2.2 problem 2

Internal problem ID [11358]
Internal file name [OUTPUT/10341_Wednesday_May_17_2023_07_49_28_PM_50385180/index.tex]

Book: A First Course in Differential Equations by J. David Logan. Third Edition. Springer-Verlag, NY. 2015.
Section: Chapter 1, First order differential equations. Section 1.1.3 Geometric. Exercises page 15
Problem number: 2.
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]]

\[ \boxed {x^{\prime }-x^{2}=t^{2}} \]

2.2.1 Solving as riccati ode

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

This is a Riccati ODE. Comparing the ODE to solve \[ x' = t^{2}+x^{2} \] With Riccati ODE standard form \[ x' = f_0(t)+ f_1(t)x+f_2(t)x^{2} \] Shows that \(f_0(t)=t^{2}\), \(f_1(t)=0\) and \(f_2(t)=1\). Let \begin {align*} x &= \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 &=t^{2} \end {align*}

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

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

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

\[ x = -\frac {t \left (\operatorname {BesselY}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right )+\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right ) c_{3} \right )}{\operatorname {BesselJ}\left (\frac {1}{4}, \frac {t^{2}}{2}\right ) c_{3} +\operatorname {BesselY}\left (\frac {1}{4}, \frac {t^{2}}{2}\right )} \]

Summary

The solution(s) found are the following \begin{align*} \tag{1} x &= -\frac {t \left (\operatorname {BesselY}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right )+\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right ) c_{3} \right )}{\operatorname {BesselJ}\left (\frac {1}{4}, \frac {t^{2}}{2}\right ) c_{3} +\operatorname {BesselY}\left (\frac {1}{4}, \frac {t^{2}}{2}\right )} \\ \end{align*}

Figure 26: Slope field plot

Verification of solutions

\[ x = -\frac {t \left (\operatorname {BesselY}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right )+\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right ) c_{3} \right )}{\operatorname {BesselJ}\left (\frac {1}{4}, \frac {t^{2}}{2}\right ) c_{3} +\operatorname {BesselY}\left (\frac {1}{4}, \frac {t^{2}}{2}\right )} \] Verified OK.

2.2.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime }-x^{2}=t^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=x^{2}+t^{2} \end {array} \]

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: 0.015 (sec). Leaf size: 43

dsolve(diff(x(t),t)=x(t)^2+t^2,x(t), singsol=all)
 

\[ x \left (t \right ) = -\frac {t \left (\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {3}{4}, \frac {t^{2}}{2}\right )\right )}{c_{1} \operatorname {BesselJ}\left (\frac {1}{4}, \frac {t^{2}}{2}\right )+\operatorname {BesselY}\left (\frac {1}{4}, \frac {t^{2}}{2}\right )} \]

✓ Solution by Mathematica

Time used: 0.203 (sec). Leaf size: 169

DSolve[x'[t]==x[t]^2+t^2,x[t],t,IncludeSingularSolutions -> True]
 

\begin{align*} x(t)\to \frac {t^2 \left (-2 \operatorname {BesselJ}\left (-\frac {3}{4},\frac {t^2}{2}\right )+c_1 \left (\operatorname {BesselJ}\left (\frac {3}{4},\frac {t^2}{2}\right )-\operatorname {BesselJ}\left (-\frac {5}{4},\frac {t^2}{2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{4},\frac {t^2}{2}\right )}{2 t \left (\operatorname {BesselJ}\left (\frac {1}{4},\frac {t^2}{2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{4},\frac {t^2}{2}\right )\right )} \\ x(t)\to -\frac {t^2 \operatorname {BesselJ}\left (-\frac {5}{4},\frac {t^2}{2}\right )-t^2 \operatorname {BesselJ}\left (\frac {3}{4},\frac {t^2}{2}\right )+\operatorname {BesselJ}\left (-\frac {1}{4},\frac {t^2}{2}\right )}{2 t \operatorname {BesselJ}\left (-\frac {1}{4},\frac {t^2}{2}\right )} \\ \end{align*}