Internal problem ID [10641]
Internal file name [OUTPUT/9589_Monday_June_06_2022_03_11_07_PM_128411/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 1.2.9. Some Transformations
Problem number: 7.
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]
\[ \boxed {y^{\prime }-y^{2}={\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\lambda ^{2}}{4}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= y^{2}+{\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\lambda ^{2}}{4} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+{\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\lambda ^{2}}{4} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)={\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\lambda ^{2}}{4}\), \(f_1(x)=0\) 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 &=0\\ f_2^2 f_0 &={\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\lambda ^{2}}{4} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\left ({\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\lambda ^{2}}{4}\right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\textit {\_Y} \left (x \right ) {\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\textit {\_Y} \left (x \right ) \lambda ^{2}}{4}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}={\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\lambda ^{2}}{4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+{\mathrm e}^{2 \lambda x} f \left ({\mathrm e}^{\lambda x}\right )-\frac {\lambda ^{2}}{4} \end {array} \]
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 inverse_Riccati trying 1st order ODE linearizable_by_differentiation --- Trying Lie symmetry methods, 1st order --- `, `-> Computing symmetries using: way = 4 `, `-> Computing symmetries using: way = 2 `, `-> Computing symmetries using: way = 6 trying symmetry patterns for 1st order ODEs -> trying a symmetry pattern of the form [F(x)*G(y), 0] -> trying a symmetry pattern of the form [0, F(x)*G(y)] -> trying symmetry patterns of the forms [F(x),G(y)] and [G(y),F(x)] `, `-> Computing symmetries using: way = HINT -> Calling odsolve with the ODE`, diff(y(x), x) = -8*y(x)*x/((lambda-2*x)*(2*x+lambda)), y(x)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful -> Calling odsolve with the ODE`, diff(y(x), x)+4*y(x)*lambda*(2*exp(2*lambda*x)*f(exp(lambda*x))+exp(3*lambda*x)*(D(f))(exp(lamb Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful -> Calling odsolve with the ODE`, diff(y(x), x)-2*lambda*K[1], y(x)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful -> Calling odsolve with the ODE`, diff(y(x), x)+8*y(x)*x/((lambda-2*x)*(2*x+lambda)), y(x)` *** Sublevel 2 *** Methods for first order ODEs: --- Trying classification methods --- trying a quadrature trying 1st order linear <- 1st order linear successful -> trying a symmetry pattern of the form [F(x),G(x)] -> trying a symmetry pattern of the form [F(y),G(y)] -> trying a symmetry pattern of the form [F(x)+G(y), 0] -> trying a symmetry pattern of the form [0, F(x)+G(y)] -> trying a symmetry pattern of the form [F(x),G(x)*y+H(x)] -> trying a symmetry pattern of conformal type`
✗ Solution by Maple
dsolve(diff(y(x),x)=y(x)^2+exp(2*lambda*x)*f(exp(lambda*x))-1/4*lambda^2,y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[y'[x]==y[x]^2+Exp[2*\[Lambda]*x]*f[Exp[\[Lambda]*x]]-1/4*\[Lambda]^2,y[x],x,IncludeSingularSolutions -> True]
Not solved