Internal problem ID [10489]
Internal file name [OUTPUT/9437_Monday_June_06_2022_02_32_25_PM_74432929/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: 15.
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]
\[ \boxed {y^{\prime }-a \ln \left (x \right )^{k} \left (y-b \,x^{n}-c \right )^{2}=b n \,x^{-1+n}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \ln \left (x \right )^{k} x^{2 n} a \,b^{2}+2 \ln \left (x \right )^{k} x^{n} a b c -2 \ln \left (x \right )^{k} x^{n} a b y +\ln \left (x \right )^{k} a \,c^{2}-2 \ln \left (x \right )^{k} a c y +\ln \left (x \right )^{k} a \,y^{2}+b n \,x^{-1+n} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = \ln \left (x \right )^{k} x^{2 n} a \,b^{2}+2 \ln \left (x \right )^{k} x^{n} a b c -2 \ln \left (x \right )^{k} x^{n} a b y +\ln \left (x \right )^{k} a \,c^{2}-2 \ln \left (x \right )^{k} a c y +\ln \left (x \right )^{k} a \,y^{2}+\frac {b n \,x^{n}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\ln \left (x \right )^{k} x^{2 n} a \,b^{2}+2 \ln \left (x \right )^{k} x^{n} a b c +\ln \left (x \right )^{k} a \,c^{2}+b n \,x^{-1+n}\), \(f_1(x)=-2 \ln \left (x \right )^{k} x^{n} a b -2 \ln \left (x \right )^{k} a c\) and \(f_2(x)=a \ln \left (x \right )^{k}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a \ln \left (x \right )^{k} 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' &=\frac {a \ln \left (x \right )^{k} k}{x \ln \left (x \right )}\\ f_1 f_2 &=\left (-2 \ln \left (x \right )^{k} x^{n} a b -2 \ln \left (x \right )^{k} a c \right ) a \ln \left (x \right )^{k}\\ f_2^2 f_0 &=a^{2} \ln \left (x \right )^{2 k} \left (\ln \left (x \right )^{k} x^{2 n} a \,b^{2}+2 \ln \left (x \right )^{k} x^{n} a b c +\ln \left (x \right )^{k} a \,c^{2}+b n \,x^{-1+n}\right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a \ln \left (x \right )^{k} u^{\prime \prime }\left (x \right )-\left (\frac {a \ln \left (x \right )^{k} k}{x \ln \left (x \right )}+\left (-2 \ln \left (x \right )^{k} x^{n} a b -2 \ln \left (x \right )^{k} a c \right ) a \ln \left (x \right )^{k}\right ) u^{\prime }\left (x \right )+a^{2} \ln \left (x \right )^{2 k} \left (\ln \left (x \right )^{k} x^{2 n} a \,b^{2}+2 \ln \left (x \right )^{k} x^{n} a b c +\ln \left (x \right )^{k} a \,c^{2}+b n \,x^{-1+n}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \frac {\left (c_{2} \left (x \left (-\ln \left (x \right )\right )^{k}+\Gamma \left (k , -\ln \left (x \right )\right ) k -\Gamma \left (k +1\right )\right ) \ln \left (x \right )^{\frac {k}{2}}+\ln \left (x \right )^{-\frac {k}{2}} c_{1} \left (-\ln \left (x \right )\right )^{k}\right ) \left (-\ln \left (x \right )\right )^{-k} {\mathrm e}^{-\frac {\left (\int \left (2 \ln \left (x \right )^{k} x^{n} a b +2 \ln \left (x \right )^{k} a c -\frac {1}{x}-\frac {k}{x \ln \left (x \right )}\right )d x \right )}{2}}}{\sqrt {x}} \] The above shows that \[ u^{\prime }\left (x \right ) = -\frac {\left (-\ln \left (x \right )\right )^{-k} \left (c_{2} \left (\left (x^{1+n} b +c x \right ) \left (-\ln \left (x \right )\right )^{k}-\left (b \,x^{n}+c \right ) \left (-\Gamma \left (k , -\ln \left (x \right )\right ) k +\Gamma \left (k +1\right )\right )\right ) a \ln \left (x \right )^{\frac {3 k}{2}}+\left (-\ln \left (x \right )\right )^{k} \ln \left (x \right )^{\frac {k}{2}} \left (x^{n} c_{1} a b +c_{1} c a -c_{2} \right )\right ) {\mathrm e}^{-\frac {\left (\int \left (2 \ln \left (x \right )^{k} x^{n} a b +2 \ln \left (x \right )^{k} a c -\frac {1}{x}-\frac {k}{x \ln \left (x \right )}\right )d x \right )}{2}}}{\sqrt {x}} \] Using the above in (1) gives the solution \[ y = \frac {\left (c_{2} \left (\left (x^{1+n} b +c x \right ) \left (-\ln \left (x \right )\right )^{k}-\left (b \,x^{n}+c \right ) \left (-\Gamma \left (k , -\ln \left (x \right )\right ) k +\Gamma \left (k +1\right )\right )\right ) a \ln \left (x \right )^{\frac {3 k}{2}}+\left (-\ln \left (x \right )\right )^{k} \ln \left (x \right )^{\frac {k}{2}} \left (x^{n} c_{1} a b +c_{1} c a -c_{2} \right )\right ) \ln \left (x \right )^{-k}}{a \left (c_{2} \left (x \left (-\ln \left (x \right )\right )^{k}+\Gamma \left (k , -\ln \left (x \right )\right ) k -\Gamma \left (k +1\right )\right ) \ln \left (x \right )^{\frac {k}{2}}+\ln \left (x \right )^{-\frac {k}{2}} c_{1} \left (-\ln \left (x \right )\right )^{k}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\ln \left (x \right )^{-k} \left (\left (\left (x^{1+n} b +c x \right ) \left (-\ln \left (x \right )\right )^{k}-\left (b \,x^{n}+c \right ) \left (-\Gamma \left (k , -\ln \left (x \right )\right ) k +\Gamma \left (k +1\right )\right )\right ) a \ln \left (x \right )^{\frac {3 k}{2}}+\left (-\ln \left (x \right )\right )^{k} \ln \left (x \right )^{\frac {k}{2}} \left (x^{n} c_{3} a b +c_{3} c a -1\right )\right )}{\left (\left (x \left (-\ln \left (x \right )\right )^{k}+\Gamma \left (k , -\ln \left (x \right )\right ) k -\Gamma \left (k +1\right )\right ) \ln \left (x \right )^{\frac {k}{2}}+\ln \left (x \right )^{-\frac {k}{2}} c_{3} \left (-\ln \left (x \right )\right )^{k}\right ) a} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\ln \left (x \right )^{-k} \left (\left (\left (x^{1+n} b +c x \right ) \left (-\ln \left (x \right )\right )^{k}-\left (b \,x^{n}+c \right ) \left (-\Gamma \left (k , -\ln \left (x \right )\right ) k +\Gamma \left (k +1\right )\right )\right ) a \ln \left (x \right )^{\frac {3 k}{2}}+\left (-\ln \left (x \right )\right )^{k} \ln \left (x \right )^{\frac {k}{2}} \left (x^{n} c_{3} a b +c_{3} c a -1\right )\right )}{\left (\left (x \left (-\ln \left (x \right )\right )^{k}+\Gamma \left (k , -\ln \left (x \right )\right ) k -\Gamma \left (k +1\right )\right ) \ln \left (x \right )^{\frac {k}{2}}+\ln \left (x \right )^{-\frac {k}{2}} c_{3} \left (-\ln \left (x \right )\right )^{k}\right ) a} \\ \end{align*}
Verification of solutions
\[ y = \frac {\ln \left (x \right )^{-k} \left (\left (\left (x^{1+n} b +c x \right ) \left (-\ln \left (x \right )\right )^{k}-\left (b \,x^{n}+c \right ) \left (-\Gamma \left (k , -\ln \left (x \right )\right ) k +\Gamma \left (k +1\right )\right )\right ) a \ln \left (x \right )^{\frac {3 k}{2}}+\left (-\ln \left (x \right )\right )^{k} \ln \left (x \right )^{\frac {k}{2}} \left (x^{n} c_{3} a b +c_{3} c a -1\right )\right )}{\left (\left (x \left (-\ln \left (x \right )\right )^{k}+\Gamma \left (k , -\ln \left (x \right )\right ) k -\Gamma \left (k +1\right )\right ) \ln \left (x \right )^{\frac {k}{2}}+\ln \left (x \right )^{-\frac {k}{2}} c_{3} \left (-\ln \left (x \right )\right )^{k}\right ) a} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-a \ln \left (x \right )^{k} \left (y-b \,x^{n}-c \right )^{2}=b n \,x^{-1+n} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a \ln \left (x \right )^{k} \left (y-b \,x^{n}-c \right )^{2}+b n \,x^{-1+n} \end {array} \]
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: 0.016 (sec). Leaf size: 24
dsolve(diff(y(x),x)=a*(ln(x))^k*(y(x)-b*x^n-c)^2+b*n*x^(n-1),y(x), singsol=all)
\[ y \left (x \right ) = b \,x^{n}+c +\frac {1}{c_{1} -a \left (\int \ln \left (x \right )^{k}d x \right )} \]
✓ Solution by Mathematica
Time used: 1.572 (sec). Leaf size: 51
DSolve[y'[x]==a*(Log[x])^k*(y[x]-b*x^n-c)^2+b*n*x^(n-1),y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {1}{-a (-\log (x))^{-k} \log ^k(x) \Gamma (k+1,-\log (x))+c_1}+b x^n+c \\ y(x)\to b x^n+c \\ \end{align*}