Internal problem ID [7482]
Internal file name [OUTPUT/6449_Sunday_June_19_2022_05_02_59_PM_96072028/index.tex
]
Book: Second order enumerated odes
Section: section 2
Problem number: 41.
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.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }+y^{2}=x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=x -y^{2} \end {array} \]
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(y(x),x)=x-y(x)^2,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: 0.118 (sec). Leaf size: 223
DSolve[y'[x]==x-y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {-i x^{3/2} \left (2 \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2}{3} i x^{3/2}\right )+c_1 \left (\operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} i x^{3/2}\right )-\operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} i x^{3/2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )}{2 x \left (\operatorname {BesselJ}\left (\frac {1}{3},\frac {2}{3} i x^{3/2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )\right )} \\ y(x)\to \frac {i x^{3/2} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} i x^{3/2}\right )-i x^{3/2} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} i x^{3/2}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )}{2 x \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} i x^{3/2}\right )} \\ \end{align*}