Internal problem ID [3322]
Book: Ordinary differential equations and their solutions. By George Moseley Murphy.
1960
Section: Various 3
Problem number: 58.
ODE order: 1.
ODE degree: 1.
CAS Maple gives this as type [[_Riccati, _special]]
\[ \boxed {y^{\prime }-b y^{2}=x^{2} a} \]
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 71
dsolve(diff(y(x),x) = a*x^2+b*y(x)^2,y(x), singsol=all)
\[ y \left (x \right ) = -\frac {\sqrt {a b}\, x \left (\operatorname {BesselJ}\left (-\frac {3}{4}, \frac {\sqrt {a b}\, x^{2}}{2}\right ) c_{1} +\operatorname {BesselY}\left (-\frac {3}{4}, \frac {\sqrt {a b}\, x^{2}}{2}\right )\right )}{b \left (c_{1} \operatorname {BesselJ}\left (\frac {1}{4}, \frac {\sqrt {a b}\, x^{2}}{2}\right )+\operatorname {BesselY}\left (\frac {1}{4}, \frac {\sqrt {a b}\, x^{2}}{2}\right )\right )} \]
✓ Solution by Mathematica
Time used: 0.143 (sec). Leaf size: 305
DSolve[y'[x]==a x^2+b y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*} y(x)\to \frac {\sqrt {a} \sqrt {b} x^2 \left (-2 \operatorname {BesselJ}\left (-\frac {3}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )+c_1 \left (\operatorname {BesselJ}\left (\frac {3}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )-\operatorname {BesselJ}\left (-\frac {5}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )}{2 b x \left (\operatorname {BesselJ}\left (\frac {1}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )\right )} y(x)\to -\frac {\sqrt {a} \sqrt {b} x^2 \operatorname {BesselJ}\left (-\frac {5}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )-\sqrt {a} \sqrt {b} x^2 \operatorname {BesselJ}\left (\frac {3}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )+\operatorname {BesselJ}\left (-\frac {1}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )}{2 b x \operatorname {BesselJ}\left (-\frac {1}{4},\frac {1}{2} \sqrt {a} \sqrt {b} x^2\right )} \end{align*}