1.20 problem 20

Internal problem ID [7984]
Book : First order enumerated odes
Section : section 1
Problem number : 20
Date solved : Monday, October 21, 2024 at 04:40:18 PM
CAS classification : [[_Riccati, _special]]

Solve

\begin{align*} c y^{\prime }&=\frac {a x +b y^{2}}{r} \end{align*}

1.20.1 Solved as first order ode of type Riccati

Time used: 0.115 (sec)

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= \frac {b \,y^{2}+a x}{r c} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = \frac {a x}{r c}+\frac {b \,y^{2}}{r c} \]

With Riccati ODE standard form

\[ y' = f_0(x)+ f_1(x)y+f_2(x)y^{2} \]

Shows that \(f_0(x)=\frac {a x}{r c}\), \(f_1(x)=0\) and \(f_2(x)=\frac {b}{c r}\). Let

\begin{align*} y &= \frac {-u'}{f_2 u} \\ &= \frac {-u'}{\frac {u b}{c r}} \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 &=\frac {b^{2} a x}{c^{3} r^{3}} \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} \frac {b u^{\prime \prime }\left (x \right )}{c r}+\frac {b^{2} a x u \left (x \right )}{c^{3} r^{3}} = 0 \end{align*}

This is Airy ODE. It has the general form

\[ a \frac {d^{2}u}{d x^{2}} + b \frac {d u}{d x} + c u x = F(x) \]

Where in this case

\begin{align*} a &= \frac {b}{c r}\\ b &= 0\\ c &= \frac {b^{2} a}{c^{3} r^{3}}\\ F &= 0 \end{align*}

Therefore the solution to the homogeneous Airy ODE becomes

\[ u = c_1 \operatorname {AiryAi}\left (-\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right )+c_2 \operatorname {AiryBi}\left (-\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right ) \]

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = -c_1 \left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right )-c_2 \left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right ) \]

Doing change of constants, the solution becomes

\[ y = -\frac {\left (-c_3 \left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} \operatorname {AiryAi}\left (1, -\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right )-\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} \operatorname {AiryBi}\left (1, -\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right )\right ) c r}{b \left (c_3 \operatorname {AiryAi}\left (-\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right )+\operatorname {AiryBi}\left (-\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right )\right )} \]

1.20.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & c y^{\prime }=\frac {a x +b y^{2}}{r} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & y^{\prime } \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\frac {a x +b y^{2}}{r c} \end {array} \]

1.20.3 Maple trace

Methods for first order ODEs:
 

1.20.4 Maple dsolve solution

Solving time : 0.002 (sec)
Leaf size : 91

dsolve(c*diff(y(x),x) = (a*x+b*y(x)^2)/r, 
       y(x),singsol=all)
 

\[ y = \frac {\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} \left (\operatorname {AiryAi}\left (1, -\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right ) c_1 +\operatorname {AiryBi}\left (1, -\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right )\right ) c r}{b \left (c_1 \operatorname {AiryAi}\left (-\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right )+\operatorname {AiryBi}\left (-\left (\frac {a b}{r^{2} c^{2}}\right )^{{1}/{3}} x \right )\right )} \]

1.20.5 Mathematica DSolve solution

Solving time : 0.21 (sec)
Leaf size : 517

DSolve[{c*D[y[x],x]==(a*x+b*y[x]^2)/r,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to \frac {c r \left (x^{3/2} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} \left (-2 \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )+c_1 \left (\operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )-\operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )\right )}{2 b x \left (\operatorname {BesselJ}\left (\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )\right )} \\ y(x)\to -\frac {c r \left (x^{3/2} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )-x^{3/2} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )\right )}{2 b x \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {\frac {a}{c r}} \sqrt {\frac {b}{c r}} x^{3/2}\right )} \\ \end{align*}