Internal problem ID [8425]
Internal file name [OUTPUT/7358_Sunday_June_05_2022_10_53_40_PM_97111455/index.tex]
Book: Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section: Chapter 1, linear first order
Problem number: 88.
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]
\[ \boxed {2 y^{\prime }-3 y^{2}-4 y a=b +c \,{\mathrm e}^{-2 x a}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {3 y^{2}}{2}+2 a y +\frac {b}{2}+\frac {c \,{\mathrm e}^{-2 x a}}{2} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {3 y^{2}}{2}+2 a y +\frac {b}{2}+\frac {c \,{\mathrm e}^{-2 x a}}{2} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {b}{2}+\frac {c \,{\mathrm e}^{-2 x a}}{2}\), \(f_1(x)=2 a\) and \(f_2(x)={\frac {3}{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {3 u}{2}} \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' &=0\\ f_1 f_2 &=3 a\\ f_2^2 f_0 &=\frac {9 b}{8}+\frac {9 c \,{\mathrm e}^{-2 x a}}{8} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {3 u^{\prime \prime }\left (x \right )}{2}-3 a u^{\prime }\left (x \right )+\left (\frac {9 b}{8}+\frac {9 c \,{\mathrm e}^{-2 x a}}{8}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{x a} \left (\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} \right ) \left (\sqrt {4 a^{2}-3 b}+2 a \right ) {\mathrm e}^{x a}}{2}+\frac {\left (\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} +\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} \right ) \sqrt {3}\, \sqrt {c}}{2} \] Using the above in (1) gives the solution \[ y = -\frac {2 \left (\frac {\left (\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} \right ) \left (\sqrt {4 a^{2}-3 b}+2 a \right ) {\mathrm e}^{x a}}{2}+\frac {\left (\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} +\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} \right ) \sqrt {3}\, \sqrt {c}}{2}\right ) {\mathrm e}^{-x a}}{3 \left (\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\left (\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} +\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} \right ) \sqrt {3}\, {\mathrm e}^{-x a} \sqrt {c}-\left (\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} \right ) \left (\sqrt {4 a^{2}-3 b}+2 a \right )}{3 \operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} +3 \operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\left (\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} +\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} \right ) \sqrt {3}\, {\mathrm e}^{-x a} \sqrt {c}-\left (\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} \right ) \left (\sqrt {4 a^{2}-3 b}+2 a \right )}{3 \operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} +3 \operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1}} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\left (\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} +\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} \right ) \sqrt {3}\, {\mathrm e}^{-x a} \sqrt {c}-\left (\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} \right ) \left (\sqrt {4 a^{2}-3 b}+2 a \right )}{3 \operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{2} +3 \operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-x a}}{2 a}\right ) c_{1}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 y^{\prime }-3 y^{2}-4 y a =b +c \,{\mathrm e}^{-2 x a} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {3 y^{2}}{2}+2 y a +\frac {b}{2}+\frac {c \,{\mathrm e}^{-2 x a}}{2} \end {array} \]
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) = 2*(diff(y(x), x))*a+(-(3/4)*b-(3/4)*c*exp(-2*a*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)/a] Linear ODE actually solved: (3*b*t^2+3*c)*u(t)-4*t^3*a^2*diff(u(t),t)+4*t^4*a^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: 256
dsolve(2*diff(y(x),x) - 3*y(x)^2 - 4*a*y(x) - b - c*exp(-2*a*x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-\sqrt {3}\, \left (\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-a x}}{2 a}\right ) c_{1} +\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}-2 a}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-a x}}{2 a}\right )\right ) {\mathrm e}^{-a x} \sqrt {c}-\left (\sqrt {4 a^{2}-3 b}+2 a \right ) \left (\operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-a x}}{2 a}\right ) c_{1} +\operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-a x}}{2 a}\right )\right )}{3 \operatorname {BesselY}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-a x}}{2 a}\right ) c_{1} +3 \operatorname {BesselJ}\left (-\frac {\sqrt {4 a^{2}-3 b}}{2 a}, \frac {\sqrt {3}\, \sqrt {c}\, {\mathrm e}^{-a x}}{2 a}\right )} \]
✓ Solution by Mathematica
Time used: 2.009 (sec). Leaf size: 2746
DSolve[2*y'[x] - 3*y[x]^2 - 4*a*y[x] - b - c*Exp[-2*a*x]==0,y[x],x,IncludeSingularSolutions -> True]
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