Internal problem ID [10483]
Internal file name [OUTPUT/9431_Monday_June_06_2022_02_32_14_PM_30986494/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: 9.
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]
\[ \boxed {x^{2} \ln \left (x a \right ) \left (y^{\prime }-y^{2}\right )=1} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\ln \left (x a \right ) y^{2} x^{2}+1}{x^{2} \ln \left (x a \right )} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+\frac {1}{x^{2} \ln \left (x a \right )} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {1}{x^{2} \ln \left (x a \right )}\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {1}{x^{2} \ln \left (x a \right )} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\frac {u \left (x \right )}{x^{2} \ln \left (x a \right )} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = -\ln \left (x a \right ) \operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right ) c_{2} -c_{2} a x +c_{1} \ln \left (x a \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {-\operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right ) c_{2} +c_{1}}{x} \] Using the above in (1) gives the solution \[ y = -\frac {-\operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right ) c_{2} +c_{1}}{x \left (-\ln \left (x a \right ) \operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right ) c_{2} -c_{2} a x +c_{1} \ln \left (x a \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right )+c_{3}}{x \left (\operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right ) \ln \left (x a \right )-c_{3} \ln \left (x a \right )+x a \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right )+c_{3}}{x \left (\operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right ) \ln \left (x a \right )-c_{3} \ln \left (x a \right )+x a \right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right )+c_{3}}{x \left (\operatorname {expIntegral}_{1}\left (-\ln \left (x a \right )\right ) \ln \left (x a \right )-c_{3} \ln \left (x a \right )+x a \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} \ln \left (x a \right ) \left (y^{\prime }-y^{2}\right )=1 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\ln \left (x a \right ) y^{2} x^{2}+1}{x^{2} \ln \left (x a \right )} \end {array} \]
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 Special trying Riccati sub-methods: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = -y(x)/(x^2*ln(a*x)), 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)] <- linear_1 successful Change of variables used: [x = exp(t)/a] Linear ODE actually solved: u(t)-t*diff(u(t),t)+t*diff(diff(u(t),t),t) = 0 <- change of variables successful <- Riccati to 2nd Order successful`
✓ Solution by Maple
Time used: 0.015 (sec). Leaf size: 45
dsolve(x^2*ln(a*x)*(diff(y(x),x)-y(x)^2)=1,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-c_{1} \operatorname {expIntegral}_{1}\left (-\ln \left (a x \right )\right )+1}{x \left (\left (c_{1} \operatorname {expIntegral}_{1}\left (-\ln \left (a x \right )\right )-1\right ) \ln \left (a x \right )+c_{1} a x \right )} \]
✓ Solution by Mathematica
Time used: 0.616 (sec). Leaf size: 74
DSolve[x^2*Log[a*x]*(y'[x]-y[x]^2)==1,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {a+c_1 \operatorname {LogIntegral}(a x)}{-c_1 x \operatorname {LogIntegral}(a x) \log (a x)+a c_1 x^2-a x \log (a x)} \\ y(x)\to \frac {\operatorname {LogIntegral}(a x)}{a x^2-x \operatorname {LogIntegral}(a x) \log (a x)} \\ \end{align*}