Internal problem ID [7327]
Internal file name [OUTPUT/6308_Sunday_June_05_2022_04_39_26_PM_47091222/index.tex
]
Book: First order enumerated odes
Section: section 1
Problem number: 11.
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]]
\[ \boxed {y^{\prime }-b y^{2}=a x} \]
In canonical form the ODE is \begin {align*} y' &= F(x,y)\\ &= b \,y^{2}+a x \end {align*}
This is a Riccati ODE. Comparing the ODE to solve \[ y' = b \,y^{2}+a x \] With Riccati ODE standard form \[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \] Shows that \(f_0(x)=a x\), \(f_1(x)=0\) and \(f_2(x)=b\). Let \begin {align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{b 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 &=a x \,b^{2} \end {align*}
Substituting the above terms back in equation (2) gives \begin {align*} b u^{\prime \prime }\left (x \right )+a x \,b^{2} u \left (x \right ) &=0 \end {align*}
Solving the above ODE (this ode solved using Maple, not this program), gives
\[ u \left (x \right ) = c_{1} \operatorname {AiryAi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )+c_{2} \operatorname {AiryBi}\left (-\left (a b \right )^{\frac {1}{3}} x \right ) \] The above shows that \[ u^{\prime }\left (x \right ) = \left (-\operatorname {AiryBi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right ) c_{2} -\operatorname {AiryAi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right ) c_{1} \right ) \left (a b \right )^{\frac {1}{3}} \] Using the above in (1) gives the solution \[ y = -\frac {\left (-\operatorname {AiryBi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right ) c_{2} -\operatorname {AiryAi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right ) c_{1} \right ) \left (a b \right )^{\frac {1}{3}}}{b \left (c_{1} \operatorname {AiryAi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )+c_{2} \operatorname {AiryBi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )\right )} \] Dividing both numerator and denominator by \(c_{1}\) gives, after renaming the constant \(\frac {c_{2}}{c_{1}}=c_{3}\) the following solution
\[ y = \frac {\left (\operatorname {AiryAi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right ) c_{3} +\operatorname {AiryBi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right )\right ) \left (a b \right )^{\frac {1}{3}}}{b \left (c_{3} \operatorname {AiryAi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )+\operatorname {AiryBi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )\right )} \]
The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\left (\operatorname {AiryAi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right ) c_{3} +\operatorname {AiryBi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right )\right ) \left (a b \right )^{\frac {1}{3}}}{b \left (c_{3} \operatorname {AiryAi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )+\operatorname {AiryBi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )\right )} \\ \end{align*}
Verification of solutions
\[ y = \frac {\left (\operatorname {AiryAi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right ) c_{3} +\operatorname {AiryBi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right )\right ) \left (a b \right )^{\frac {1}{3}}}{b \left (c_{3} \operatorname {AiryAi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )+\operatorname {AiryBi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )\right )} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime }-b y^{2}=a x \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=a x +b y^{2} \end {array} \]
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: 59
dsolve(diff(y(x),x)=a*x+b*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = \frac {\left (a b \right )^{\frac {1}{3}} \left (\operatorname {AiryAi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right ) c_{1} +\operatorname {AiryBi}\left (1, -\left (a b \right )^{\frac {1}{3}} x \right )\right )}{b \left (c_{1} \operatorname {AiryAi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )+\operatorname {AiryBi}\left (-\left (a b \right )^{\frac {1}{3}} x \right )\right )} \]
✓ Solution by Mathematica
Time used: 0.163 (sec). Leaf size: 331
DSolve[y'[x]==a*x+b*y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sqrt {a} \sqrt {b} x^{3/2} \left (-2 \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )+c_1 \left (\operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )-\operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )}{2 b x \left (\operatorname {BesselJ}\left (\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )\right )} \\ y(x)\to -\frac {\sqrt {a} \sqrt {b} x^{3/2} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )-\sqrt {a} \sqrt {b} x^{3/2} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )}{2 b x \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )} \\ \end{align*}