57.1.11 problem 11
Internal
problem
ID
[8995]
Book
:
First
order
enumerated
odes
Section
:
section
1
Problem
number
:
11
Date
solved
:
Monday, January 27, 2025 at 05:26:43 PM
CAS
classification
:
[[_Riccati, _special]]
\begin{align*} y^{\prime }&=a x +b y^{2} \end{align*}
✓ Solution by Maple
Time used: 0.003 (sec). Leaf size: 59
dsolve(diff(y(x),x)=a*x+b*y(x)^2,y(x), singsol=all)
\[
y = \frac {\left (a b \right )^{{1}/{3}} \left (\operatorname {AiryAi}\left (1, -\left (a b \right )^{{1}/{3}} x \right ) c_{1} +\operatorname {AiryBi}\left (1, -\left (a b \right )^{{1}/{3}} x \right )\right )}{b \left (c_{1} \operatorname {AiryAi}\left (-\left (a b \right )^{{1}/{3}} x \right )+\operatorname {AiryBi}\left (-\left (a b \right )^{{1}/{3}} x \right )\right )}
\]
✓ Solution by Mathematica
Time used: 0.165 (sec). Leaf size: 331
DSolve[D[y[x],x]==a*x+b*y[x]^2,y[x],x,IncludeSingularSolutions -> True]
\begin{align*}
y(x)\to \frac {\sqrt {a} \sqrt {b} x^{3/2} \left (-2 \operatorname {BesselJ}\left (-\frac {2}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )+c_1 \left (\operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )-\operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )\right )\right )-c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )}{2 b x \left (\operatorname {BesselJ}\left (\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )+c_1 \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )\right )} \\
y(x)\to -\frac {\sqrt {a} \sqrt {b} x^{3/2} \operatorname {BesselJ}\left (-\frac {4}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )-\sqrt {a} \sqrt {b} x^{3/2} \operatorname {BesselJ}\left (\frac {2}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )+\operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )}{2 b x \operatorname {BesselJ}\left (-\frac {1}{3},\frac {2}{3} \sqrt {a} \sqrt {b} x^{3/2}\right )} \\
\end{align*}