Internal problem ID [10497]
Internal file name [OUTPUT/9445_Monday_June_06_2022_02_32_43_PM_25188570/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: 23.
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]
\[ \boxed {\left (a \ln \left (x \right )+b \right ) y^{\prime }-\ln \left (x \right )^{n} y^{2}-y c=-\lambda ^{2} \ln \left (x \right )^{n}+c \lambda } \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {\ln \left (x \right )^{n} y^{2}+y c -\lambda ^{2} \ln \left (x \right )^{n}+c \lambda }{a \ln \left (x \right )+b} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {\lambda ^{2} \ln \left (x \right )^{n}}{a \ln \left (x \right )+b}+\frac {\ln \left (x \right )^{n} y^{2}}{a \ln \left (x \right )+b}+\frac {c \lambda }{a \ln \left (x \right )+b}+\frac {y c}{a \ln \left (x \right )+b} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {-\lambda ^{2} \ln \left (x \right )^{n}+c \lambda }{a \ln \left (x \right )+b}\), \(f_1(x)=\frac {c}{a \ln \left (x \right )+b}\) and \(f_2(x)=\frac {\ln \left (x \right )^{n}}{a \ln \left (x \right )+b}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {\ln \left (x \right )^{n} u}{a \ln \left (x \right )+b}} \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 {\ln \left (x \right )^{n} n}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )}-\frac {\ln \left (x \right )^{n} a}{\left (a \ln \left (x \right )+b \right )^{2} x}\\ f_1 f_2 &=\frac {c \ln \left (x \right )^{n}}{\left (a \ln \left (x \right )+b \right )^{2}}\\ f_2^2 f_0 &=\frac {\ln \left (x \right )^{2 n} \left (-\lambda ^{2} \ln \left (x \right )^{n}+c \lambda \right )}{\left (a \ln \left (x \right )+b \right )^{3}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {\ln \left (x \right )^{n} u^{\prime \prime }\left (x \right )}{a \ln \left (x \right )+b}-\left (\frac {\ln \left (x \right )^{n} n}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )}-\frac {\ln \left (x \right )^{n} a}{\left (a \ln \left (x \right )+b \right )^{2} x}+\frac {c \ln \left (x \right )^{n}}{\left (a \ln \left (x \right )+b \right )^{2}}\right ) u^{\prime }\left (x \right )+\frac {\ln \left (x \right )^{2 n} \left (-\lambda ^{2} \ln \left (x \right )^{n}+c \lambda \right ) u \left (x \right )}{\left (a \ln \left (x \right )+b \right )^{3}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )-\left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime }\left (x \right ) \left (\frac {n}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )}-\frac {a}{\left (a \ln \left (x \right )+b \right )^{2} x}+\frac {c}{\left (a \ln \left (x \right )+b \right )^{2}}\right )+\frac {\textit {\_Y} \left (x \right ) \left (-\ln \left (x \right )^{2 n} \lambda ^{2}+\lambda c \ln \left (x \right )^{n}\right )}{\left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )-\left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime }\left (x \right ) \left (\frac {n}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )}-\frac {a}{\left (a \ln \left (x \right )+b \right )^{2} x}+\frac {c}{\left (a \ln \left (x \right )+b \right )^{2}}\right )+\frac {\textit {\_Y} \left (x \right ) \left (-\ln \left (x \right )^{2 n} \lambda ^{2}+\lambda c \ln \left (x \right )^{n}\right )}{\left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )-\left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime }\left (x \right ) \left (\frac {n}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )}-\frac {a}{\left (a \ln \left (x \right )+b \right )^{2} x}+\frac {c}{\left (a \ln \left (x \right )+b \right )^{2}}\right )+\frac {\textit {\_Y} \left (x \right ) \left (-\ln \left (x \right )^{2 n} \lambda ^{2}+\lambda c \ln \left (x \right )^{n}\right )}{\left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) \ln \left (x \right )^{-n} \left (a \ln \left (x \right )+b \right )}{\operatorname {DESol}\left (\left \{\textit {\_Y}^{\prime \prime }\left (x \right )-\left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime }\left (x \right ) \left (\frac {n}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )}-\frac {a}{\left (a \ln \left (x \right )+b \right )^{2} x}+\frac {c}{\left (a \ln \left (x \right )+b \right )^{2}}\right )+\frac {\textit {\_Y} \left (x \right ) \left (-\ln \left (x \right )^{2 n} \lambda ^{2}+\lambda c \ln \left (x \right )^{n}\right )}{\left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {-x \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+2 n} \lambda ^{2}+c x \lambda \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+n}+\left (a \ln \left (x \right )+b \right ) \left (x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime \prime }\left (x \right )-\left (\left (a \left (-1+n \right )+c x \right ) \ln \left (x \right )+b n \right ) \textit {\_Y}^{\prime }\left (x \right )\right )}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) \ln \left (x \right )^{-n} \left (a \ln \left (x \right )+b \right )}{\operatorname {DESol}\left (\left \{\frac {-x \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+2 n} \lambda ^{2}+c x \lambda \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+n}+\left (a \ln \left (x \right )+b \right ) \left (x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime \prime }\left (x \right )-\left (\left (a \left (-1+n \right )+c x \right ) \ln \left (x \right )+b n \right ) \textit {\_Y}^{\prime }\left (x \right )\right )}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {-x \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+2 n} \lambda ^{2}+c x \lambda \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+n}+\left (a \ln \left (x \right )+b \right ) \left (x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime \prime }\left (x \right )-\left (\left (a \left (-1+n \right )+c x \right ) \ln \left (x \right )+b n \right ) \textit {\_Y}^{\prime }\left (x \right )\right )}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) \ln \left (x \right )^{-n} \left (a \ln \left (x \right )+b \right )}{\operatorname {DESol}\left (\left \{\frac {-x \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+2 n} \lambda ^{2}+c x \lambda \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+n}+\left (a \ln \left (x \right )+b \right ) \left (x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime \prime }\left (x \right )-\left (\left (a \left (-1+n \right )+c x \right ) \ln \left (x \right )+b n \right ) \textit {\_Y}^{\prime }\left (x \right )\right )}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\left (\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {-x \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+2 n} \lambda ^{2}+c x \lambda \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+n}+\left (a \ln \left (x \right )+b \right ) \left (x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime \prime }\left (x \right )-\left (\left (a \left (-1+n \right )+c x \right ) \ln \left (x \right )+b n \right ) \textit {\_Y}^{\prime }\left (x \right )\right )}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )\right ) \ln \left (x \right )^{-n} \left (a \ln \left (x \right )+b \right )}{\operatorname {DESol}\left (\left \{\frac {-x \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+2 n} \lambda ^{2}+c x \lambda \textit {\_Y} \left (x \right ) \ln \left (x \right )^{1+n}+\left (a \ln \left (x \right )+b \right ) \left (x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right ) \textit {\_Y}^{\prime \prime }\left (x \right )-\left (\left (a \left (-1+n \right )+c x \right ) \ln \left (x \right )+b n \right ) \textit {\_Y}^{\prime }\left (x \right )\right )}{x \ln \left (x \right ) \left (a \ln \left (x \right )+b \right )^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left (a \ln \left (x \right )+b \right ) y^{\prime }-\ln \left (x \right )^{n} y^{2}-y c =-\lambda ^{2} \ln \left (x \right )^{n}+c \lambda \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {\ln \left (x \right )^{n} y^{2}+y c -\lambda ^{2} \ln \left (x \right )^{n}+c \lambda }{a \ln \left (x \right )+b} \end {array} \]
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: trying Riccati_symmetries trying Riccati to 2nd Order -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (ln(x)*a*n+c*x*ln(x)-ln(x)*a+b*n)*(diff(y(x), x))/(x*ln(x)*(ln(x)*a+b) 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 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients trying 2nd order, integrating factor of the form mu(x,y) 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 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 a symmetry of the form [xi=0, eta=F(x)] trying 2nd order exact linear trying symmetries linear in x and y(x) trying to convert to a linear ODE with constant coefficients <- unable to find a useful change of variables trying a symmetry of the form [xi=0, eta=F(x)] trying to convert to an ODE of Bessel type -> Trying a change of variables to reduce to Bernoulli -> Calling odsolve with the ODE`, diff(y(x), x)-(ln(x)^n*y(x)^2/(ln(x)*a+b)+y(x)+c*y(x)*x/(ln(x)*a+b)+x^2*(-ln(x)^n*lambda^2/(ln( 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: trying Riccati_symmetries trying inverse_Riccati trying 1st order ODE linearizable_by_differentiation -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] <- symmetry pattern of the form [0, F(x)*G(y)] successful <- Riccati with symmetry pattern of the form [0,F(x)*G(y)] successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 124
dsolve((a*ln(x)+b)*diff(y(x),x)=(ln(x))^n*y(x)^2+c*y(x)-lambda^2*(ln(x))^n+c*lambda,y(x), singsol=all)
\[ y \left (x \right ) = \frac {-\lambda c_{1} -\left (\int \frac {\ln \left (x \right )^{n} {\mathrm e}^{-\left (\int \frac {2 \ln \left (x \right )^{n} \lambda -c}{a \ln \left (x \right )+b}d x \right )}}{a \ln \left (x \right )+b}d x \right ) \lambda -{\mathrm e}^{-\left (\int \frac {2 \ln \left (x \right )^{n} \lambda -c}{a \ln \left (x \right )+b}d x \right )}}{c_{1} +\int \frac {\ln \left (x \right )^{n} {\mathrm e}^{-\left (\int \frac {2 \ln \left (x \right )^{n} \lambda -c}{a \ln \left (x \right )+b}d x \right )}}{a \ln \left (x \right )+b}d x} \]
✓ Solution by Mathematica
Time used: 5.137 (sec). Leaf size: 286
DSolve[(a*Log[x]+b)*y'[x]==(Log[x])^n*y[x]^2+c*y[x]-\[Lambda]^2*(Log[x])^n+c*\[Lambda],y[x],x,IncludeSingularSolutions -> True]
\[ \text {Solve}\left [\int _1^x\frac {\exp \left (-\int _1^{K[2]}-\frac {c-2 \lambda \log ^n(K[1])}{b+a \log (K[1])}dK[1]\right ) \left (-\lambda \log ^n(K[2])+y(x) \log ^n(K[2])+c\right )}{c n (b+a \log (K[2])) (\lambda +y(x))}dK[2]+\int _1^{y(x)}\left (-\int _1^x\left (\frac {\exp \left (-\int _1^{K[2]}-\frac {c-2 \lambda \log ^n(K[1])}{b+a \log (K[1])}dK[1]\right ) \log ^n(K[2])}{c n (\lambda +K[3]) (b+a \log (K[2]))}-\frac {\exp \left (-\int _1^{K[2]}-\frac {c-2 \lambda \log ^n(K[1])}{b+a \log (K[1])}dK[1]\right ) \left (-\lambda \log ^n(K[2])+K[3] \log ^n(K[2])+c\right )}{c n (\lambda +K[3])^2 (b+a \log (K[2]))}\right )dK[2]-\frac {\exp \left (-\int _1^x-\frac {c-2 \lambda \log ^n(K[1])}{b+a \log (K[1])}dK[1]\right )}{c n (\lambda +K[3])^2}\right )dK[3]=c_1,y(x)\right ] \]