60.5.27 problem 1562

Internal problem ID [11561]
Book : Differential Gleichungen, E. Kamke, 3rd ed. Chelsea Pub. NY, 1948
Section : Chapter 4, linear fourth order
Problem number : 1562
Date solved : Tuesday, January 28, 2025 at 06:06:51 PM
CAS classification : [[_high_order, _with_linear_symmetries]]

\begin{align*} x^{4} y^{\prime \prime \prime \prime }+4 x^{3} y^{\prime \prime \prime }-\left (4 n^{2}-1\right ) x^{2} y^{\prime \prime }+\left (4 n^{2}-1\right ) x y^{\prime }-4 y x^{4}&=0 \end{align*}

✓ Solution by Maple

Time used: 0.036 (sec). Leaf size: 65

dsolve(x^4*diff(diff(diff(diff(y(x),x),x),x),x)+4*x^3*diff(diff(diff(y(x),x),x),x)-(4*n^2-1)*x^2*diff(diff(y(x),x),x)+(4*n^2-1)*x*diff(y(x),x)-4*y(x)*x^4=0,y(x), singsol=all)
 

\[ y = \left (\operatorname {BesselY}\left (n , \left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {2}\, x \right ) c_3 +\operatorname {BesselJ}\left (n , \left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {2}\, x \right ) c_{1} \right ) \operatorname {BesselJ}\left (n , \left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, x \right )+\operatorname {BesselY}\left (n , \left (\frac {1}{2}+\frac {i}{2}\right ) \sqrt {2}\, x \right ) \left (c_4 \operatorname {BesselY}\left (n , \left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {2}\, x \right )+c_{2} \operatorname {BesselJ}\left (n , \left (\frac {1}{2}-\frac {i}{2}\right ) \sqrt {2}\, x \right )\right ) \]

✓ Solution by Mathematica

Time used: 0.509 (sec). Leaf size: 140

DSolve[-4*x^4*y[x] + (-1 + 4*n^2)*x*D[y[x],x] - (-1 + 4*n^2)*x^2*D[y[x],{x,2}] + 4*x^3*Derivative[3][y][x] + x^4*Derivative[4][y][x] == 0,y[x],x,IncludeSingularSolutions -> True]
 

\[ y(x)\to c_1 \, _0F_3\left (;\frac {1}{2},1-\frac {n}{2},\frac {n}{2}+1;\frac {x^4}{64}\right )+\frac {1}{8} i c_2 x^2 \, _0F_3\left (;\frac {3}{2},\frac {3}{2}-\frac {n}{2},\frac {n}{2}+\frac {3}{2};\frac {x^4}{64}\right )+c_3 \left (\frac {i}{2}\right )^{-n} \operatorname {Gamma}(1-n)^2 \left (\text {ber}_{-n}(x){}^2+\text {bei}_{-n}(x){}^2\right )+c_4 \left (\frac {i}{2}\right )^n \operatorname {Gamma}(n+1)^2 \left (\text {ber}_n(x){}^2+\text {bei}_n(x){}^2\right ) \]