Internal problem ID [10353]
Internal file name [OUTPUT/9301_Monday_June_06_2022_01_49_47_PM_70081015/index.tex]
Book: Handbook of exact solutions for ordinary differential equations. By Polyanin and Zaitsev.
Second edition
Section: Chapter 1, section 1.2. Riccati Equation. 1.2.2. Equations Containing Power
Functions
Problem number: 24.
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]
\[ \boxed {-a y^{2}+y^{\prime }-y b=c x +k} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= a \,y^{2}+y b +c x +k \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = a \,y^{2}+y b +c x +k \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=c x +k\), \(f_1(x)=b\) and \(f_2(x)=a\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{a 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 &=a b\\ f_2^2 f_0 &=a^{2} \left (c x +k \right ) \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} a u^{\prime \prime }\left (x \right )-a b u^{\prime }\left (x \right )+a^{2} \left (c x +k \right ) u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = {\mathrm e}^{\frac {b x}{2}} \left (\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{1} +\operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{2} \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \frac {\left (-2 \left (c a \right )^{\frac {1}{3}} \operatorname {AiryAi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{1} -2 \left (c a \right )^{\frac {1}{3}} \operatorname {AiryBi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{2} +b \left (\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{1} +\operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{2} \right )\right ) {\mathrm e}^{\frac {b x}{2}}}{2} \] Using the above in (1) gives the solution \[ y = -\frac {-2 \left (c a \right )^{\frac {1}{3}} \operatorname {AiryAi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{1} -2 \left (c a \right )^{\frac {1}{3}} \operatorname {AiryBi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{2} +b \left (\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{1} +\operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{2} \right )}{2 a \left (\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{1} +\operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{2} \right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {-\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{3} b -b \operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right )+2 \left (c a \right )^{\frac {1}{3}} \left (\operatorname {AiryAi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{3} +\operatorname {AiryBi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right )\right )}{2 a \left (\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{3} +\operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {-\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{3} b -b \operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right )+2 \left (c a \right )^{\frac {1}{3}} \left (\operatorname {AiryAi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{3} +\operatorname {AiryBi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right )\right )}{2 a \left (\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{3} +\operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {-\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{3} b -b \operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right )+2 \left (c a \right )^{\frac {1}{3}} \left (\operatorname {AiryAi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{3} +\operatorname {AiryBi}\left (1, -\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right )\right )}{2 a \left (\operatorname {AiryAi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right ) c_{3} +\operatorname {AiryBi}\left (-\frac {\left (\left (c x +k \right ) a -\frac {b^{2}}{4}\right ) \left (c a \right )^{\frac {1}{3}}}{a c}\right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & -a y^{2}+y^{\prime }-y b =c x +k \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a y^{2}+y b +c x +k \end {array} \]
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 sub-methods: <- Abel AIR successful: ODE belongs to the 0F1 0-parameter (Airy type) class`
✓ Solution by Maple
Time used: 0.016 (sec). Leaf size: 194
dsolve(diff(y(x),x)=a*y(x)^2+b*y(x)+c*x+k,y(x), singsol=all)
\[ y \left (x \right ) = \frac {2 \sqrt {a}\, \left (\frac {c}{\sqrt {a}}\right )^{\frac {1}{3}} \left (\operatorname {AiryAi}\left (1, -\frac {a \left (c x +k \right )-\frac {b^{2}}{4}}{\left (\frac {c}{\sqrt {a}}\right )^{\frac {2}{3}} a}\right ) c_{1} +\operatorname {AiryBi}\left (1, -\frac {a \left (c x +k \right )-\frac {b^{2}}{4}}{\left (\frac {c}{\sqrt {a}}\right )^{\frac {2}{3}} a}\right )\right )-b \left (c_{1} \operatorname {AiryAi}\left (-\frac {a \left (c x +k \right )-\frac {b^{2}}{4}}{\left (\frac {c}{\sqrt {a}}\right )^{\frac {2}{3}} a}\right )+\operatorname {AiryBi}\left (-\frac {a \left (c x +k \right )-\frac {b^{2}}{4}}{\left (\frac {c}{\sqrt {a}}\right )^{\frac {2}{3}} a}\right )\right )}{2 a \left (c_{1} \operatorname {AiryAi}\left (-\frac {a \left (c x +k \right )-\frac {b^{2}}{4}}{\left (\frac {c}{\sqrt {a}}\right )^{\frac {2}{3}} a}\right )+\operatorname {AiryBi}\left (-\frac {a \left (c x +k \right )-\frac {b^{2}}{4}}{\left (\frac {c}{\sqrt {a}}\right )^{\frac {2}{3}} a}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.51 (sec). Leaf size: 359
DSolve[y'[x]==a*y[x]^2+b*y[x]+c*x+k,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to -\frac {c \left (-b (-a c)^{2/3} \operatorname {AiryBi}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )+2 a c \operatorname {AiryBiPrime}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )+c_1 \left (2 a c \operatorname {AiryAiPrime}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )-b (-a c)^{2/3} \operatorname {AiryAi}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )\right )\right )}{2 (-a c)^{5/3} \left (\operatorname {AiryBi}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )+c_1 \operatorname {AiryAi}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )\right )} \\ y(x)\to -\frac {\frac {2 \sqrt [3]{-a c} \operatorname {AiryAiPrime}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )}{\operatorname {AiryAi}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )}+b}{2 a} \\ y(x)\to -\frac {\frac {2 \sqrt [3]{-a c} \operatorname {AiryAiPrime}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )}{\operatorname {AiryAi}\left (\frac {b^2-4 a (k+c x)}{4 (-a c)^{2/3}}\right )}+b}{2 a} \\ \end{align*}