17.4 problem section 9.1, problem 5(b) 2

Internal problem ID [2110]
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) 2
Date solved : Sunday, October 06, 2024 at 03:11:40 AM
CAS classification : [[_3rd_order, _exact, _linear, _homogeneous]]

Solve

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

17.4.1 Solved as higher order Euler type ode

Time used: 0.113 (sec)

This is Euler ODE of higher order. Let \(y = x^{\lambda }\). Hence

\begin{align*} y^{\prime } &= \lambda \,x^{\lambda -1}\\ y^{\prime \prime } &= \lambda \left (\lambda -1\right ) x^{\lambda -2}\\ y^{\prime \prime \prime } &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3} \end{align*}

Substituting these back into

\[ x^{3} y^{\prime \prime \prime }-x^{2} y^{\prime \prime }-2 x y^{\prime }+6 y = 0 \]

gives

\[ -2 x \lambda \,x^{\lambda -1}-x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+6 x^{\lambda } = 0 \]

Which simplifies to

\[ -2 \lambda \,x^{\lambda }-\lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+6 x^{\lambda } = 0 \]

And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes

\[ -2 \lambda -\lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )+6 = 0 \]

Simplifying gives the characteristic equation as

\[ \lambda ^{3}-4 \lambda ^{2}+\lambda +6 = 0 \]

Solving the above gives the following roots

\begin{align*} \lambda _1 &= 2\\ \lambda _2 &= 3\\ \lambda _3 &= -1 \end{align*}

This table summarises the result

root	multiplicity	type of root
\(-1\)	\(1\)	real root
\(2\)	\(1\)	real root
\(3\)	\(1\)	real root

The solution is generated by going over the above table. For each real root \(\lambda \) of multiplicity one generates a \(c_1x^{\lambda }\) basis solution. Each real root of multiplicty two, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) basis solutions. Each real root of multiplicty three, generates \(c_1x^{\lambda }\) and \(c_2x^{\lambda } \ln \left (x \right )\) and \(c_3x^{\lambda } \ln \left (x \right )^{2}\) basis solutions, and so on. Each complex root \(\alpha \pm i \beta \) of multiplicity one generates \(x^{\alpha } \left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity two generates \(\ln \left (x \right ) x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And each complex root \(\alpha \pm i \beta \) of multiplicity three generates \(\ln \left (x \right )^{2} x^{\alpha }\left (c_1\cos (\beta \ln \left (x \right ))+c_2\sin (\beta \ln \left (x \right ))\right )\) basis solutions. And so on. Using the above show that the solution is

\[ y = \frac {c_1}{x}+c_2 \,x^{2}+c_3 \,x^{3} \]

The fundamental set of solutions for the homogeneous solution are the following

\begin{align*} y_1 &= \frac {1}{x} \\ y_2 &= x^{2} \\ y_3 &= x^{3} \\ \end{align*}

Solving for constants of integration using given initial conditions, the solution becomes

\begin{align*} y &= -\frac {1}{3 x}+\frac {x^{2}}{3} \\ \end{align*}

Figure 692: Solution plot
\(y = -\frac {1}{3 x}+\frac {x^{2}}{3}\)

17.4.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & \left [x^{3} \left (\frac {d^{3}}{d x^{3}}y \left (x \right )\right )-x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-2 x \left (\frac {d}{d x}y \left (x \right )\right )+6 y \left (x \right )=0, y \left (1\right )=0, \left (\frac {d}{d x}y \left (x \right )\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1, \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0\right ] \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )=-\frac {6 y \left (x \right )}{x^{3}}+\frac {\left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right ) x +2 \frac {d}{d x}y \left (x \right )}{x^{2}} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y \left (x \right )\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )-\frac {\frac {d^{2}}{d x^{2}}y \left (x \right )}{x}-\frac {2 \left (\frac {d}{d x}y \left (x \right )\right )}{x^{2}}+\frac {6 y \left (x \right )}{x^{3}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d^{3}}{d x^{3}}y \left (x \right )\right )-x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-2 x \left (\frac {d}{d x}y \left (x \right )\right )+6 y \left (x \right )=0 \\ \bullet & {} & \textrm {Make a change of variables}\hspace {3pt} \\ {} & {} & t =\ln \left (x \right ) \\ \square & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {1st}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\left (\frac {d}{d t}y \left (t \right )\right ) \left (\frac {d}{d x}t \left (x \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=\frac {\frac {d}{d t}y \left (t \right )}{x} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {2nd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) \left (\frac {d}{d x}t \left (x \right )\right )^{2}+\left (\frac {d^{2}}{d x^{2}}t \left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {3rd}\hspace {3pt} \hspace {3pt}\textrm {derivative of}\hspace {3pt} \hspace {3pt}\textrm {y}\hspace {3pt} \hspace {3pt}\textrm {with respect to}\hspace {3pt} \hspace {3pt}\textrm {x}\hspace {3pt} \hspace {3pt}\textrm {, using the chain rule}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )=\left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right ) \left (\frac {d}{d x}t \left (x \right )\right )^{3}+3 \left (\frac {d}{d x}t \left (x \right )\right ) \left (\frac {d^{2}}{d x^{2}}t \left (x \right )\right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+\left (\frac {d^{3}}{d x^{3}}t \left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d x^{3}}y \left (x \right )=\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{3}}+\frac {2 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{3}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{3}}+\frac {2 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{3}}\right )-x^{2} \left (\frac {\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right )-2 \frac {d}{d t}y \left (t \right )+6 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d t^{3}}y \left (t \right )-4 \frac {d^{2}}{d t^{2}}y \left (t \right )+\frac {d}{d t}y \left (t \right )+6 y \left (t \right )=0 \\ \square & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{1}\left (t \right ) \\ {} & {} & y_{1}\left (t \right )=y \left (t \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{2}\left (t \right ) \\ {} & {} & y_{2}\left (t \right )=\frac {d}{d t}y \left (t \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{3}\left (t \right ) \\ {} & {} & y_{3}\left (t \right )=\frac {d^{2}}{d t^{2}}y \left (t \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d t}y_{3}\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y_{3}\left (t \right )=4 y_{3}\left (t \right )-y_{2}\left (t \right )-6 y_{1}\left (t \right ) \\ & {} & \textrm {Convert linear ODE into a system of first order ODEs}\hspace {3pt} \\ {} & {} & \left [y_{2}\left (t \right )=\frac {d}{d t}y_{1}\left (t \right ), y_{3}\left (t \right )=\frac {d}{d t}y_{2}\left (t \right ), \frac {d}{d t}y_{3}\left (t \right )=4 y_{3}\left (t \right )-y_{2}\left (t \right )-6 y_{1}\left (t \right )\right ] \\ \bullet & {} & \textrm {Define vector}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{c} y_{1}\left (t \right ) \\ y_{2}\left (t \right ) \\ y_{3}\left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {System to solve}\hspace {3pt} \\ {} & {} & \frac {d}{d t}{\moverset {\rightarrow }{y}}\left (t \right )=\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -1 & 4 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{ccc} 0 & 1 & 0 \\ 0 & 0 & 1 \\ -6 & -1 & 4 \end {array}\right ] \\ \bullet & {} & \textrm {Rewrite the system as}\hspace {3pt} \\ {} & {} & \frac {d}{d t}{\moverset {\rightarrow }{y}}\left (t \right )=A \cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {To solve the system, find the eigenvalues and eigenvectors of}\hspace {3pt} A \\ \bullet & {} & \textrm {Eigenpairs of}\hspace {3pt} A \\ {} & {} & \left [\left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ], \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-t}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [2, \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [3, \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}={\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\mathit {C1} {\moverset {\rightarrow }{y}}_{1}+\mathit {C2} {\moverset {\rightarrow }{y}}_{2}+\mathit {C3} {\moverset {\rightarrow }{y}}_{3} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\mathit {C1} \,{\mathrm e}^{-t}\cdot \left [\begin {array}{c} 1 \\ -1 \\ 1 \end {array}\right ]+\mathit {C2} \,{\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ]+\mathit {C3} \,{\mathrm e}^{3 t}\cdot \left [\begin {array}{c} \frac {1}{9} \\ \frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\mathit {C1} \,{\mathrm e}^{-t}+\frac {\mathit {C2} \,{\mathrm e}^{2 t}}{4}+\frac {\mathit {C3} \,{\mathrm e}^{3 t}}{9} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {\mathit {C1}}{x}+\frac {\mathit {C2} \,x^{2}}{4}+\frac {\mathit {C3} \,x^{3}}{9} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} y \left (1\right )=0 \\ {} & {} & 0=\mathit {C1} +\frac {\mathit {C2}}{4}+\frac {\mathit {C3}}{9} \\ \bullet & {} & \textrm {Calculate the 1st derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y \left (x \right )=-\frac {\mathit {C1}}{x^{2}}+\frac {\mathit {C2} x}{2}+\frac {\mathit {C3} \,x^{2}}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d}{d x}y \left (x \right )\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=1 \\ {} & {} & 1=-\mathit {C1} +\frac {\mathit {C2}}{2}+\frac {\mathit {C3}}{3} \\ \bullet & {} & \textrm {Calculate the 2nd derivative of the solution}\hspace {3pt} \\ {} & {} & \frac {d^{2}}{d x^{2}}y \left (x \right )=\frac {2 \mathit {C1}}{x^{3}}+\frac {\mathit {C2}}{2}+\frac {2 \mathit {C3} x}{3} \\ \bullet & {} & \textrm {Use the initial condition}\hspace {3pt} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )\bigg | {\mstack {}{_{\left \{x \hiderel {=}1\right \}}}}=0 \\ {} & {} & 0=2 \mathit {C1} +\frac {\mathit {C2}}{2}+\frac {2 \mathit {C3}}{3} \\ \bullet & {} & \textrm {Solve for the unknown coefficients}\hspace {3pt} \\ {} & {} & \left \{\mathit {C1} =-\frac {1}{3}, \mathit {C2} =\frac {4}{3}, \mathit {C3} =0, x =x \right \} \\ \bullet & {} & \textrm {Solution to the IVP}\hspace {3pt} \\ {} & {} & y \left (x \right )=\frac {x^{3}-1}{3 x} \end {array} \]

17.4.3 Maple trace

`Methods for third order ODEs: 
--- Trying classification methods --- 
trying a quadrature 
checking if the LODE has constant coefficients 
checking if the LODE is of Euler type 
<- LODE of Euler type successful`
 

17.4.4 Maple dsolve solution

Solving time : 0.013 (sec)
Leaf size : 14

dsolve([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, 
       op([y(1) = 0, D(y)(1) = 1, (D@@2)(y)(1) = 0])],y(x),singsol=all)
 

\[ y = \frac {x^{3}-1}{3 x} \]

17.4.5 Mathematica DSolve solution

Solving time : 0.004 (sec)
Leaf size : 17

DSolve[{x^3*D[y[x],{x,3}]-x^2*D[y[x],{x,2}]-2*x*D[y[x],x]+6*y[x]==0,{y[1]==0,Derivative[1][y][1]==1,Derivative[2][y][1]==0}}, 
       y[x],x,IncludeSingularSolutions->True]
 

\[ y(x)\to \frac {x^3-1}{3 x} \]