76.12.1 problem 1

Internal problem ID [17557]
Book : Differential equations. An introduction to modern methods and applications. James Brannan, William E. Boyce. Third edition. Wiley 2015
Section : Chapter 4. Second order linear equations. Section 4.2 (Theory of second order linear homogeneous equations). Problems at page 226
Problem number : 1
Date solved : Tuesday, January 28, 2025 at 10:43:43 AM
CAS classification : [[_2nd_order, _with_linear_symmetries]]

\begin{align*} t y^{\prime \prime }+3 y&=t \end{align*}

With initial conditions

\begin{align*} y \left (1\right )&=1\\ y^{\prime }\left (1\right )&=2 \end{align*}

✓ Solution by Maple

Time used: 0.296 (sec). Leaf size: 82

dsolve([t*diff(y(t),t$2)+3*y(t)=t,y(1) = 1, D(y)(1) = 2],y(t), singsol=all)
 

\[ y = \frac {2 \sqrt {t}\, \left (\sqrt {3}\, \operatorname {BesselY}\left (0, 2 \sqrt {3}\right )-\frac {5 \operatorname {BesselY}\left (1, 2 \sqrt {3}\right )}{2}\right ) \pi \operatorname {BesselJ}\left (1, 2 \sqrt {3}\, \sqrt {t}\right )}{3}-\frac {2 \sqrt {t}\, \left (\operatorname {BesselJ}\left (0, 2 \sqrt {3}\right ) \sqrt {3}-\frac {5 \operatorname {BesselJ}\left (1, 2 \sqrt {3}\right )}{2}\right ) \pi \operatorname {BesselY}\left (1, 2 \sqrt {3}\, \sqrt {t}\right )}{3}+\frac {t}{3} \]

✓ Solution by Mathematica

Time used: 0.228 (sec). Leaf size: 673

DSolve[{t*D[y[t],{t,2}]+3*y[t]==t,{y[1]==1,Derivative[1][y][1]==2}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to \frac {\sqrt {t} \left (\operatorname {BesselJ}\left (1,2 \sqrt {3} \sqrt {t}\right ) \left (\pi \operatorname {BesselY}\left (1,2 \sqrt {3}\right ) \left (\sqrt {3} \left (\operatorname {BesselJ}\left (2,2 \sqrt {3}\right )-\operatorname {BesselJ}\left (0,2 \sqrt {3}\right )\right ) G_{2,4}^{2,1}\left (3\left | \begin {array}{c} 0,-\frac {1}{2} \\ 0,1,-1,-\frac {1}{2} \\ \end {array} \right .\right )+6 (2 \operatorname {Hypergeometric0F1Regularized}(3,-3)-3 \operatorname {Hypergeometric0F1Regularized}(4,-3)) \operatorname {BesselY}\left (1,2 \sqrt {3}\right )\right )+\sqrt {3} \pi \operatorname {BesselJ}\left (1,2 \sqrt {3}\right ) \left (\left (\operatorname {BesselY}\left (0,2 \sqrt {3}\right )-\operatorname {BesselY}\left (2,2 \sqrt {3}\right )\right ) G_{2,4}^{2,1}\left (3\left | \begin {array}{c} 0,-\frac {1}{2} \\ 0,1,-1,-\frac {1}{2} \\ \end {array} \right .\right )-2 \operatorname {BesselY}\left (1,2 \sqrt {3}\right )^2\right )+3 \left (\operatorname {BesselY}\left (0,2 \sqrt {3}\right )-\sqrt {3} \operatorname {BesselY}\left (1,2 \sqrt {3}\right )-\operatorname {BesselY}\left (2,2 \sqrt {3}\right )\right )\right )+\sqrt {3} \pi t \operatorname {BesselJ}\left (1,2 \sqrt {3} \sqrt {t}\right ) \left (\left (\operatorname {BesselJ}\left (0,2 \sqrt {3}\right )-\operatorname {BesselJ}\left (2,2 \sqrt {3}\right )\right ) \operatorname {BesselY}\left (1,2 \sqrt {3}\right )+\operatorname {BesselJ}\left (1,2 \sqrt {3}\right ) \left (\operatorname {BesselY}\left (2,2 \sqrt {3}\right )-\operatorname {BesselY}\left (0,2 \sqrt {3}\right )\right )\right ) G_{2,4}^{2,1}\left (\sqrt {3} \sqrt {t},\frac {1}{2}| \begin {array}{c} 0,-\frac {1}{2} \\ 0,1,-1,-\frac {1}{2} \\ \end {array} \right )+\operatorname {BesselY}\left (1,2 \sqrt {3} \sqrt {t}\right ) \left (3 \sqrt {3} \pi t^2 \operatorname {Hypergeometric0F1Regularized}(3,-3 t) \left (\left (\operatorname {BesselJ}\left (2,2 \sqrt {3}\right )-\operatorname {BesselJ}\left (0,2 \sqrt {3}\right )\right ) \operatorname {BesselY}\left (1,2 \sqrt {3}\right )+\operatorname {BesselJ}\left (1,2 \sqrt {3}\right ) \left (\operatorname {BesselY}\left (0,2 \sqrt {3}\right )-\operatorname {BesselY}\left (2,2 \sqrt {3}\right )\right )\right )+\pi \left (3 \sqrt {3} \operatorname {Hypergeometric0F1Regularized}(3,-3) \left (\operatorname {BesselJ}\left (0,2 \sqrt {3}\right )-\operatorname {BesselJ}\left (2,2 \sqrt {3}\right )\right ) \operatorname {BesselY}\left (1,2 \sqrt {3}\right )-3 \operatorname {BesselJ}\left (1,2 \sqrt {3}\right ) \left (\sqrt {3} \operatorname {Hypergeometric0F1Regularized}(3,-3) \operatorname {BesselY}\left (0,2 \sqrt {3}\right )-\sqrt {3} \operatorname {Hypergeometric0F1Regularized}(3,-3) \operatorname {BesselY}\left (2,2 \sqrt {3}\right )+(4 \operatorname {Hypergeometric0F1Regularized}(3,-3)-6 \operatorname {Hypergeometric0F1Regularized}(4,-3)) \operatorname {BesselY}\left (1,2 \sqrt {3}\right )\right )+2 \sqrt {3} \operatorname {BesselJ}\left (1,2 \sqrt {3}\right )^2 \operatorname {BesselY}\left (1,2 \sqrt {3}\right )\right )+3 \left (-\operatorname {BesselJ}\left (0,2 \sqrt {3}\right )+\sqrt {3} \operatorname {BesselJ}\left (1,2 \sqrt {3}\right )+\operatorname {BesselJ}\left (2,2 \sqrt {3}\right )\right )\right )\right )}{3 \left (\left (\operatorname {BesselJ}\left (2,2 \sqrt {3}\right )-\operatorname {BesselJ}\left (0,2 \sqrt {3}\right )\right ) \operatorname {BesselY}\left (1,2 \sqrt {3}\right )+\operatorname {BesselJ}\left (1,2 \sqrt {3}\right ) \left (\operatorname {BesselY}\left (0,2 \sqrt {3}\right )-\operatorname {BesselY}\left (2,2 \sqrt {3}\right )\right )\right )} \]