76.12.2 problem 2

Internal problem ID [17558]
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 : 2
Date solved : Tuesday, January 28, 2025 at 08:27:44 PM
CAS classification : [[_2nd_order, _linear, _nonhomogeneous]]

\begin{align*} \left (t -1\right ) y^{\prime \prime }-3 y^{\prime } t +4 y&=\sin \left (t \right ) \end{align*}

With initial conditions

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

✓ Solution by Maple

Time used: 7.805 (sec). Leaf size: 420

dsolve([(t-1)*diff(y(t),t$2)-3*t*diff(y(t),t)+4*y(t)=sin(t),y(-2) = 2, D(y)(-2) = 1],y(t), singsol=all)
 

\[ \text {Expression too large to display} \]

✓ Solution by Mathematica

Time used: 0.524 (sec). Leaf size: 782

DSolve[{(t-1)*D[y[t],{t,2}]-3*t*D[y[t],t]+4*y[t]==Sin[t],{y[-2]==2,Derivative[1][y][-2]==1}},y[t],t,IncludeSingularSolutions -> True]
 

\[ y(t)\to -\frac {(t-1)^4 \left (243 \left (3 \operatorname {HypergeometricU}\left (\frac {8}{3},5,-9\right ) L_{-\frac {11}{3}}^5(-9)-8 \operatorname {HypergeometricU}\left (\frac {11}{3},6,-9\right ) L_{-\frac {8}{3}}^4(-9)\right ) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 t-3\right ) \int _1^{-2}-\frac {243 L_{-\frac {8}{3}}^4(3 K[1]-3) \sin (K[1])}{8 \left (\operatorname {Hypergeometric1F1}\left (\frac {11}{3},6,3 K[1]-3\right ) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[1]-3\right )+5 \operatorname {Hypergeometric1F1}\left (\frac {8}{3},5,3 K[1]-3\right ) \operatorname {HypergeometricU}\left (\frac {11}{3},6,3 K[1]-3\right )\right ) (K[1]-1)^5}dK[1]-243 \left (3 \operatorname {HypergeometricU}\left (\frac {8}{3},5,-9\right ) L_{-\frac {11}{3}}^5(-9)-8 \operatorname {HypergeometricU}\left (\frac {11}{3},6,-9\right ) L_{-\frac {8}{3}}^4(-9)\right ) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 t-3\right ) \int _1^t-\frac {243 L_{-\frac {8}{3}}^4(3 K[1]-3) \sin (K[1])}{8 \left (\operatorname {Hypergeometric1F1}\left (\frac {11}{3},6,3 K[1]-3\right ) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[1]-3\right )+5 \operatorname {Hypergeometric1F1}\left (\frac {8}{3},5,3 K[1]-3\right ) \operatorname {HypergeometricU}\left (\frac {11}{3},6,3 K[1]-3\right )\right ) (K[1]-1)^5}dK[1]-1944 \operatorname {HypergeometricU}\left (\frac {11}{3},6,-9\right ) L_{-\frac {8}{3}}^4(-9) L_{-\frac {8}{3}}^4(3 t-3) \int _1^{-2}\frac {243 \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[2]-3\right ) \sin (K[2])}{8 \left (\operatorname {Hypergeometric1F1}\left (\frac {11}{3},6,3 K[2]-3\right ) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[2]-3\right )+5 \operatorname {Hypergeometric1F1}\left (\frac {8}{3},5,3 K[2]-3\right ) \operatorname {HypergeometricU}\left (\frac {11}{3},6,3 K[2]-3\right )\right ) (K[2]-1)^5}dK[2]+729 \operatorname {HypergeometricU}\left (\frac {8}{3},5,-9\right ) L_{-\frac {11}{3}}^5(-9) L_{-\frac {8}{3}}^4(3 t-3) \int _1^{-2}\frac {243 \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[2]-3\right ) \sin (K[2])}{8 \left (\operatorname {Hypergeometric1F1}\left (\frac {11}{3},6,3 K[2]-3\right ) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[2]-3\right )+5 \operatorname {Hypergeometric1F1}\left (\frac {8}{3},5,3 K[2]-3\right ) \operatorname {HypergeometricU}\left (\frac {11}{3},6,3 K[2]-3\right )\right ) (K[2]-1)^5}dK[2]+1944 \operatorname {HypergeometricU}\left (\frac {11}{3},6,-9\right ) L_{-\frac {8}{3}}^4(-9) L_{-\frac {8}{3}}^4(3 t-3) \int _1^t\frac {243 \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[2]-3\right ) \sin (K[2])}{8 \left (\operatorname {Hypergeometric1F1}\left (\frac {11}{3},6,3 K[2]-3\right ) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[2]-3\right )+5 \operatorname {Hypergeometric1F1}\left (\frac {8}{3},5,3 K[2]-3\right ) \operatorname {HypergeometricU}\left (\frac {11}{3},6,3 K[2]-3\right )\right ) (K[2]-1)^5}dK[2]-729 \operatorname {HypergeometricU}\left (\frac {8}{3},5,-9\right ) L_{-\frac {11}{3}}^5(-9) L_{-\frac {8}{3}}^4(3 t-3) \int _1^t\frac {243 \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[2]-3\right ) \sin (K[2])}{8 \left (\operatorname {Hypergeometric1F1}\left (\frac {11}{3},6,3 K[2]-3\right ) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 K[2]-3\right )+5 \operatorname {Hypergeometric1F1}\left (\frac {8}{3},5,3 K[2]-3\right ) \operatorname {HypergeometricU}\left (\frac {11}{3},6,3 K[2]-3\right )\right ) (K[2]-1)^5}dK[2]-11 L_{-\frac {8}{3}}^4(-9) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 t-3\right )-18 L_{-\frac {11}{3}}^5(-9) \operatorname {HypergeometricU}\left (\frac {8}{3},5,3 t-3\right )+48 \operatorname {HypergeometricU}\left (\frac {11}{3},6,-9\right ) L_{-\frac {8}{3}}^4(3 t-3)+11 \operatorname {HypergeometricU}\left (\frac {8}{3},5,-9\right ) L_{-\frac {8}{3}}^4(3 t-3)\right )}{243 \left (3 \operatorname {HypergeometricU}\left (\frac {8}{3},5,-9\right ) L_{-\frac {11}{3}}^5(-9)-8 \operatorname {HypergeometricU}\left (\frac {11}{3},6,-9\right ) L_{-\frac {8}{3}}^4(-9)\right )} \]