1.2 problem 1. direct method

Internal problem ID [6545]
Internal file name [OUTPUT/5793_Sunday_June_05_2022_03_54_47_PM_50830833/index.tex]

Book: A course in Ordinary Differential Equations. by Stephen A. Wirkus, Randall J. Swift. CRC Press NY. 2015. 2nd Edition
Section: Chapter 8. Series Methods. section 8.2. The Power Series Method. Problems Page 603
Problem number: 1. direct method.
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, _special]]

\[ \boxed {y^{\prime }-y^{2}=-x} \] With initial conditions \begin {align*} [y \left (0\right ) = 1] \end {align*}

1.2.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(x,y)\\ &= y^{2}-x \end {align*}

The \(x\) domain of \(f(x,y)\) when \(y=1\) is \[ \{-\infty

The \(y\) domain of \(\frac {\partial f}{\partial y}\) when \(x=0\) is \[ \{-\infty

1.2.2 Solving as riccati ode

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

This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}-x \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=-x\), \(f_1(x)=0\) and \(f_2(x)=1\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{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' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=-x \end {align*}

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

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

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

\[ y = \frac {-c_{3} \operatorname {AiryAi}\left (1, x\right )-\operatorname {AiryBi}\left (1, x\right )}{c_{3} \operatorname {AiryAi}\left (x \right )+\operatorname {AiryBi}\left (x \right )} \] Initial conditions are used to solve for \(c_{3}\). Substituting \(x=0\) and \(y=1\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 1 = \frac {3 \Gamma \left (\frac {2}{3}\right )^{2} c_{3} 3^{\frac {1}{6}}-3 \Gamma \left (\frac {2}{3}\right )^{2} 3^{\frac {2}{3}}}{2 \,3^{\frac {5}{6}} \pi +2 \pi c_{3} 3^{\frac {1}{3}}} \end {align*}

The solutions are \begin {align*} c_{3} = -\frac {2 \,3^{\frac {5}{6}} \pi +3 \Gamma \left (\frac {2}{3}\right )^{2} 3^{\frac {2}{3}}}{-3 \,3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}+2 \pi 3^{\frac {1}{3}}} \end {align*}

Trying the constant \begin {align*} c_{3} = -\frac {2 \,3^{\frac {5}{6}} \pi +3 \Gamma \left (\frac {2}{3}\right )^{2} 3^{\frac {2}{3}}}{-3 \,3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}+2 \pi 3^{\frac {1}{3}}} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {2 \pi \operatorname {AiryAi}\left (1, x\right ) 3^{\frac {5}{6}}+3 \,3^{\frac {2}{3}} \operatorname {AiryAi}\left (1, x\right ) \Gamma \left (\frac {2}{3}\right )^{2}+3 \operatorname {AiryBi}\left (1, x\right ) 3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}-2 \pi 3^{\frac {1}{3}} \operatorname {AiryBi}\left (1, x\right )}{-2 \pi \operatorname {AiryAi}\left (x \right ) 3^{\frac {5}{6}}-3 \,3^{\frac {2}{3}} \operatorname {AiryAi}\left (x \right ) \Gamma \left (\frac {2}{3}\right )^{2}-3 \operatorname {AiryBi}\left (x \right ) 3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}+2 \pi 3^{\frac {1}{3}} \operatorname {AiryBi}\left (x \right )} \end {align*}

The constant \(c_{3} = -\frac {2 \,3^{\frac {5}{6}} \pi +3 \Gamma \left (\frac {2}{3}\right )^{2} 3^{\frac {2}{3}}}{-3 \,3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}+2 \pi 3^{\frac {1}{3}}}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {2 \pi \operatorname {AiryAi}\left (1, x\right ) 3^{\frac {5}{6}}+3 \,3^{\frac {2}{3}} \operatorname {AiryAi}\left (1, x\right ) \Gamma \left (\frac {2}{3}\right )^{2}+3 \operatorname {AiryBi}\left (1, x\right ) 3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}-2 \pi 3^{\frac {1}{3}} \operatorname {AiryBi}\left (1, x\right )}{-2 \pi \operatorname {AiryAi}\left (x \right ) 3^{\frac {5}{6}}-3 \,3^{\frac {2}{3}} \operatorname {AiryAi}\left (x \right ) \Gamma \left (\frac {2}{3}\right )^{2}-3 \operatorname {AiryBi}\left (x \right ) 3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}+2 \pi 3^{\frac {1}{3}} \operatorname {AiryBi}\left (x \right )} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {2 \pi \operatorname {AiryAi}\left (1, x\right ) 3^{\frac {5}{6}}+3 \,3^{\frac {2}{3}} \operatorname {AiryAi}\left (1, x\right ) \Gamma \left (\frac {2}{3}\right )^{2}+3 \operatorname {AiryBi}\left (1, x\right ) 3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}-2 \pi 3^{\frac {1}{3}} \operatorname {AiryBi}\left (1, x\right )}{-2 \pi \operatorname {AiryAi}\left (x \right ) 3^{\frac {5}{6}}-3 \,3^{\frac {2}{3}} \operatorname {AiryAi}\left (x \right ) \Gamma \left (\frac {2}{3}\right )^{2}-3 \operatorname {AiryBi}\left (x \right ) 3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}+2 \pi 3^{\frac {1}{3}} \operatorname {AiryBi}\left (x \right )} \] Verified OK.

1.2.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{2}=-x , y \left (0\right )=1\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=-x +y^{2} \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=1 \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]

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 Special 
<- Riccati Special successful`
 

✓ Solution by Maple

Time used: 0.219 (sec). Leaf size: 90

dsolve([diff(y(x),x)=y(x)^2-x,y(0) = 1],y(x), singsol=all)
 

\[ y \left (x \right ) = \frac {-2 \operatorname {AiryAi}\left (1, x\right ) 3^{\frac {5}{6}} \pi -3 \operatorname {AiryAi}\left (1, x\right ) \Gamma \left (\frac {2}{3}\right )^{2} 3^{\frac {2}{3}}-3 \operatorname {AiryBi}\left (1, x\right ) 3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}+2 \operatorname {AiryBi}\left (1, x\right ) 3^{\frac {1}{3}} \pi }{2 \operatorname {AiryAi}\left (x \right ) 3^{\frac {5}{6}} \pi +3 \operatorname {AiryAi}\left (x \right ) \Gamma \left (\frac {2}{3}\right )^{2} 3^{\frac {2}{3}}+3 \operatorname {AiryBi}\left (x \right ) 3^{\frac {1}{6}} \Gamma \left (\frac {2}{3}\right )^{2}-2 \operatorname {AiryBi}\left (x \right ) 3^{\frac {1}{3}} \pi } \]

✓ Solution by Mathematica

Time used: 7.282 (sec). Leaf size: 164

DSolve[{y'[x]==y[x]^2-x,{y[0]==1}},y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to \frac {\sqrt [3]{-3} \operatorname {Gamma}\left (\frac {2}{3}\right ) \left (i x^{3/2} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} i x^{3/2}\right )-i x^{3/2} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} i x^{3/2}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )\right )-2 i x^{3/2} \operatorname {Gamma}\left (\frac {1}{3}\right ) \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2}{3} i x^{3/2}\right )}{2 x \left (\operatorname {Gamma}\left (\frac {1}{3}\right ) \operatorname {BesselJ}\left (\frac {1}{3},\frac {2}{3} i x^{3/2}\right )-\sqrt [3]{-3} \operatorname {Gamma}\left (\frac {2}{3}\right ) \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )\right )} \]