1.28 problem Problem 40

Internal problem ID [12138]
Internal file name [OUTPUT/10791_Tuesday_September_12_2023_08_53_19_AM_27305349/index.tex]

Book: Differential equations and the calculus of variations by L. ElSGOLTS. MIR PUBLISHERS, MOSCOW, Third printing 1977.
Section: Chapter 1, First-Order Differential Equations. Problems page 88
Problem number: Problem 40.
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 (1\right ) = 0] \end {align*}

1.28.1 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 \(y=0\) is \[ \{-\infty

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

1.28.2 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 \(x=1\) and \(y=0\) in the above solution gives an equation to solve for the constant of integration. \begin {align*} 0 = \frac {c_{3} \operatorname {AiryAi}\left (1, 1\right )+\operatorname {AiryBi}\left (1, 1\right )}{c_{3} \operatorname {AiryAi}\left (1\right )+\operatorname {AiryBi}\left (1\right )} \end {align*}

The solutions are \begin {align*} c_{3} = -\frac {\operatorname {AiryBi}\left (1, 1\right )}{\operatorname {AiryAi}\left (1, 1\right )} \end {align*}

Trying the constant \begin {align*} c_{3} = -\frac {\operatorname {AiryBi}\left (1, 1\right )}{\operatorname {AiryAi}\left (1, 1\right )} \end {align*}

Substituting this in the general solution gives \begin {align*} y&=\frac {\operatorname {AiryAi}\left (1, x\right ) \operatorname {AiryBi}\left (1, 1\right )-\operatorname {AiryBi}\left (1, x\right ) \operatorname {AiryAi}\left (1, 1\right )}{\operatorname {AiryAi}\left (x \right ) \operatorname {AiryBi}\left (1, 1\right )-\operatorname {AiryBi}\left (x \right ) \operatorname {AiryAi}\left (1, 1\right )} \end {align*}

The constant \(c_{3} = -\frac {\operatorname {AiryBi}\left (1, 1\right )}{\operatorname {AiryAi}\left (1, 1\right )}\) gives valid solution.

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\operatorname {AiryAi}\left (1, x\right ) \operatorname {AiryBi}\left (1, 1\right )-\operatorname {AiryBi}\left (1, x\right ) \operatorname {AiryAi}\left (1, 1\right )}{\operatorname {AiryAi}\left (x \right ) \operatorname {AiryBi}\left (1, 1\right )-\operatorname {AiryBi}\left (x \right ) \operatorname {AiryAi}\left (1, 1\right )} \\ \end{align*}

(a) Solution plot

(b) Slope field plot

Verification of solutions

\[ y = \frac {\operatorname {AiryAi}\left (1, x\right ) \operatorname {AiryBi}\left (1, 1\right )-\operatorname {AiryBi}\left (1, x\right ) \operatorname {AiryAi}\left (1, 1\right )}{\operatorname {AiryAi}\left (x \right ) \operatorname {AiryBi}\left (1, 1\right )-\operatorname {AiryBi}\left (x \right ) \operatorname {AiryAi}\left (1, 1\right )} \] Verified OK.

1.28.3 Maple step by step solution

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

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: 0.125 (sec). Leaf size: 37

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

\[ y \left (x \right ) = \frac {\operatorname {AiryBi}\left (1, 1\right ) \operatorname {AiryAi}\left (1, x\right )-\operatorname {AiryBi}\left (1, x\right ) \operatorname {AiryAi}\left (1, 1\right )}{\operatorname {AiryBi}\left (1, 1\right ) \operatorname {AiryAi}\left (x \right )-\operatorname {AiryBi}\left (x \right ) \operatorname {AiryAi}\left (1, 1\right )} \]

✓ Solution by Mathematica

Time used: 0.206 (sec). Leaf size: 229

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

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