5.1 problem 1

5.1.1 Solving as riccati ode
5.1.2 Maple step by step solution

Internal problem ID [4400]
Internal file name [OUTPUT/3893_Sunday_June_05_2022_11_36_39_AM_95241541/index.tex]

Book: Differential Equations, By George Boole F.R.S. 1865
Section: Chapter 6
Problem number: 1.
ODE order: 1.
ODE degree: 1.

The type(s) of ODE detected by this program :

Maple gives the following as the ode type

[_rational, _Riccati]

\[ \boxed {y^{\prime } x -a y+y^{2}=x^{-2 a}} \]

5.1.1 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= -\frac {-a y +y^{2}-x^{-2 a}}{x} \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = \frac {a y}{x}-\frac {y^{2}}{x}+\frac {x^{-2 a}}{x} \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=\frac {x^{-2 a}}{x}\), \(f_1(x)=\frac {a}{x}\) and \(f_2(x)=-\frac {1}{x}\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-\frac {u}{x}} \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 {1}{x^{2}}\\ f_1 f_2 &=-\frac {a}{x^{2}}\\ f_2^2 f_0 &=\frac {x^{-2 a}}{x^{3}} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} -\frac {u^{\prime \prime }\left (x \right )}{x}-\left (\frac {1}{x^{2}}-\frac {a}{x^{2}}\right ) u^{\prime }\left (x \right )+\frac {x^{-2 a} u \left (x \right )}{x^{3}} &=0 \end {align*}

Solving the above ODE (this ode solved using Maple, not this program), gives

\[ u \left (x \right ) = x^{a} \left (c_{1} \sinh \left (\frac {x^{-a}}{a}\right )+c_{2} \cosh \left (\frac {x^{-a}}{a}\right )\right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (a c_{2} x^{a}-c_{1} \right ) \cosh \left (\frac {x^{-a}}{a}\right )+\sinh \left (\frac {x^{-a}}{a}\right ) \left (a c_{1} x^{a}-c_{2} \right )}{x} \] Using the above in (1) gives the solution \[ y = \frac {\left (\left (a c_{2} x^{a}-c_{1} \right ) \cosh \left (\frac {x^{-a}}{a}\right )+\sinh \left (\frac {x^{-a}}{a}\right ) \left (a c_{1} x^{a}-c_{2} \right )\right ) x^{-a}}{c_{1} \sinh \left (\frac {x^{-a}}{a}\right )+c_{2} \cosh \left (\frac {x^{-a}}{a}\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = \frac {\left (\left (a \,x^{a}-c_{3} \right ) \cosh \left (\frac {x^{-a}}{a}\right )+\sinh \left (\frac {x^{-a}}{a}\right ) \left (a c_{3} x^{a}-1\right )\right ) x^{-a}}{c_{3} \sinh \left (\frac {x^{-a}}{a}\right )+\cosh \left (\frac {x^{-a}}{a}\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\left (a \,x^{a}-c_{3} \right ) \cosh \left (\frac {x^{-a}}{a}\right )+\sinh \left (\frac {x^{-a}}{a}\right ) \left (a c_{3} x^{a}-1\right )\right ) x^{-a}}{c_{3} \sinh \left (\frac {x^{-a}}{a}\right )+\cosh \left (\frac {x^{-a}}{a}\right )} \\ \end{align*}

Verification of solutions

\[ y = \frac {\left (\left (a \,x^{a}-c_{3} \right ) \cosh \left (\frac {x^{-a}}{a}\right )+\sinh \left (\frac {x^{-a}}{a}\right ) \left (a c_{3} x^{a}-1\right )\right ) x^{-a}}{c_{3} \sinh \left (\frac {x^{-a}}{a}\right )+\cosh \left (\frac {x^{-a}}{a}\right )} \] Verified OK.

5.1.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime } x -a y+y^{2}=x^{-2 a} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a y-y^{2}+x^{-2 a}}{x} \end {array} \]

Maple trace Kovacic algorithm successful

`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) = (a-1)*(diff(y(x), x))/x+x^(-2*a-1)*y(x)/x, y(x)`      *** Sublevel 2 * 
      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 
      -> Trying an equivalence, under non-integer power transformations, 
         to LODEs admitting Liouvillian solutions. 
         -> Trying a Liouvillian solution using Kovacics algorithm 
            A Liouvillian solution exists 
            Group is reducible or imprimitive 
         <- Kovacics algorithm successful 
      <- Equivalence, under non-integer power transformations successful 
   <- Riccati to 2nd Order successful`
 

Solution by Maple

Time used: 0.016 (sec). Leaf size: 74

dsolve(x*diff(y(x),x)-a*y(x)+y(x)^2=x^(-2*a),y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {\left (-x^{-a} c_{1} +a \right ) \sinh \left (\frac {x^{-a}}{a}\right )+\left (c_{1} a -x^{-a}\right ) \cosh \left (\frac {x^{-a}}{a}\right )}{\cosh \left (\frac {x^{-a}}{a}\right ) c_{1} +\sinh \left (\frac {x^{-a}}{a}\right )} \]

Solution by Mathematica

Time used: 0.393 (sec). Leaf size: 112

DSolve[x*y'[x]-a*y[x]+y[x]^2==x^(-2*a),y[x],x,IncludeSingularSolutions -> True]
 

\begin{align*} y(x)\to \frac {x^{-a} \left (\left (a x^a+i c_1\right ) \cosh \left (\frac {x^{-a}}{a}\right )-i \left (a c_1 x^a-i\right ) \sinh \left (\frac {x^{-a}}{a}\right )\right )}{\cosh \left (\frac {x^{-a}}{a}\right )-i c_1 \sinh \left (\frac {x^{-a}}{a}\right )} \\ y(x)\to a-x^{-a} \coth \left (\frac {x^{-a}}{a}\right ) \\ \end{align*}