Internal problem ID [10423]
Internal file name [OUTPUT/9371_Monday_June_06_2022_02_19_38_PM_36966880/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. subsection 1.2.3. Equations Containing
Exponential Functions
Problem number: 16.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_Riccati]
\[ \boxed {y^{\prime }-a \,{\mathrm e}^{k x} y^{2}-y b=c \,{\mathrm e}^{s x}+d \,{\mathrm e}^{-k x}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,{\mathrm e}^{k x} y^{2}+y b +c \,{\mathrm e}^{s x}+d \,{\mathrm e}^{-k x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,{\mathrm e}^{k x} y^{2}+y b +c \,{\mathrm e}^{s x}+d \,{\mathrm e}^{-k x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c \,{\mathrm e}^{s x}+d \,{\mathrm e}^{-k x}\), \(f_1(x)=b\) and \(f_2(x)=a \,{\mathrm e}^{k x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a \,{\mathrm e}^{k x} u} \tag {1} \end {align*}
Using the above substitution in the given ODE results (after some simplification)in a second order ODE to solve for \(u(x)\) which is \begin {align*} f_2 u''(x) -\left ( f_2' + f_1 f_2 \right ) u'(x) + f_2^2 f_0 u(x) &= 0 \tag {2} \end {align*}
But \begin {align*} f_2' &=a k \,{\mathrm e}^{k x}\\ f_1 f_2 &=b a \,{\mathrm e}^{k x}\\ f_2^2 f_0 &=a^{2} {\mathrm e}^{2 k x} \left (c \,{\mathrm e}^{s x}+d \,{\mathrm e}^{-k x}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \,{\mathrm e}^{k x} u^{\prime \prime }\left (x \right )-\left (b a \,{\mathrm e}^{k x}+a k \,{\mathrm e}^{k x}\right ) u^{\prime }\left (x \right )+a^{2} {\mathrm e}^{2 k x} \left (c \,{\mathrm e}^{s x}+d \,{\mathrm e}^{-k x}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {\left (b +k \right ) x}{2}} \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = -\sqrt {c}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{2} \right ) \sqrt {a}\, {\mathrm e}^{\frac {x \left (b +s +2 k \right )}{2}}+\frac {{\mathrm e}^{\frac {\left (b +k \right ) x}{2}} \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{2} \right ) \left (\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+b +k \right )}{2} \] Using the above in (1) gives the solution \[ y = -\frac {\left (-\sqrt {c}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{2} \right ) \sqrt {a}\, {\mathrm e}^{\frac {x \left (b +s +2 k \right )}{2}}+\frac {{\mathrm e}^{\frac {\left (b +k \right ) x}{2}} \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{2} \right ) \left (\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+b +k \right )}{2}\right ) {\mathrm e}^{-k x} {\mathrm e}^{-\frac {\left (b +k \right ) x}{2}}}{a \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{1} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {{\mathrm e}^{-\frac {x \left (3 k +b \right )}{2}} \left (-2 \sqrt {c}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right ) \sqrt {a}\, {\mathrm e}^{\frac {x \left (b +s +2 k \right )}{2}}+{\mathrm e}^{\frac {\left (b +k \right ) x}{2}} \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right ) \left (\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+b +k \right )\right )}{2 a \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {{\mathrm e}^{-\frac {x \left (3 k +b \right )}{2}} \left (-2 \sqrt {c}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right ) \sqrt {a}\, {\mathrm e}^{\frac {x \left (b +s +2 k \right )}{2}}+{\mathrm e}^{\frac {\left (b +k \right ) x}{2}} \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right ) \left (\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+b +k \right )\right )}{2 a \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {{\mathrm e}^{-\frac {x \left (3 k +b \right )}{2}} \left (-2 \sqrt {c}\, \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+k +s}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right ) \sqrt {a}\, {\mathrm e}^{\frac {x \left (b +s +2 k \right )}{2}}+{\mathrm e}^{\frac {\left (b +k \right ) x}{2}} \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right ) \left (\sqrt {-4 a d +b^{2}+2 b k +k^{2}}+b +k \right )\right )}{2 a \left (\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right ) c_{3} +\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 b k +k^{2}}}{k +s}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (k +s \right )}{2}}}{k +s}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \,{\mathrm e}^{k x} y^{2}-y b =c \,{\mathrm e}^{s x}+d \,{\mathrm e}^{-k x} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \,{\mathrm e}^{k x} y^{2}+y b +c \,{\mathrm e}^{s x}+d \,{\mathrm e}^{-k x} \end {array} \]
Maple trace
`Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear trying Bernoulli trying separable trying inverse linear trying homogeneous types: trying Chini differential order: 1; looking for linear symmetries trying exact Looking for potential symmetries trying Riccati trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (b+k)*(diff(y(x), x))-a*exp(k*x)*(c*exp(s*x)+d*exp(-k*x))*y(x), y(x)` 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)/(s+k)] Linear ODE actually solved: (a*c*t+a*d)*u(t)+(-b*k*t-b*s*t+k*s*t+s^2*t)*diff(u(t),t)+(k^2*t^2+2*k*s*t^2+s^2*t^2)*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 332
dsolve(diff(y(x),x)=a*exp(k*x)*y(x)^2+b*y(x)+c*exp(s*x)+d*exp(-k*x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\sqrt {c}\, a \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 k b +k^{2}}+s +k}{s +k}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 k b +k^{2}}+s +k}{s +k}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right )\right ) {\mathrm e}^{\frac {x \left (s +k \right )}{2}}-\frac {\sqrt {a}\, \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 k b +k^{2}}}{s +k}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 k b +k^{2}}}{s +k}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right )\right ) \left (\sqrt {-4 a d +b^{2}+2 k b +k^{2}}+b +k \right )}{2}\right ) {\mathrm e}^{-k x}}{a^{\frac {3}{2}} \left (\operatorname {BesselY}\left (\frac {\sqrt {-4 a d +b^{2}+2 k b +k^{2}}}{s +k}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right ) c_{1} +\operatorname {BesselJ}\left (\frac {\sqrt {-4 a d +b^{2}+2 k b +k^{2}}}{s +k}, \frac {2 \sqrt {c}\, \sqrt {a}\, {\mathrm e}^{\frac {x \left (s +k \right )}{2}}}{s +k}\right )\right )} \]
✓ Solution by Mathematica
Time used: 18.386 (sec). Leaf size: 1636
DSolve[y'[x]==a*Exp[k*x]*y[x]^2+b*y[x]+c*Exp[s*x]+d*Exp[-k*x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {e^{-k x} \left (-\left ((b+k) K_{\frac {\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4}}\left (2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )\right )+(-1)^{\frac {k^4+4 s k^3+6 s^2 k^2+4 s^3 k+s^4+\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4}} (b+k) \operatorname {BesselI}\left (\frac {\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4},2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right ) c_1+(k+s) \left (K_{-\frac {k^4+4 s k^3+6 s^2 k^2+4 s^3 k+s^4-\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4}}\left (2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+K_{\frac {k^4+4 s k^3+6 s^2 k^2+4 s^3 k+s^4+\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4}}\left (2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+(-1)^{\frac {k^4+4 s k^3+6 s^2 k^2+4 s^3 k+s^4+\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4}} \left (\operatorname {BesselI}\left (-\frac {k^4+4 s k^3+6 s^2 k^2+4 s^3 k+s^4-\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4},2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+\operatorname {BesselI}\left (\frac {k^4+4 s k^3+6 s^2 k^2+4 s^3 k+s^4+\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4},2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )\right ) c_1\right ) \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )}{2 a \left (K_{\frac {\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4}}\left (2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )-(-1)^{\frac {k^4+4 s k^3+6 s^2 k^2+4 s^3 k+s^4+\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4}} \operatorname {BesselI}\left (\frac {\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4},2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right ) c_1\right )} \\ y(x)\to \frac {e^{-k x} \left (-(b+k) (k+s)^3 \sqrt {-\frac {a c \log ^2\left (e^{k+s}\right ) \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}}}{(k+s)^4}} \operatorname {BesselI}\left (\frac {\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4},2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+a c \log ^2\left (e^{k+s}\right ) \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \operatorname {BesselI}\left (\frac {k^4+4 s k^3+6 s^2 k^2+4 s^3 k+s^4+\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4},2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )+a c \log ^2\left (e^{k+s}\right ) \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \operatorname {BesselI}\left (\frac {k^4+4 s k^3+6 s^2 k^2+4 s^3 k+s^4+\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4}-2,2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )\right )}{2 a (k+s)^3 \sqrt {-\frac {a c \log ^2\left (e^{k+s}\right ) \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}}}{(k+s)^4}} \operatorname {BesselI}\left (\frac {\sqrt {\left (b^2+2 k b+k^2-4 a d\right ) (k+s)^4 \log ^2\left (e^{k+s}\right )}}{(k+s)^4},2 \sqrt {-\frac {a c \left (\left (e^{k+s}\right )^x\right )^{\frac {k+s}{\log \left (e^{k+s}\right )}} \log ^2\left (e^{k+s}\right )}{(k+s)^4}}\right )} \\ \end{align*}