2.1.81 Internal
problem
ID
[10067]Book
:
Own
collection
of
miscellaneous
problems Section
:
section
1.0 Problem
number
:
80Date
solved
:
Thursday, November 27, 2025 at 10:10:11 AMCAS
classification
:
[[_Riccati, _special]]
Time used: 0.046 (sec)
Solve
\begin{align*}
y^{\prime }&=x^{2}+y^{2} \\
\end{align*}
This is reduced Riccati ode of the form
\begin{align*} y^{\prime }&=a \,x^{n}+b y^{2} \end{align*}
Comparing the given ode to the above shows that
\begin{align*} a &= 1\\ b &= 1\\ n &= 2 \end{align*}
Since \(n\neq -2\) then the solution of the reduced Riccati ode is given by
\begin{align*} w & =\sqrt {x}\left \{ \begin {array}[c]{cc} c_{1}\operatorname {BesselJ}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab} x^{k}\right ) +c_{2}\operatorname {BesselY}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {ab}x^{k}\right ) & ab>0\\ c_{1}\operatorname {BesselI}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-ab}x^{k}\right ) +c_{2}\operatorname {BesselK}\left ( \frac {1}{2k},\frac {1}{k}\sqrt {-ab}x^{k}\right ) & ab<0 \end {array} \right . \tag {1}\\ y & =-\frac {1}{b}\frac {w^{\prime }}{w}\nonumber \\ k &=1+\frac {n}{2}\nonumber \end{align*}
Since \(ab>0\) then EQ(1) gives
\begin{align*} k &= 2\\ w &= \sqrt {x}\, \left (c_1 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+c_2 \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )\right ) \end{align*}
Therefore the solution becomes
\begin{align*} y & =-\frac {1}{b}\frac {w^{\prime }}{w} \end{align*}
Substituting the value of \(b,w\) found above and simplifying gives
\[
y = \frac {\left (-\operatorname {BesselY}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_2 -\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_1 \right ) x}{c_1 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+c_2 \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}
\]
Letting
\(c_2 = 1\) the above becomes
\[
y = \frac {\left (-\operatorname {BesselY}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right )-\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_1 \right ) x}{c_1 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+\operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}
\]
Simplifying the above gives
\begin{align*}
y &= -\frac {\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 ) x}{c_1 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+\operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \\
\end{align*}
Figure 2.164: Slope field \(y^{\prime } = x^{2}+y^{2}\) Summary of solutions found
\begin{align*}
y &= -\frac {\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 ) x}{c_1 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+\operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \\
\end{align*}
Time used: 0.131 (sec)
Solve
\begin{align*}
y^{\prime }&=x^{2}+y^{2} \\
\end{align*}
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
\[
u^{\prime \prime }\left (x \right )+x^{2} u \left (x \right ) = 0
\]
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*}
Taking derivative gives
\begin{equation}
\tag{4} 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 )
\end{equation}
Substituting equations (3,4) into (1) results in
\begin{align*}
y &= \frac {-u'}{f_2 u} \\
y &= \frac {-u'}{u} \\
y &= \frac {\left (-\operatorname {BesselY}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_2 -\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_1 \right ) x}{c_1 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+c_2 \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \\
\end{align*}
Doing change of constants,
the above solution becomes
\[
y = -\frac {\frac {\operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 \sqrt {x}}+x^{{3}/{2}} \left (-\operatorname {BesselJ}\left (\frac {5}{4}, \frac {x^{2}}{2}\right )+\frac {\operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 x^{2}}\right )+\frac {c_3 \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 \sqrt {x}}+c_3 \,x^{{3}/{2}} \left (-\operatorname {BesselY}\left (\frac {5}{4}, \frac {x^{2}}{2}\right )+\frac {\operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}{2 x^{2}}\right )}{\sqrt {x}\, \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+c_3 \sqrt {x}\, \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}
\]
Simplifying the above gives
\begin{align*}
y &= \frac {\left (-\operatorname {BesselY}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_3 -\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right )\right ) x}{c_3 \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+\operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \\
\end{align*}
Figure 2.165: Slope field \(y^{\prime } = x^{2}+y^{2}\) Summary of solutions found
\begin{align*}
y &= \frac {\left (-\operatorname {BesselY}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_3 -\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right )\right ) x}{c_3 \operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+\operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )} \\
\end{align*}
✓ Time used: 0.003 (sec). Leaf size: 43 ode := diff ( y ( x ), x ) = x^2+y(x)^2;
dsolve ( ode , y ( x ), singsol=all);
\[
y = -\frac {x \left (\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right ) c_1 +\operatorname {BesselY}\left (-\frac {3}{4}, \frac {x^{2}}{2}\right )\right )}{c_1 \operatorname {BesselJ}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )+\operatorname {BesselY}\left (\frac {1}{4}, \frac {x^{2}}{2}\right )}
\]
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
Maple step by step
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=x^{2}+y \left (x \right )^{2} \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 1 \\ {} & {} & \frac {d}{d x}y \left (x \right ) \\ \bullet & {} & \textrm {Solve for the highest derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=x^{2}+y \left (x \right )^{2} \end {array} \]
✓ Time used: 0.087 (sec). Leaf size: 169 ode = D [ y [ x ], x ]== x ^2+ y [ x ]^2; ic ={}; DSolve [{ ode , ic }, 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*}
✗ from sympy import *
x = symbols("x")
y = Function("y")
ode = Eq(-x**2 - y(x)**2 + Derivative(y(x), x),0)
ics = {}
dsolve ( ode , func = y ( x ), ics = ics ) TypeError : bad operand type for unary -: list