Internal problem ID [10476]
Internal file name [OUTPUT/9424_Monday_June_06_2022_02_31_55_PM_71859029/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 1.2.5-1. Equations Containing
Logarithmic Functions
Problem number: 2.
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]
\[ \boxed {y^{\prime } x -a y^{2}=b \ln \left (x \right )+c} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,y^{2}+b \ln \left (x \right )+c}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,y^{2}}{x}+\frac {b \ln \left (x \right )}{x}+\frac {c}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {b \ln \left (x \right )+c}{x}\), \(f_1(x)=0\) and \(f_2(x)=\frac {a}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {a u}{x}} \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' &=-\frac {a}{x^{2}}\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {a^{2} \left (b \ln \left (x \right )+c \right )}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {a u^{\prime \prime }\left (x \right )}{x}+\frac {a u^{\prime }\left (x \right )}{x^{2}}+\frac {a^{2} \left (b \ln \left (x \right )+c \right ) u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+c_{2} \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (-\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) c_{1} -\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) c_{2} \right ) \left (a b \right )^{\frac {1}{3}}}{x} \] Using the above in (1) gives the solution \[ y = -\frac {\left (-\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) c_{1} -\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) c_{2} \right ) \left (a b \right )^{\frac {1}{3}}}{a \left (c_{1} \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+c_{2} \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) c_{3} +\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right ) \left (a b \right )^{\frac {1}{3}}}{a \left (c_{3} \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+\operatorname {AiryBi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) c_{3} +\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right ) \left (a b \right )^{\frac {1}{3}}}{a \left (c_{3} \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+\operatorname {AiryBi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) c_{3} +\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right ) \left (a b \right )^{\frac {1}{3}}}{a \left (c_{3} \operatorname {AiryAi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+\operatorname {AiryBi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a y^{2}=b \ln \left (x \right )+c \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a y^{2}+b \ln \left (x \right )+c}{x} \end {array} \]
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) = -(diff(y(x), x))/x-a*(ln(x)*b+c)*y(x)/x^2, y(x)` *** Sublevel 2 * 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 = exp(t)] Linear ODE actually solved: (a*b*t+a*c)*u(t)+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: 91
dsolve(x*diff(y(x),x)=a*y(x)^2+b*ln(x)+c,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (a b \right )^{\frac {1}{3}} \left (\operatorname {AiryBi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right ) c_{1} +\operatorname {AiryAi}\left (1, -\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )}{a \left (c_{1} \operatorname {AiryBi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )+\operatorname {AiryAi}\left (-\frac {\left (a b \right )^{\frac {1}{3}} \left (b \ln \left (x \right )+c \right )}{b}\right )\right )} \]
✓ Solution by Mathematica
Time used: 1.682 (sec). Leaf size: 149
DSolve[x*y'[x]==a*y[x]^2+b*Log[x]+c,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {b \left (\operatorname {AiryBiPrime}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )+c_1 \operatorname {AiryAiPrime}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )\right )}{(-a b)^{2/3} \left (\operatorname {AiryBi}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )+c_1 \operatorname {AiryAi}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )\right )} \\ y(x)\to \frac {b \operatorname {AiryAiPrime}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )}{(-a b)^{2/3} \operatorname {AiryAi}\left (-\frac {a (c+b \log (x))}{(-a b)^{2/3}}\right )} \\ \end{align*}