6.17.6 problem section 9.1, problem 5(b)

Internal problem ID [2112]
Book : Elementary differential equations with boundary value problems. William F. Trench. Brooks/Cole 2001
Section : Chapter 9 Introduction to Linear Higher Order Equations. Section 9.1. Page 471
Problem number : section 9.1, problem 5(b)
Date solved : Tuesday, September 30, 2025 at 05:24:10 AM
CAS classification : [[_3rd_order, _exact, _linear, _homogeneous]]

\begin{align*} x^{3} y^{\prime \prime \prime }-x^{2} y^{\prime \prime }-2 x y^{\prime }+6 y&=0 \end{align*}

With initial conditions

\begin{align*} y \left (1\right )&=k_{0} \\ y^{\prime }\left (1\right )&=k_{1} \\ y^{\prime \prime }\left (1\right )&=k_{2} \\ \end{align*}

✓ Maple. Time used: 0.032 (sec). Leaf size: 40

ode:=x^3*diff(diff(diff(y(x),x),x),x)-x^2*diff(diff(y(x),x),x)-2*x*diff(y(x),x)+6*y(x) = 0; 
ic:=[y(1) = k__0, D(y)(1) = k__1, (D@@2)(y)(1) = k__2]; 
dsolve([ode,op(ic)],y(x), singsol=all);
 

\[ y = \frac {3 \left (k_{2} -2 k_{0} \right ) x^{4}+4 \left (k_{1} +3 k_{0} -k_{2} \right ) x^{3}+6 k_{0} -4 k_{1} +k_{2}}{12 x} \]

✓ Mathematica. Time used: 0.003 (sec). Leaf size: 47

ode=x^3*D[y[x],{x,3}]-x^2*D[y[x],{x,2}]-2*x*D[y[x],x]+6*y[x]==0; 
ic={y[1]==k0,Derivative[1][y][1]==k1,Derivative[2][y][1]==k2}; 
DSolve[{ode,ic},y[x],x,IncludeSingularSolutions->True]
 

\begin{align*} y(x)&\to \frac {-6 \text {k0} \left (x^4-2 x^3-1\right )+4 \text {k1} \left (x^3-1\right )+3 \text {k2} x^4-4 \text {k2} x^3+\text {k2}}{12 x} \end{align*}

✓ Sympy. Time used: 0.137 (sec). Leaf size: 36

from sympy import * 
x = symbols("x") 
y = Function("y") 
ode = Eq(x**3*Derivative(y(x), (x, 3)) - x**2*Derivative(y(x), (x, 2)) - 2*x*Derivative(y(x), x) + 6*y(x),0) 
ics = {y(1): k__0, Subs(Derivative(y(x), x), x, 1): k__1, Subs(Derivative(y(x), (x, 2)), x, 1): k__2} 
dsolve(ode,func=y(x),ics=ics)
 

\[ y{\left (x \right )} = x^{3} \left (- \frac {k^{0}}{2} + \frac {k^{2}}{4}\right ) + x^{2} \left (k^{0} + \frac {k^{1}}{3} - \frac {k^{2}}{3}\right ) + \frac {\frac {k^{0}}{2} - \frac {k^{1}}{3} + \frac {k^{2}}{12}}{x} \]