3.3 problem 3

Internal problem ID [12930]
Internal file name [OUTPUT/11583_Tuesday_November_07_2023_11_27_21_PM_53982265/index.tex]

Book: DIFFERENTIAL EQUATIONS by Paul Blanchard, Robert L. Devaney, Glen R. Hall. 4th edition. Brooks/Cole. Boston, USA. 2012
Section: Chapter 1. First-Order Differential Equations. Exercises section 1.4 page 61
Problem number: 3.
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]]

\[ \boxed {y^{\prime }-y^{2}=-4 t} \] With initial conditions \begin {align*} \left [y \left (0\right ) = {\frac {1}{2}}\right ] \end {align*}

3.3.1 Existence and uniqueness analysis

This is non linear first order ODE. In canonical form it is written as \begin {align*} y^{\prime } &= f(t,y)\\ &= y^{2}-4 t \end {align*}

The \(t\) domain of \(f(t,y)\) when \(y={\frac {1}{2}}\) is \[ \{-\infty

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

3.3.2 Solving as riccati ode

In canonical form the ODE is \begin {align*} y' &= F(t,y)\\ &= y^{2}-4 t \end {align*}

This is a Riccati ODE. Comparing the ODE to solve \[ y' = y^{2}-4 t \] With Riccati ODE standard form \[ y' = f_0(t)+ f_1(t)y+f_2(t)y^{2} \] Shows that \(f_0(t)=-4 t\), \(f_1(t)=0\) and \(f_2(t)=1\). 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''(t) -\left ( f_2' + f_1 f_2 \right ) u'(t) + f_2^2 f_0 u(t) &= 0 \tag {2} \end {align*}

But \begin {align*} f_2' &=0\\ f_1 f_2 &=0\\ f_2^2 f_0 &=-4 t \end {align*}

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

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

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

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

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

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

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

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

Summary

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

(a) Solution plot

(b) Slope field plot

Verification of solutions

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

3.3.3 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [y^{\prime }-y^{2}=-4 t , y \left (0\right )=\frac {1}{2}\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}-4 t \\ \bullet & {} & \textrm {Use initial condition}\hspace {3pt} y \left (0\right )=\frac {1}{2} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solve for}\hspace {3pt} 0 \\ {} & {} & 0=0 \\ \bullet & {} & \textrm {Substitute}\hspace {3pt} 0=0\hspace {3pt}\textrm {into general solution and simplify}\hspace {3pt} \\ {} & {} & 0 \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & 0 \end {array} \]

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: 0.109 (sec). Leaf size: 115

dsolve([diff(y(t),t)=y(t)^2-4*t,y(0) = 1/2],y(t), singsol=all)
 

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

✓ Solution by Mathematica

Time used: 10.151 (sec). Leaf size: 193

DSolve[{y'[t]==y[t]^2-4*t,{y[0]==1/2}},y[t],t,IncludeSingularSolutions -> True]
 

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