Internal problem ID [10494]
Internal file name [OUTPUT/9442_Monday_June_06_2022_02_32_35_PM_83494421/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: 20.
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]
\[ \boxed {y^{\prime } x -a \,x^{2 n} \ln \left (x \right ) y^{2}-\left (b \,x^{n} \ln \left (x \right )-n \right ) y=c \ln \left (x \right )} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,x^{2 n} \ln \left (x \right ) y^{2}+x^{n} \ln \left (x \right ) b y +c \ln \left (x \right )-n y}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a \,x^{2 n} \ln \left (x \right ) y^{2}}{x}+\frac {x^{n} \ln \left (x \right ) b y}{x}+\frac {c \ln \left (x \right )}{x}-\frac {n y}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {c \ln \left (x \right )}{x}\), \(f_1(x)=\frac {b \,x^{n} \ln \left (x \right )-n}{x}\) and \(f_2(x)=\frac {\ln \left (x \right ) a \,x^{2 n}}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {\ln \left (x \right ) a \,x^{2 n} 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 n}}{x^{2}}-\frac {\ln \left (x \right ) a \,x^{2 n}}{x^{2}}+\frac {2 \ln \left (x \right ) a \,x^{2 n} n}{x^{2}}\\ f_1 f_2 &=\frac {\left (b \,x^{n} \ln \left (x \right )-n \right ) \ln \left (x \right ) a \,x^{2 n}}{x^{2}}\\ f_2^2 f_0 &=\frac {\ln \left (x \right )^{3} a^{2} x^{4 n} c}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {\ln \left (x \right ) a \,x^{2 n} u^{\prime \prime }\left (x \right )}{x}-\left (\frac {a \,x^{2 n}}{x^{2}}-\frac {\ln \left (x \right ) a \,x^{2 n}}{x^{2}}+\frac {2 \ln \left (x \right ) a \,x^{2 n} n}{x^{2}}+\frac {\left (b \,x^{n} \ln \left (x \right )-n \right ) \ln \left (x \right ) a \,x^{2 n}}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {\ln \left (x \right )^{3} a^{2} x^{4 n} c u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \left (c_{1} \operatorname {BesselJ}\left (\frac {\sqrt {3}\, \sqrt {-c a}}{8 b}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {c a \,x^{2 n}}\, x^{-n}}{8 b}\right )+c_{2} \operatorname {BesselY}\left (\frac {\sqrt {3}\, \sqrt {-c a}}{8 b}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {c a \,x^{2 n}}\, x^{-n}}{8 b}\right )\right ) \sqrt {\ln \left (x \right )}\, x^{\frac {b \,x^{n}+3 n^{2}}{2 n}} {\mathrm e}^{-\frac {b \,x^{n}}{2 n^{2}}} \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {x^{\frac {-2 n +b \,x^{n}+3 n^{2}}{2 n}} {\mathrm e}^{-\frac {b \,x^{n}}{2 n^{2}}} \left (1+\ln \left (x \right )^{2} x^{n} b +3 n \ln \left (x \right )\right ) \left (c_{1} \operatorname {BesselJ}\left (\frac {\sqrt {3}\, \sqrt {-c a}}{8 b}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {c a \,x^{2 n}}\, x^{-n}}{8 b}\right )+c_{2} \operatorname {BesselY}\left (\frac {\sqrt {3}\, \sqrt {-c a}}{8 b}, \frac {\sqrt {3}\, \sqrt {2}\, \sqrt {c a \,x^{2 n}}\, x^{-n}}{8 b}\right )\right )}{2 \sqrt {\ln \left (x \right )}} \] Using the above in (1) gives the solution \[ y = -\frac {x^{\frac {-2 n +b \,x^{n}+3 n^{2}}{2 n}} \left (1+\ln \left (x \right )^{2} x^{n} b +3 n \ln \left (x \right )\right ) x \,x^{-2 n} x^{-\frac {b \,x^{n}+3 n^{2}}{2 n}}}{2 \ln \left (x \right )^{2} a} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {x^{-2 n} \left (1+\ln \left (x \right )^{2} x^{n} b +3 n \ln \left (x \right )\right )}{2 a \ln \left (x \right )^{2}} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {x^{-2 n} \left (1+\ln \left (x \right )^{2} x^{n} b +3 n \ln \left (x \right )\right )}{2 a \ln \left (x \right )^{2}} \\ \end{align*}
Verification of solutions
\[ y = -\frac {x^{-2 n} \left (1+\ln \left (x \right )^{2} x^{n} b +3 n \ln \left (x \right )\right )}{2 a \ln \left (x \right )^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a \,x^{2 n} \ln \left (x \right ) y^{2}-\left (b \,x^{n} \ln \left (x \right )-n \right ) y=c \ln \left (x \right ) \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,x^{2 n} \ln \left (x \right ) y^{2}+\left (b \,x^{n} \ln \left (x \right )-n \right ) y+c \ln \left (x \right )}{x} \end {array} \]
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 <- Chini successful`
✓ Solution by Maple
Time used: 0.032 (sec). Leaf size: 80
dsolve(x*diff(y(x),x)=a*x^(2*n)*ln(x)*y(x)^2+(b*x^n*ln(x)-n)*y(x)+c*ln(x),y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (\tan \left (\frac {\left (b \left (n \ln \left (x \right )-1\right ) x^{n}+c_{1} n^{2}\right ) \sqrt {4 a \,b^{2} c -b^{4}}}{2 b^{2} n^{2}}\right ) \sqrt {4 a \,b^{2} c -b^{4}}-b^{2}\right ) x^{-n}}{2 a b} \]
✓ Solution by Mathematica
Time used: 1.872 (sec). Leaf size: 130
DSolve[x*y'[x]==a*x^(2*n)*Log[x]*y[x]^2+(b*x^n*Log[x]-n)*y[x]+c*Log[x],y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^{-n} \left (-b+\frac {\sqrt {b^2-4 a c} \left (-e^{\frac {x^n \sqrt {b^2-4 a c} (n \log (x)-1)}{n^2}}+c_1\right )}{e^{\frac {x^n \sqrt {b^2-4 a c} (n \log (x)-1)}{n^2}}+c_1}\right )}{2 a} \\ y(x)\to \frac {x^{-n} \left (\sqrt {b^2-4 a c}-b\right )}{2 a} \\ \end{align*}