Internal problem ID [11400]
Internal file name [OUTPUT/10383_Wednesday_May_17_2023_08_10_22_PM_96590617/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 1.4.1. Integrating factors. Exercises
page 41
Problem number: 1(c).
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]]
\[ \boxed {x^{\prime }+x^{2}=t} \]
In canonical form the ODE is \begin {align*} x' &= F(t,x)\\ &= -x^{2}+t \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ x' = -x^{2}+t \] With Riccati ODE standard form \[ x' = f_0(t)+ f_1(t)x+f_2(t)x^{2} \] Shows that \(f_0(t)=t\), \(f_1(t)=0\) 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 \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} -u^{\prime \prime }\left (t \right )+t u \left (t \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (t \right ) = c_{1} \operatorname {AiryAi}\left (t \right )+c_{2} \operatorname {AiryBi}\left (t \right ) \] The above shows that \[ u^{\prime }\left (t \right ) = c_{1} \operatorname {AiryAi}\left (1, t\right )+c_{2} \operatorname {AiryBi}\left (1, t\right ) \] Using the above in (1) gives the solution \[ x = \frac {c_{1} \operatorname {AiryAi}\left (1, t\right )+c_{2} \operatorname {AiryBi}\left (1, t\right )}{c_{1} \operatorname {AiryAi}\left (t \right )+c_{2} \operatorname {AiryBi}\left (t \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ x = \frac {c_{3} \operatorname {AiryAi}\left (1, t\right )+\operatorname {AiryBi}\left (1, t\right )}{c_{3} \operatorname {AiryAi}\left (t \right )+\operatorname {AiryBi}\left (t \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} x &= \frac {c_{3} \operatorname {AiryAi}\left (1, t\right )+\operatorname {AiryBi}\left (1, t\right )}{c_{3} \operatorname {AiryAi}\left (t \right )+\operatorname {AiryBi}\left (t \right )} \\ \end{align*}
Verification of solutions
\[ x = \frac {c_{3} \operatorname {AiryAi}\left (1, t\right )+\operatorname {AiryBi}\left (1, t\right )}{c_{3} \operatorname {AiryAi}\left (t \right )+\operatorname {AiryBi}\left (t \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{\prime }+x^{2}=t \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & x^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & x^{\prime }=t -x^{2} \end {array} \]
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: 0.0 (sec). Leaf size: 23
dsolve(diff(x(t),t)=t-x(t)^2,x(t), singsol=all)
\[ x \left (t \right ) = \frac {c_{1} \operatorname {AiryAi}\left (1, t\right )+\operatorname {AiryBi}\left (1, t\right )}{c_{1} \operatorname {AiryAi}\left (t \right )+\operatorname {AiryBi}\left (t \right )} \]
✓ Solution by Mathematica
Time used: 0.221 (sec). Leaf size: 223
DSolve[x'[t]==t-x[t]^2,x[t],t,IncludeSingularSolutions -> True]
\begin{align*} x(t)\to -\frac {-i t^{3/2} \left (2 \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2}{3} i t^{3/2}\right )+c_1 \left (\operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} i t^{3/2}\right )-\operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} i t^{3/2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i t^{3/2}\right )}{2 t \left (\operatorname {BesselJ}\left (\frac {1}{3},\frac {2}{3} i t^{3/2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i t^{3/2}\right )\right )} \\ x(t)\to \frac {i t^{3/2} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} i t^{3/2}\right )-i t^{3/2} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} i t^{3/2}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i t^{3/2}\right )}{2 t \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i t^{3/2}\right )} \\ \end{align*}