Internal problem ID [13291]
Internal file name [OUTPUT/12463_Wednesday_February_14_2024_02_06_26_AM_76958906/index.tex]
Book: Ordinary Differential Equations. An introduction to the fundamentals. Kenneth B. Howell.
second edition. CRC Press. FL, USA. 2020
Section: Chapter 3. Some basics about First order equations. Additional exercises. page
63
Problem number: 3.4 e.
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]]
\[ \boxed {y^{\prime }-y^{2}=x} \]
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
\[ y = \frac {c_{3} \operatorname {AiryAi}\left (1, -x \right )+\operatorname {AiryBi}\left (1, -x \right )}{c_{3} \operatorname {AiryAi}\left (-x \right )+\operatorname {AiryBi}\left (-x \right )} \]
The solution(s) found are the following \begin{align*} \tag{1} 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 )} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{3} \operatorname {AiryAi}\left (1, -x \right )+\operatorname {AiryBi}\left (1, -x \right )}{c_{3} \operatorname {AiryAi}\left (-x \right )+\operatorname {AiryBi}\left (-x \right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-y^{2}=x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=y^{2}+x \end {array} \]
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: 31
dsolve(diff(y(x),x)-y(x)^2=x,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{1} \operatorname {AiryAi}\left (1, -x \right )+\operatorname {AiryBi}\left (1, -x \right )}{c_{1} \operatorname {AiryAi}\left (-x \right )+\operatorname {AiryBi}\left (-x \right )} \]
✓ Solution by Mathematica
Time used: 0.114 (sec). Leaf size: 195
DSolve[y'[x]-y[x]^2==x,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {x^{3/2} \left (-2 \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2 x^{3/2}}{3}\right )+c_1 \left (\operatorname {BesselJ}\left (\frac {2}{3},\frac {2 x^{3/2}}{3}\right )-\operatorname {BesselJ}\left (-\frac {4}{3},\frac {2 x^{3/2}}{3}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2 x^{3/2}}{3}\right )}{2 x \left (\operatorname {BesselJ}\left (\frac {1}{3},\frac {2 x^{3/2}}{3}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2 x^{3/2}}{3}\right )\right )} \\ y(x)\to -\frac {x^{3/2} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2 x^{3/2}}{3}\right )-x^{3/2} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2 x^{3/2}}{3}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2 x^{3/2}}{3}\right )}{2 x \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2 x^{3/2}}{3}\right )} \\ \end{align*}