Internal problem ID [10639]
Internal file name [OUTPUT/9587_Monday_June_06_2022_03_11_03_PM_91047867/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: 5.
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]
\[ \boxed {x^{2} y^{\prime }-y^{2} x^{4}=x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4}} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {4 y^{2} x^{4}+4 x^{2 n} f \left (a \,x^{n}+b \right )-n^{2}+1}{4 x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{2} y^{2}+\frac {x^{2 n} f \left (a \,x^{n}+b \right )}{x^{2}}-\frac {n^{2}}{4 x^{2}}+\frac {1}{4 x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {4 x^{2 n} f \left (a \,x^{n}+b \right )-n^{2}+1}{4 x^{2}}\), \(f_1(x)=0\) and \(f_2(x)=x^{2}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{x^{2} 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' &=2 x\\ f_1 f_2 &=0\\ f_2^2 f_0 &=\frac {x^{2} \left (4 x^{2 n} f \left (a \,x^{n}+b \right )-n^{2}+1\right )}{4} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} x^{2} u^{\prime \prime }\left (x \right )-2 x u^{\prime }\left (x \right )+\frac {x^{2} \left (4 x^{2 n} f \left (a \,x^{n}+b \right )-n^{2}+1\right ) u \left (x \right )}{4} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = \operatorname {DESol}\left (\left \{\left (x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4}\right ) \textit {\_Y} \left (x \right )-\frac {2 \textit {\_Y}^{\prime }\left (x \right )}{x}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\left (x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4}\right ) \textit {\_Y} \left (x \right )-\frac {2 \textit {\_Y}^{\prime }\left (x \right )}{x}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right ) \] Using the above in (1) gives the solution \[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\left (x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4}\right ) \textit {\_Y} \left (x \right )-\frac {2 \textit {\_Y}^{\prime }\left (x \right )}{x}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{x^{2} \operatorname {DESol}\left (\left \{\left (x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4}\right ) \textit {\_Y} \left (x \right )-\frac {2 \textit {\_Y}^{\prime }\left (x \right )}{x}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\left (x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4}\right ) \textit {\_Y} \left (x \right )-\frac {2 \textit {\_Y}^{\prime }\left (x \right )}{x}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{x^{2} \operatorname {DESol}\left (\left \{\frac {4 x^{1+2 n} \textit {\_Y} \left (x \right ) f \left (a \,x^{n}+b \right )+4 \textit {\_Y}^{\prime \prime }\left (x \right ) x -8 \textit {\_Y}^{\prime }\left (x \right )+x \left (-n^{2}+1\right ) \textit {\_Y} \left (x \right )}{4 x}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\left (x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4}\right ) \textit {\_Y} \left (x \right )-\frac {2 \textit {\_Y}^{\prime }\left (x \right )}{x}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{x^{2} \operatorname {DESol}\left (\left \{\frac {4 x^{1+2 n} \textit {\_Y} \left (x \right ) f \left (a \,x^{n}+b \right )+4 \textit {\_Y}^{\prime \prime }\left (x \right ) x -8 \textit {\_Y}^{\prime }\left (x \right )+x \left (-n^{2}+1\right ) \textit {\_Y} \left (x \right )}{4 x}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \\ \end{align*}
Verification of solutions
\[ y = -\frac {\frac {\partial }{\partial x}\operatorname {DESol}\left (\left \{\left (x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4}\right ) \textit {\_Y} \left (x \right )-\frac {2 \textit {\_Y}^{\prime }\left (x \right )}{x}+\textit {\_Y}^{\prime \prime }\left (x \right )\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )}{x^{2} \operatorname {DESol}\left (\left \{\frac {4 x^{1+2 n} \textit {\_Y} \left (x \right ) f \left (a \,x^{n}+b \right )+4 \textit {\_Y}^{\prime \prime }\left (x \right ) x -8 \textit {\_Y}^{\prime }\left (x \right )+x \left (-n^{2}+1\right ) \textit {\_Y} \left (x \right )}{4 x}\right \}, \left \{\textit {\_Y} \left (x \right )\right \}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{2} y^{\prime }-y^{2} x^{4}=x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {y^{2} x^{4}+x^{2 n} f \left (a \,x^{n}+b \right )-\frac {n^{2}}{4}+\frac {1}{4}}{x^{2}} \end {array} \]
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^4*y(x)^2+x^(2*n)*f(a*x^n+b)+1/4*(1-n^2),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^4*y[x]^2+x^(2*n)*f[a*x^n+b]+1/4*(1-n^2),y[x],x,IncludeSingularSolutions -> True]
Not solved