Internal problem ID [10645]
Internal file name [OUTPUT/9593_Monday_June_06_2022_03_11_31_PM_11209688/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: 11.
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]
\[ \boxed {x^{2} y^{\prime }-x^{2} y^{2}=f \left (a \ln \left (x \right )+b \right )+\frac {1}{4}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {4 x^{2} y^{2}+4 f \left (a \ln \left (x \right )+b \right )+1}{4 x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}+\frac {f \left (a \ln \left (x \right )+b \right )}{x^{2}}+\frac {1}{4 x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {4 f \left (a \ln \left (x \right )+b \right )+1}{4 x^{2}}\), \(f_1(x)=0\) 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 &=\frac {4 f \left (a \ln \left (x \right )+b \right )+1}{4 x^{2}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} u^{\prime \prime }\left (x \right )+\frac {\left (4 f \left (a \ln \left (x \right )+b \right )+1\right ) u \left (x \right )}{4 x^{2}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{\frac {\left (4 f \left (a \ln \left (x \right )+b \right )+1\right ) \textit {\_Y} \left (x \right )}{4 x^{2}}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {\left (4 f \left (a \ln \left (x \right )+b \right )+1\right ) \textit {\_Y} \left (x \right )}{4 x^{2}}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {\left (4 f \left (a \ln \left (x \right )+b \right )+1\right ) \textit {\_Y} \left (x \right )}{4 x^{2}}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\frac {\left (4 f \left (a \ln \left (x \right )+b \right )+1\right ) \textit {\_Y} \left (x \right )}{4 x^{2}}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {4 \textit {\_Y}^{\prime \prime }\left (x \right ) x^{2}+4 \textit {\_Y} \left (x \right ) f \left (a \ln \left (x \right )+b \right )+\textit {\_Y} \left (x \right )}{4 x^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\frac {4 \textit {\_Y}^{\prime \prime }\left (x \right ) x^{2}+4 \textit {\_Y} \left (x \right ) f \left (a \ln \left (x \right )+b \right )+\textit {\_Y} \left (x \right )}{4 x^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {4 \textit {\_Y}^{\prime \prime }\left (x \right ) x^{2}+4 \textit {\_Y} \left (x \right ) f \left (a \ln \left (x \right )+b \right )+\textit {\_Y} \left (x \right )}{4 x^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\frac {4 \textit {\_Y}^{\prime \prime }\left (x \right ) x^{2}+4 \textit {\_Y} \left (x \right ) f \left (a \ln \left (x \right )+b \right )+\textit {\_Y} \left (x \right )}{4 x^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\frac {4 \textit {\_Y}^{\prime \prime }\left (x \right ) x^{2}+4 \textit {\_Y} \left (x \right ) f \left (a \ln \left (x \right )+b \right )+\textit {\_Y} \left (x \right )}{4 x^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{\operatorname {DESol}\left (\left \{\frac {4 \textit {\_Y}^{\prime \prime }\left (x \right ) x^{2}+4 \textit {\_Y} \left (x \right ) f \left (a \ln \left (x \right )+b \right )+\textit {\_Y} \left (x \right )}{4 x^{2}}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-x^{2} y^{2}=f \left (a \ln \left (x \right )+b \right )+\frac {1}{4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {x^{2} y^{2}+f \left (a \ln \left (x \right )+b \right )+\frac {1}{4}}{x^{2}} \end {array} \]
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 -> 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(x^2*diff(y(x),x)=x^2*y(x)^2+f(a*ln(x)+b)+1/4,y(x), singsol=all)
\[ \text {No solution found} \]
✗ Solution by Mathematica
Time used: 0.0 (sec). Leaf size: 0
DSolve[x^2*y'[x]==x^2*y[x]^2+f[a*Log[x]+b]+1/4,y[x],x,IncludeSingularSolutions -> True]
Not solved