2.1.37 problem 38

Internal problem ID [8175]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 38
Date solved : Tuesday, November 12, 2024 at 11:07:22 PM
CAS classification : [_rational, _Riccati]

Solve

\begin{align*} x y^{\prime }-y+y^{2}&=x^{{2}/{3}} \end{align*}

Solved as first order ode of type Riccati

Time used: 0.335 (sec)

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= -\frac {y^{2}-x^{{2}/{3}}-y}{x} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = -\frac {y^{2}}{x}+\frac {1}{x^{{1}/{3}}}+\frac {y}{x} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=\frac {1}{x^{{1}/{3}}}\), \(f_1(x)=\frac {1}{x}\) 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 {1}{x^{2}}\\ f_2^2 f_0 &=\frac {1}{x^{{7}/{3}}} \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} -\frac {u^{\prime \prime }\left (x \right )}{x}+\frac {u \left (x \right )}{x^{{7}/{3}}} = 0 \end{align*}

Writing the ode as

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )-x^{{2}/{3}} u = 0\tag {1} \end{align*}

Bessel ode has the form

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )+\left (\frac {d u}{d x}\right ) x +\left (-n^{2}+x^{2}\right ) u = 0\tag {2} \end{align*}

The generalized form of Bessel ode is given by Bowman (1958) as the following

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )+\left (1-2 \alpha \right ) x \left (\frac {d u}{d x}\right )+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) u = 0\tag {3} \end{align*}

With the standard solution

\begin{align*} u&=x^{\alpha } \left (c_1 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_2 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}

Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives

\begin{align*} \alpha &= {\frac {1}{2}}\\ \beta &= 3 i\\ n &= {\frac {3}{2}}\\ \gamma &= {\frac {1}{3}} \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} u = -\frac {c_1 \,x^{{1}/{6}} \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{9 \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {i c_2 \,x^{{1}/{6}} \sqrt {2}\, \sqrt {3}\, \left (3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\cosh \left (3 x^{{1}/{3}}\right )\right )}{9 \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}} \end{align*}

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = -\frac {c_1 \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{54 x^{{5}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {c_1 \sqrt {2}\, \sqrt {3}\, \sinh \left (3 x^{{1}/{3}}\right )}{3 x^{{1}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}+\frac {i c_1 \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{54 \sqrt {x}\, \sqrt {\pi }\, \left (i x^{{1}/{3}}\right )^{{3}/{2}}}-\frac {i c_2 \sqrt {2}\, \sqrt {3}\, \left (3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\cosh \left (3 x^{{1}/{3}}\right )\right )}{54 x^{{5}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {i c_2 \sqrt {2}\, \sqrt {3}\, \cosh \left (3 x^{{1}/{3}}\right )}{3 x^{{1}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {c_2 \sqrt {2}\, \sqrt {3}\, \left (3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\cosh \left (3 x^{{1}/{3}}\right )\right )}{54 \sqrt {x}\, \sqrt {\pi }\, \left (i x^{{1}/{3}}\right )^{{3}/{2}}} \]

Doing change of constants, the solution becomes

\[ y = \frac {\left (-\frac {c_3 \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{54 x^{{5}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {c_3 \sqrt {2}\, \sqrt {3}\, \sinh \left (3 x^{{1}/{3}}\right )}{3 x^{{1}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}+\frac {i c_3 \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{54 \sqrt {x}\, \sqrt {\pi }\, \left (i x^{{1}/{3}}\right )^{{3}/{2}}}-\frac {i \sqrt {2}\, \sqrt {3}\, \left (3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\cosh \left (3 x^{{1}/{3}}\right )\right )}{54 x^{{5}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {i \sqrt {2}\, \sqrt {3}\, \cosh \left (3 x^{{1}/{3}}\right )}{3 x^{{1}/{6}} \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {\sqrt {2}\, \sqrt {3}\, \left (3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\cosh \left (3 x^{{1}/{3}}\right )\right )}{54 \sqrt {x}\, \sqrt {\pi }\, \left (i x^{{1}/{3}}\right )^{{3}/{2}}}\right ) x}{-\frac {c_3 \,x^{{1}/{6}} \sqrt {2}\, \sqrt {3}\, \left (3 \cosh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\sinh \left (3 x^{{1}/{3}}\right )\right )}{9 \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}-\frac {i x^{{1}/{6}} \sqrt {2}\, \sqrt {3}\, \left (3 \sinh \left (3 x^{{1}/{3}}\right ) x^{{1}/{3}}-\cosh \left (3 x^{{1}/{3}}\right )\right )}{9 \sqrt {\pi }\, \sqrt {i x^{{1}/{3}}}}} \]

Figure 2.85: Slope field plot
\(x y^{\prime }-y+y^{2} = x^{{2}/{3}}\)

Summary of solutions found

\begin{align*} y &= \frac {3 x^{{2}/{3}} \left (i \sinh \left (3 x^{{1}/{3}}\right ) c_3 -\cosh \left (3 x^{{1}/{3}}\right )\right )}{\left (-i c_3 -3 x^{{1}/{3}}\right ) \sinh \left (3 x^{{1}/{3}}\right )+\left (3 i x^{{1}/{3}} c_3 +1\right ) \cosh \left (3 x^{{1}/{3}}\right )} \\ \end{align*}

Maple step by step solution

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

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 Riccati sub-methods: 
   trying Riccati_symmetries 
   trying Riccati to 2nd Order 
   -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = y(x)/x^(4/3), 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`
 

Maple dsolve solution

Solving time : 0.006 (sec)
Leaf size : 72

dsolve(diff(y(x),x)*x-y(x)+y(x)^2 = x^(2/3), 
       y(x),singsol=all)
 

\[ y = \frac {x^{{1}/{3}} \left (c_{1} {\mathrm e}^{6 x^{{1}/{3}}} \operatorname {abs}\left (1, 3 x^{{1}/{3}}-1\right )+c_{1} {\mathrm e}^{6 x^{{1}/{3}}} {| 3 x^{{1}/{3}}-1|}-3 x^{{1}/{3}}\right )}{c_{1} {\mathrm e}^{6 x^{{1}/{3}}} {| 3 x^{{1}/{3}}-1|}+3 x^{{1}/{3}}+1} \]

Mathematica DSolve solution

Solving time : 0.193 (sec)
Leaf size : 131

DSolve[{x*D[y[x],x]-y[x]+y[x]^2==x^(2/3),{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to \frac {3 x^{2/3} \left (c_1 \cosh \left (3 \sqrt [3]{x}\right )-i \sinh \left (3 \sqrt [3]{x}\right )\right )}{\left (-3 i \sqrt [3]{x}-c_1\right ) \cosh \left (3 \sqrt [3]{x}\right )+\left (3 c_1 \sqrt [3]{x}+i\right ) \sinh \left (3 \sqrt [3]{x}\right )} \\ y(x)\to \frac {3 x^{2/3} \cosh \left (3 \sqrt [3]{x}\right )}{3 \sqrt [3]{x} \sinh \left (3 \sqrt [3]{x}\right )-\cosh \left (3 \sqrt [3]{x}\right )} \\ \end{align*}