Internal problem ID [10485]
Internal file name [OUTPUT/9433_Monday_June_06_2022_02_32_17_PM_25607459/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 1.2.5-2
Problem number: 11.
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]
\[ \boxed {y^{\prime }-y^{2}-a x \ln \left (b x \right )^{m} y=a \ln \left (b x \right )^{m}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+a x \ln \left (b x \right )^{m} y +a \ln \left (b x \right )^{m} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+a x \ln \left (b x \right )^{m} y +a \ln \left (b x \right )^{m} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a \ln \left (b x \right )^{m}\), \(f_1(x)=\ln \left (b x \right )^{m} a x\) 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 &=\ln \left (b x \right )^{m} a x\\ f_2^2 f_0 &=a \ln \left (b x \right )^{m} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )-\ln \left (b x \right )^{m} a x u^{\prime }\left (x \right )+a \ln \left (b x \right )^{m} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = x \left (\left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{1} +c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{1} +c_{2} +x \,{\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x} c_{1} \] Using the above in (1) gives the solution \[ y = -\frac {\left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{1} +c_{2} +x \,{\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x} c_{1}}{x \left (\left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{1} +c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{3} -1-x \,{\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x} c_{3}}{x \left (\left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{3} +1\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{3} -1-x \,{\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x} c_{3}}{x \left (\left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{3} +1\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{3} -1-x \,{\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x} c_{3}}{x \left (\left (\int {\mathrm e}^{\int \frac {\ln \left (b x \right )^{m} a \,x^{2}-2}{x}d x}d x \right ) c_{3} +1\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}-a x \ln \left (b x \right )^{m} y=a \ln \left (b x \right )^{m} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+a x \ln \left (b x \right )^{m} y+a \ln \left (b x \right )^{m} \end {array} \]
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 found: 2 potential symmetries. Proceeding with integration step`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 85
dsolve(diff(y(x),x)=y(x)^2+a*x*(ln(b*x))^m*y(x)+a*(ln(b*x))^m,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-{\mathrm e}^{\int \frac {a \ln \left (b x \right )^{m} x^{2}-2}{x}d x} x -\left (\int {\mathrm e}^{\int \frac {a \ln \left (b x \right )^{m} x^{2}-2}{x}d x}d x \right )+c_{1}}{\left (-c_{1} +\int {\mathrm e}^{\int \frac {a \ln \left (b x \right )^{m} x^{2}-2}{x}d x}d x \right ) x} \]
✓ Solution by Mathematica
Time used: 3.589 (sec). Leaf size: 181
DSolve[y'[x]==y[x]^2+a*x*(Log[b*x])^m*y[x]+a*(Log[b*x])^m,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {x \int _1^x\frac {\exp \left (\frac {2^{-m-1} a \Gamma (m+1,-2 \log (b K[1])) (-\log (b K[1]))^{-m} \log ^m(b K[1])}{b^2}\right )}{K[1]^2}dK[1]+\exp \left (\frac {a 2^{-m-1} (-\log (b x))^{-m} \log ^m(b x) \Gamma (m+1,-2 \log (b x))}{b^2}\right )+c_1 x}{x^2 \left (\int _1^x\frac {\exp \left (\frac {2^{-m-1} a \Gamma (m+1,-2 \log (b K[1])) (-\log (b K[1]))^{-m} \log ^m(b K[1])}{b^2}\right )}{K[1]^2}dK[1]+c_1\right )} \\ y(x)\to -\frac {1}{x} \\ \end{align*}