x}{2}} \operatorname {BesselY}\left (\frac {\sqrt {a^{2}-4 c}}{\lambda }, \frac {2 \sqrt {b}\, {\mathrm e}^{\frac {\lambda x}{2}}}{\lambda }\right ) \] Verified OK.
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 <- Bessel successful <- special function solution successful Change of variables used: [x = ln(t)/lambda] Linear ODE actually solved: (b*t+c)*u(t)+(a*lambda*t+lambda^2*t)*diff(u(t),t)+lambda^2*t^2*diff(diff(u(t),t),t) = 0 <- change of variables successful`
✓ Solution by Maple
Time used: 0.344 (sec). Leaf size: 69
dsolve(diff(y(x),x$2)+a*diff(y(x),x)+(b*exp(lambda*x)+c)*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = {\mathrm e}^{-\frac {a x}{2}} \left (\operatorname {BesselJ}\left (\frac {\sqrt {a^{2}-4 c}}{\lambda }, \frac {2 \sqrt {b}\, {\mathrm e}^{\frac {x \lambda }{2}}}{\lambda }\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {a^{2}-4 c}}{\lambda }, \frac {2 \sqrt {b}\, {\mathrm e}^{\frac {x \lambda }{2}}}{\lambda }\right ) c_{2} \right ) \]
✓ Solution by Mathematica
Time used: 0.185 (sec). Leaf size: 123
DSolve[y''[x]+a*y'[x]+(b*Exp[\[Lambda]*x]+c)*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to e^{-\frac {a x}{2}} \left (c_1 \operatorname {Gamma}\left (1-\frac {\sqrt {a^2-4 c}}{\lambda }\right ) \operatorname {BesselJ}\left (-\frac {\sqrt {a^2-4 c}}{\lambda },\frac {2 \sqrt {b e^{x \lambda }}}{\lambda }\right )+c_2 \operatorname {Gamma}\left (\frac {\lambda +\sqrt {a^2-4 c}}{\lambda }\right ) \operatorname {BesselJ}\left (\frac {\sqrt {a^2-4 c}}{\lambda },\frac {2 \sqrt {b e^{x \lambda }}}{\lambda }\right )\right ) \]