3.11 problem 65

3.11.1 Solving as riccati ode
3.11.2 Maple step by step solution

Internal problem ID [3329]
Internal file name [OUTPUT/2821_Sunday_June_05_2022_08_41_08_AM_69568672/index.tex]

Book: Ordinary differential equations and their solutions. By George Moseley Murphy. 1960
Section: Various 3
Problem number: 65.
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]

\[ \boxed {y^{\prime }-x \left (2+y x^{2}-y^{2}\right )=0} \]

3.11.1 Solving as riccati ode

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

This is a Riccati ODE. Comparing the ODE to solve \[ y' = x^{3} y -x \,y^{2}+2 x \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=2 x\), \(f_1(x)=x^{3}\) and \(f_2(x)=-x\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{-x 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' &=-1\\ f_1 f_2 &=-x^{4}\\ f_2^2 f_0 &=2 x^{3} \end {align*}

Substituting the above terms back in equation (2) gives \begin {align*} -x u^{\prime \prime }\left (x \right )-\left (-x^{4}-1\right ) u^{\prime }\left (x \right )+2 x^{3} u \left (x \right ) &=0 \end {align*}

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

\[ u \left (x \right ) = {\mathrm e}^{\frac {x^{4}}{4}} \left (c_{1} +\operatorname {erf}\left (\frac {x^{2}}{2}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (x^{2} \sqrt {\pi }\, \left (c_{1} +\operatorname {erf}\left (\frac {x^{2}}{2}\right ) c_{2} \right ) {\mathrm e}^{\frac {x^{4}}{4}}+2 c_{2} \right ) x}{\sqrt {\pi }} \] Using the above in (1) gives the solution \[ y = \frac {\left (x^{2} \sqrt {\pi }\, \left (c_{1} +\operatorname {erf}\left (\frac {x^{2}}{2}\right ) c_{2} \right ) {\mathrm e}^{\frac {x^{4}}{4}}+2 c_{2} \right ) {\mathrm e}^{-\frac {x^{4}}{4}}}{\sqrt {\pi }\, \left (c_{1} +\operatorname {erf}\left (\frac {x^{2}}{2}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution

\[ y = \frac {\operatorname {erf}\left (\frac {x^{2}}{2}\right ) \sqrt {\pi }\, x^{2}+c_{3} \sqrt {\pi }\, x^{2}+2 \,{\mathrm e}^{-\frac {x^{4}}{4}}}{\sqrt {\pi }\, \left (c_{3} +\operatorname {erf}\left (\frac {x^{2}}{2}\right )\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\operatorname {erf}\left (\frac {x^{2}}{2}\right ) \sqrt {\pi }\, x^{2}+c_{3} \sqrt {\pi }\, x^{2}+2 \,{\mathrm e}^{-\frac {x^{4}}{4}}}{\sqrt {\pi }\, \left (c_{3} +\operatorname {erf}\left (\frac {x^{2}}{2}\right )\right )} \\ \end{align*}

Figure 122: Slope field plot

Verification of solutions

\[ y = \frac {\operatorname {erf}\left (\frac {x^{2}}{2}\right ) \sqrt {\pi }\, x^{2}+c_{3} \sqrt {\pi }\, x^{2}+2 \,{\mathrm e}^{-\frac {x^{4}}{4}}}{\sqrt {\pi }\, \left (c_{3} +\operatorname {erf}\left (\frac {x^{2}}{2}\right )\right )} \] Verified OK.

3.11.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-x \left (2+y x^{2}-y^{2}\right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=x \left (2+y x^{2}-y^{2}\right ) \end {array} \]

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 to 2nd Order 
   -> Calling odsolve with the ODE`, diff(diff(y(x), x), x) = (x^4+1)*(diff(y(x), x))/x+2*y(x)*x^2, y(x)`      *** Sublevel 2 *** 
      Methods for second order ODEs: 
      --- Trying classification methods --- 
      trying a quadrature 
      checking if the LODE has constant coefficients 
      checking if the LODE is of Euler type 
      trying a symmetry of the form [xi=0, eta=F(x)] 
      checking if the LODE is missing y 
      -> Trying a Liouvillian solution using Kovacics algorithm 
         A Liouvillian solution exists 
         Reducible group (found an exponential solution) 
         Group is reducible, not completely reducible 
      <- Kovacics algorithm successful 
   <- Riccati to 2nd Order successful`
 

Solution by Maple

Time used: 0.0 (sec). Leaf size: 51

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

\[ y \left (x \right ) = \frac {\operatorname {erf}\left (\frac {x^{2}}{2}\right ) \sqrt {\pi }\, c_{1} x^{2}+x^{2} \sqrt {\pi }+2 \,{\mathrm e}^{-\frac {x^{4}}{4}} c_{1}}{\sqrt {\pi }\, \left (\operatorname {erf}\left (\frac {x^{2}}{2}\right ) c_{1} +1\right )} \]

Solution by Mathematica

Time used: 0.298 (sec). Leaf size: 70

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

\begin{align*} y(x)\to \frac {\sqrt {\pi } x^2 \text {erf}\left (\frac {x^2}{2}\right )+2 e^{-\frac {x^4}{4}}+2 c_1 x^2}{\sqrt {\pi } \text {erf}\left (\frac {x^2}{2}\right )+2 c_1} \\ y(x)\to x^2 \\ \end{align*}