Internal problem ID [10376]
Internal file name [OUTPUT/9324_Monday_June_06_2022_01_51_35_PM_72746468/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. 1.2.2. Equations Containing Power
Functions
Problem number: 47.
ODE order: 1.
ODE degree: 1.
The type(s) of ODE detected by this program : "riccati"
Maple gives the following as the ode type
[_rational, _Riccati]
\[ \boxed {2 x^{2} y^{\prime }-2 y^{2}-x y=-2 a^{2} x} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= \frac {-2 a^{2} x +y x +2 y^{2}}{2 x^{2}} \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = -\frac {a^{2}}{x}+\frac {y}{2 x}+\frac {y^{2}}{x^{2}} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-\frac {a^{2}}{x}\), \(f_1(x)=\frac {1}{2 x}\) and \(f_2(x)=\frac {1}{x^{2}}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u}{x^{2}}} \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 {2}{x^{3}}\\ f_1 f_2 &=\frac {1}{2 x^{3}}\\ f_2^2 f_0 &=-\frac {a^{2}}{x^{5}} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} \frac {u^{\prime \prime }\left (x \right )}{x^{2}}+\frac {3 u^{\prime }\left (x \right )}{2 x^{3}}-\frac {a^{2} u \left (x \right )}{x^{5}} &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{2} \cosh \left (\frac {2 a}{\sqrt {x}}\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {a \left (-c_{1} \cosh \left (\frac {2 a}{\sqrt {x}}\right )-c_{2} \sinh \left (\frac {2 a}{\sqrt {x}}\right )\right )}{x^{\frac {3}{2}}} \] Using the above in (1) gives the solution \[ y = -\frac {a \sqrt {x}\, \left (-c_{1} \cosh \left (\frac {2 a}{\sqrt {x}}\right )-c_{2} \sinh \left (\frac {2 a}{\sqrt {x}}\right )\right )}{c_{1} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+c_{2} \cosh \left (\frac {2 a}{\sqrt {x}}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (c_{3} \cosh \left (\frac {2 a}{\sqrt {x}}\right )+\sinh \left (\frac {2 a}{\sqrt {x}}\right )\right ) a \sqrt {x}}{c_{3} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+\cosh \left (\frac {2 a}{\sqrt {x}}\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (c_{3} \cosh \left (\frac {2 a}{\sqrt {x}}\right )+\sinh \left (\frac {2 a}{\sqrt {x}}\right )\right ) a \sqrt {x}}{c_{3} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+\cosh \left (\frac {2 a}{\sqrt {x}}\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (c_{3} \cosh \left (\frac {2 a}{\sqrt {x}}\right )+\sinh \left (\frac {2 a}{\sqrt {x}}\right )\right ) a \sqrt {x}}{c_{3} \sinh \left (\frac {2 a}{\sqrt {x}}\right )+\cosh \left (\frac {2 a}{\sqrt {x}}\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & 2 x^{2} y^{\prime }-2 y^{2}-x y=-2 a^{2} x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {2 y^{2}+x y-2 a^{2} x}{2 x^{2}} \end {array} \]
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 <- Chini successful`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 25
dsolve(2*x^2*diff(y(x),x)=2*y(x)^2+x*y(x)-2*a^2*x,y(x), singsol=all)
\[ y \left (x \right ) = \tanh \left (\frac {i c_{1} \sqrt {x}+2 a}{\sqrt {x}}\right ) \sqrt {x}\, a \]
✓ Solution by Mathematica
Time used: 0.619 (sec). Leaf size: 43
DSolve[2*x^2*y'[x]==2*y[x]^2+x*y[x]-2*a^2*x,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to -\sqrt {-a^2} \sqrt {x} \tan \left (\frac {2 \sqrt {-a^2}}{\sqrt {x}}-c_1\right ) \]