15.15 problem 15

Internal problem ID [15362]
Book : INTRODUCTORY DIFFERENTIAL EQUATIONS. Martha L. Abell, James P. Braselton. Fourth edition 2014. ElScAe. 2014
Section : Chapter 4. Higher Order Equations. Exercises 4.7, page 195
Problem number : 15
Date solved : Friday, October 18, 2024 at 01:07:35 PM
CAS classification : [[_3rd_order, _with_linear_symmetries]]

Solve

\begin{align*} x^{3} y^{\prime \prime \prime }+2 x^{2} y^{\prime \prime }-4 x y^{\prime }+4 y&=0 \end{align*}

15.15.1 Solved as higher order Euler type ode

Time used: 0.101 (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

\[ x^{3} y^{\prime \prime \prime }+2 x^{2} y^{\prime \prime }-4 x y^{\prime }+4 y = 0 \]

gives

\[ -4 x \lambda \,x^{\lambda -1}+2 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+4 x^{\lambda } = 0 \]

Which simplifies to

\[ -4 \lambda \,x^{\lambda }+2 \lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+4 x^{\lambda } = 0 \]

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

\[ -4 \lambda +2 \lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )+4 = 0 \]

Simplifying gives the characteristic equation as

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

Solving the above gives the following roots

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

This table summarises the result

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

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

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

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

15.15.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {d^{3}}{d x^{3}}y \left (x \right )\right )+2 x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-4 x \left (\frac {d}{d x}y \left (x \right )\right )+4 y \left (x \right )=0 \\ \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 {4 y \left (x \right )}{x^{3}}+\frac {2 \left (-x \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+2 \frac {d}{d x}y \left (x \right )\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 {2 \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )}{x}-\frac {4 \left (\frac {d}{d x}y \left (x \right )\right )}{x^{2}}+\frac {4 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 )+2 x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )-4 x \left (\frac {d}{d x}y \left (x \right )\right )+4 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 )+2 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 )-4 \frac {d}{d t}y \left (t \right )+4 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d t^{3}}y \left (t \right )-\frac {d^{2}}{d t^{2}}y \left (t \right )-4 \frac {d}{d t}y \left (t \right )+4 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 )=y_{3}\left (t \right )+4 y_{2}\left (t \right )-4 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 )=y_{3}\left (t \right )+4 y_{2}\left (t \right )-4 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 \\ -4 & 4 & 1 \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 \\ -4 & 4 & 1 \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 [-2, \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]\right ], \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 ]\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}}_{1}={\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 [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}}_{2}={\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}}_{3}={\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 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}^{-2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ -\frac {1}{2} \\ 1 \end {array}\right ]+\mathit {C2} \,{\mathrm e}^{t}\cdot \left [\begin {array}{c} 1 \\ 1 \\ 1 \end {array}\right ]+\mathit {C3} \,{\mathrm e}^{2 t}\cdot \left [\begin {array}{c} \frac {1}{4} \\ \frac {1}{2} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {\left (\mathit {C3} \,{\mathrm e}^{4 t}+4 \mathit {C2} \,{\mathrm e}^{3 t}+\mathit {C1} \right ) {\mathrm e}^{-2 t}}{4} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {\mathit {C3} \,x^{4}+4 \mathit {C2} \,x^{3}+\mathit {C1}}{4 x^{2}} \end {array} \]

15.15.3 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`
 

15.15.4 Maple dsolve solution

Solving time : 0.004 (sec)
Leaf size : 20

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

\[ y \left (x \right ) = \frac {c_{2} x^{4}+c_{1} x^{3}+c_{3}}{x^{2}} \]

15.15.5 Mathematica DSolve solution

Solving time : 0.004 (sec)
Leaf size : 22

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

\[ y(x)\to c_3 x^2+\frac {c_1}{x^2}+c_2 x \]