Internal problem ID [10493]
Internal file name [OUTPUT/9441_Monday_June_06_2022_02_32_34_PM_36674882/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: 19.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[[_1st_order, `_with_symmetry_[F(x),G(x)]`], _Riccati]
\[ \boxed {y^{\prime } x -a \,x^{n} \left (y+b \ln \left (x \right )\right )^{2}=-b} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {a \,b^{2} x^{n} \ln \left (x \right )^{2}+2 x^{n} \ln \left (x \right ) a b y +a \,x^{n} y^{2}-b}{x} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {x^{n} \ln \left (x \right )^{2} a \,b^{2}}{x}+\frac {2 x^{n} \ln \left (x \right ) a b y}{x}+\frac {a \,x^{n} y^{2}}{x}-\frac {b}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {-b +a \,b^{2} x^{n} \ln \left (x \right )^{2}}{x}\), \(f_1(x)=\frac {2 a b \,x^{n} \ln \left (x \right )}{x}\) and \(f_2(x)=\frac {a \,x^{n}}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {a \,x^{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 {x^{n} a}{x^{2}}+\frac {x^{n} n a}{x^{2}}\\ f_1 f_2 &=\frac {2 a^{2} b \,x^{2 n} \ln \left (x \right )}{x^{2}}\\ f_2^2 f_0 &=\frac {a^{2} x^{2 n} \left (-b +a \,b^{2} x^{n} \ln \left (x \right )^{2}\right )}{x^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {a \,x^{n} u^{\prime \prime }\left (x \right )}{x}-\left (-\frac {x^{n} a}{x^{2}}+\frac {x^{n} n a}{x^{2}}+\frac {2 a^{2} b \,x^{2 n} \ln \left (x \right )}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {a^{2} x^{2 n} \left (-b +a \,b^{2} x^{n} \ln \left (x \right )^{2}\right ) u \left (x \right )}{x^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{-\frac {a b \,x^{n}}{n^{2}}} \left (x^{\frac {a b \,x^{n}}{n}} c_{1} +x^{\frac {n^{2}+b a \,x^{n}}{n}} c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (x^{\frac {2 n^{2}+b a \,x^{n}}{n}} \ln \left (x \right ) c_{2} a b +\ln \left (x \right ) x^{\frac {n^{2}+b a \,x^{n}}{n}} c_{1} a b +x^{\frac {n^{2}+b a \,x^{n}}{n}} c_{2} n \right ) {\mathrm e}^{-\frac {a b \,x^{n}}{n^{2}}}}{x} \] Using the above in (1) gives the solution \[ y = -\frac {\left (x^{\frac {2 n^{2}+b a \,x^{n}}{n}} \ln \left (x \right ) c_{2} a b +\ln \left (x \right ) x^{\frac {n^{2}+b a \,x^{n}}{n}} c_{1} a b +x^{\frac {n^{2}+b a \,x^{n}}{n}} c_{2} n \right ) x^{-n}}{a \left (x^{\frac {a b \,x^{n}}{n}} c_{1} +x^{\frac {n^{2}+b a \,x^{n}}{n}} c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left (a b \,x^{\frac {n^{2}+b a \,x^{n}}{n}} \ln \left (x \right )+x^{\frac {a b \,x^{n}}{n}} \left (\ln \left (x \right ) c_{3} a b +n \right )\right ) x^{-\frac {a b \,x^{n}}{n}}}{a \left (x^{n}+c_{3} \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (a b \,x^{\frac {n^{2}+b a \,x^{n}}{n}} \ln \left (x \right )+x^{\frac {a b \,x^{n}}{n}} \left (\ln \left (x \right ) c_{3} a b +n \right )\right ) x^{-\frac {a b \,x^{n}}{n}}}{a \left (x^{n}+c_{3} \right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (a b \,x^{\frac {n^{2}+b a \,x^{n}}{n}} \ln \left (x \right )+x^{\frac {a b \,x^{n}}{n}} \left (\ln \left (x \right ) c_{3} a b +n \right )\right ) x^{-\frac {a b \,x^{n}}{n}}}{a \left (x^{n}+c_{3} \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a \,x^{n} \left (y+b \ln \left (x \right )\right )^{2}=-b \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a \,x^{n} \left (y+b \ln \left (x \right )\right )^{2}-b}{x} \end {array} \]
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: <- Riccati particular case Kamke (d) successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 24
dsolve(x*diff(y(x),x)=a*x^n*(y(x)+b*ln(x))^2-b,y(x), singsol=all)
\[ y \left (x \right ) = -b \ln \left (x \right )+\frac {n}{c_{1} n -a \,x^{n}} \]
✓ Solution by Mathematica
Time used: 0.649 (sec). Leaf size: 35
DSolve[x*y'[x]==a*x^n*(y[x]+b*Log[x])^2-b,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -b \log (x)+\frac {n}{-a x^n+c_1 n} \\ y(x)\to -b \log (x) \\ \end{align*}