1.81 problem 80

Internal problem ID [7773]
Book : Own collection of miscellaneous problems
Section : section 1.0
Problem number : 80
Date solved : Monday, October 21, 2024 at 04:18:24 PM
CAS classification : [[_Riccati, _special]]

Solve

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

1.81.1 Solved as first order ode of type Riccati

Time used: 0.107 (sec)

In canonical form the ODE is

\begin{align*} y' &= F(x,y)\\ &= x^{2}+y^{2} \end{align*}

This is a Riccati ODE. Comparing the ODE to solve

\[ y' = x^{2}+y^{2} \]

With Riccati ODE standard form

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

Shows that \(f_0(x)=x^{2}\), \(f_1(x)=0\) 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^{2} \end{align*}

Substituting the above terms back in equation (2) gives

\begin{align*} u^{\prime \prime }\left (x \right )+x^{2} u \left (x \right ) = 0 \end{align*}

Writing the ode as

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )+x^{4} u = 0\tag {1} \end{align*}

Bessel ode has the form

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )+\left (\frac {d u}{d x}\right ) x +\left (-n^{2}+x^{2}\right ) u = 0\tag {2} \end{align*}

The generalized form of Bessel ode is given by Bowman (1958) as the following

\begin{align*} x^{2} \left (\frac {d^{2}u}{d x^{2}}\right )+\left (1-2 \alpha \right ) x \left (\frac {d u}{d x}\right )+\left (\beta ^{2} \gamma ^{2} x^{2 \gamma }-n^{2} \gamma ^{2}+\alpha ^{2}\right ) u = 0\tag {3} \end{align*}

With the standard solution

\begin{align*} u&=x^{\alpha } \left (c_1 \operatorname {BesselJ}\left (n , \beta \,x^{\gamma }\right )+c_2 \operatorname {BesselY}\left (n , \beta \,x^{\gamma }\right )\right )\tag {4} \end{align*}

Comparing (3) to (1) and solving for \(\alpha ,\beta ,n,\gamma \) gives

\begin{align*} \alpha &= {\frac {1}{2}}\\ \beta &= {\frac {1}{2}}\\ n &= {\frac {1}{4}}\\ \gamma &= 2 \end{align*}

Substituting all the above into (4) gives the solution as

\begin{align*} u = c_1 \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+c_2 \sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right ) \end{align*}

Will add steps showing solving for IC soon.

Taking derivative gives

\[ u^{\prime }\left (x \right ) = \frac {c_1 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 \sqrt {x}}+c_1 \,x^{{3}/{2}} \left (-\operatorname {BesselJ}\left (\frac {5}{4}, \frac {x^{2}}{2}\right )+\frac {\operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 x^{2}}\right )+\frac {c_2 \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 \sqrt {x}}+c_2 \,x^{{3}/{2}} \left (-\operatorname {BesselY}\left (\frac {5}{4}, \frac {x^{2}}{2}\right )+\frac {\operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 x^{2}}\right ) \]

Doing change of constants, the solution becomes

\[ y = -\frac {\frac {c_3 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 \sqrt {x}}+c_3 \,x^{{3}/{2}} \left (-\operatorname {BesselJ}\left (\frac {5}{4}, \frac {x^{2}}{2}\right )+\frac {\operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 x^{2}}\right )+\frac {\operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 \sqrt {x}}+x^{{3}/{2}} \left (-\operatorname {BesselY}\left (\frac {5}{4}, \frac {x^{2}}{2}\right )+\frac {\operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 x^{2}}\right )}{c_3 \sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+\sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \]

Figure 155: Slope field plot
\(y^{\prime } = x^{2}+y^{2}\)

1.81.2 Maple step by step solution

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

1.81.3 Maple trace

Methods for first order ODEs:
 

1.81.4 Maple dsolve solution

Solving time : 0.003 (sec)
Leaf size : 43

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

\[ y = -\frac {x \left (\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_1 +\operatorname {BesselY}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right )\right )}{c_1 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+\operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \]

1.81.5 Mathematica DSolve solution

Solving time : 0.134 (sec)
Leaf size : 169

DSolve[{D[y[x],x]==x^2+y[x]^2,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)\to \frac {x^2 \left (-2 \operatorname {BesselJ}\left (-\frac {3}{4},\frac {x^2}{2}\right )+c_1 \left (\operatorname {BesselJ}\left (\frac {3}{4},\frac {x^2}{2}\right )-\operatorname {BesselJ}\left (-\frac {5}{4},\frac {x^2}{2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{4},\frac {x^2}{2}\right )}{2 x \left (\operatorname {BesselJ}\left (\frac {1}{4},\frac {x^2}{2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{4},\frac {x^2}{2}\right )\right )} \\ y(x)\to -\frac {x^2 \operatorname {BesselJ}\left (-\frac {5}{4},\frac {x^2}{2}\right )-x^2 \operatorname {BesselJ}\left (\frac {3}{4},\frac {x^2}{2}\right )+\operatorname {BesselJ}\left (-\frac {1}{4},\frac {x^2}{2}\right )}{2 x \operatorname {BesselJ}\left (-\frac {1}{4},\frac {x^2}{2}\right )} \\ \end{align*}