2.41 problem 41

Internal problem ID [8130]
Book : Second order enumerated odes
Section : section 2
Problem number : 41
Date solved : Monday, October 21, 2024 at 04:54:08 PM
CAS classification : [[_Riccati, _special]]

Solve

\begin{align*} y^{\prime }&=x -y^{2} \end{align*}

2.41.1 Solved as first order ode of type Riccati

Time used: 0.266 (sec)

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*}

This is Airy ODE. It has the general form

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

Where in this case

\begin{align*} a &= -1\\ b &= 0\\ c &= 1\\ F &= 0 \end{align*}

Therefore the solution to the homogeneous Airy ODE becomes

\[ u = c_1 \operatorname {AiryAi}\left (-x \left (-1\right )^{{1}/{3}}\right )+c_2 \operatorname {AiryBi}\left (-x \left (-1\right )^{{1}/{3}}\right ) \]

Will add steps showing solving for IC soon.

Taking derivative gives

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

Doing change of constants, the solution becomes

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

Figure 163: Slope field plot
\(y^{\prime } = x -y^{2}\)

2.41.2 Maple step by step solution

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

2.41.3 Maple trace

Methods for first order ODEs:
 

2.41.4 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 23

dsolve(diff(y(x),x) = x-y(x)^2, 
       y(x),singsol=all)
 

\[ y = \frac {c_1 \operatorname {AiryAi}\left (1, x\right )+\operatorname {AiryBi}\left (1, x\right )}{c_1 \operatorname {AiryAi}\left (x \right )+\operatorname {AiryBi}\left (x \right )} \]

2.41.5 Mathematica DSolve solution

Solving time : 0.18 (sec)
Leaf size : 223

DSolve[{D[y[x],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*}