problem 20

Internal problem ID [14728]
Internal ﬁle name [OUTPUT/14408_Wednesday_April_03_2024_02_18_13_PM_70059996/index.tex]

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: 20.
ODE order: 4.
ODE degree: 1.

\[ \boxed {x^{3} y^{\prime \prime \prime \prime }+6 x^{2} y^{\prime \prime \prime }+7 x y^{\prime \prime }+y^{\prime }=0} \] Since \(y\) is missing from the ode then we can use the substitution \(y^{\prime } = v \left (x \right )\) to reduce the order by one. The ODE becomes \begin {align*} x^{3} v^{\prime \prime \prime }\left (x \right )+6 x^{2} v^{\prime \prime }\left (x \right )+7 v^{\prime }\left (x \right ) x +v \left (x \right ) = 0 \end {align*}

This is Euler ODE of higher order. Let \(v \left (x \right ) = x^{\lambda }\). Hence \begin {align*} v^{\prime }\left (x \right ) &= \lambda \,x^{\lambda -1}\\ v^{\prime \prime }\left (x \right ) &= \lambda \left (\lambda -1\right ) x^{\lambda -2}\\ v^{\prime \prime \prime }\left (x \right ) &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3} \end {align*}

Substituting these back into \[ x^{3} v^{\prime \prime \prime }\left (x \right )+6 x^{2} v^{\prime \prime }\left (x \right )+7 v^{\prime }\left (x \right ) x +v \left (x \right ) = 0 \] gives \[ 7 x \lambda \,x^{\lambda -1}+6 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+x^{\lambda } = 0 \] Which simpliﬁes to \[ 7 \lambda \,x^{\lambda }+6 \lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes

\[ 7 \lambda +6 \lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )+1 = 0 \] Simplifying gives the characteristic equation as \[ \left (\lambda +1\right )^{3} = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= -1\\ \lambda _2 &= -1\\ \lambda _3 &= -1 \end {align*}


root	multiplicity	type of root

\(-1\)	\(3\)	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

\[ v \left (x \right ) = \frac {c_{1}}{x}+\frac {c_{2} \ln \left (x \right )}{x}+\frac {c_{3} \ln \left (x \right )^{2}}{x} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} v_1 &= \frac {1}{x}\\ v_2 &= \frac {\ln \left (x \right )}{x}\\ v_3 &= \frac {\ln \left (x \right )^{2}}{x} \end {align*}

But since \(y^{\prime } = v \left (x \right )\) then we now need to solve the ode \(y^{\prime } = \frac {c_{1}}{x}+\frac {c_{2} \ln \left (x \right )}{x}+\frac {c_{3} \ln \left (x \right )^{2}}{x}\). Integrating both sides gives \begin {align*} y &= \int { \frac {c_{3} \ln \left (x \right )^{2}+c_{2} \ln \left (x \right )+c_{1}}{x}\,\mathop {\mathrm {d}x}}\\ &= \frac {\ln \left (x \right )^{3} c_{3}}{3}+\frac {\ln \left (x \right )^{2} c_{2}}{2}+c_{1} \ln \left (x \right )+c_{1} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {\ln \left (x \right )^{3} c_{3}}{3}+\frac {\ln \left (x \right )^{2} c_{2}}{2}+c_{1} \ln \left (x \right )+c_{1} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {\ln \left (x \right )^{3} c_{3}}{3}+\frac {\ln \left (x \right )^{2} c_{2}}{2}+c_{1} \ln \left (x \right )+c_{1} \] Veriﬁed OK.

15.20.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{3} y^{\prime \prime \prime \prime }+6 x^{2} y^{\prime \prime \prime }+7 y^{\prime \prime } x +y^{\prime }=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & y^{\prime \prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=-\frac {y^{\prime }+7 y^{\prime \prime } x +6 x^{2} y^{\prime \prime \prime }}{x^{3}} \\ \bullet & {} & \textrm {Group terms with}\hspace {3pt} y\hspace {3pt}\textrm {on the lhs of the ODE and the rest on the rhs of the ODE; ODE is linear}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }+\frac {6 y^{\prime \prime \prime }}{x}+\frac {7 y^{\prime \prime }}{x^{2}}+\frac {y^{\prime }}{x^{3}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{3} y^{\prime \prime \prime \prime }+6 x^{2} y^{\prime \prime \prime }+7 y^{\prime \prime } x +y^{\prime }=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} \\ {} & {} & y^{\prime }=\left (\frac {d}{d t}y \left (t \right )\right ) t^{\prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime }=\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} \\ {} & {} & y^{\prime \prime }=\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+t^{\prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime }=\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} \\ {} & {} & y^{\prime \prime \prime }=\left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{3}+3 t^{\prime }\left (x \right ) t^{\prime \prime }\left (x \right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+t^{\prime \prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=\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}} \\ {} & \circ & \textrm {Calculate the}\hspace {3pt} \hspace {3pt}\textrm {4th}\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} \\ {} & {} & y^{\prime \prime \prime \prime }=\left (\frac {d^{4}}{d t^{4}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{4}+3 {t^{\prime }\left (x \right )}^{2} t^{\prime \prime }\left (x \right ) \left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right )+3 {t^{\prime \prime }\left (x \right )}^{2} \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+3 \left (t^{\prime \prime \prime }\left (x \right ) \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+\left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right ) t^{\prime }\left (x \right ) t^{\prime \prime }\left (x \right )\right ) t^{\prime }\left (x \right )+t^{\prime \prime \prime \prime }\left (x \right ) \left (\frac {d}{d t}y \left (t \right )\right )+\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t^{\prime }\left (x \right ) t^{\prime \prime \prime }\left (x \right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime \prime }=\frac {\frac {d^{4}}{d t^{4}}y \left (t \right )}{x^{4}}-\frac {3 \left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right )}{x^{4}}+\frac {5 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{4}}+\frac {3 \left (\frac {2 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{3}}-\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}\right )}{x}-\frac {6 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{4}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & x^{3} \left (\frac {\frac {d^{4}}{d t^{4}}y \left (t \right )}{x^{4}}-\frac {3 \left (\frac {d^{3}}{d t^{3}}y \left (t \right )\right )}{x^{4}}+\frac {5 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{4}}+\frac {3 \left (\frac {2 \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{3}}-\frac {\frac {d^{3}}{d t^{3}}y \left (t \right )}{x^{3}}\right )}{x}-\frac {6 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{4}}\right )+6 x^{2} \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 )+7 \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 ) x +\frac {\frac {d}{d t}y \left (t \right )}{x}=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {\frac {d^{4}}{d t^{4}}y \left (t \right )}{x}=0 \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d t^{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 {Define new variable}\hspace {3pt} y_{4}\left (t \right ) \\ {} & {} & y_{4}\left (t \right )=\frac {d^{3}}{d t^{3}}y \left (t \right ) \\ {} & \circ & \textrm {Isolate for}\hspace {3pt} \frac {d}{d t}y_{4}\left (t \right )\hspace {3pt}\textrm {using original ODE}\hspace {3pt} \\ {} & {} & \frac {d}{d t}y_{4}\left (t \right )=0 \\ & {} & \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 ), y_{4}\left (t \right )=\frac {d}{d t}y_{3}\left (t \right ), \frac {d}{d t}y_{4}\left (t \right )=0\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 ) \\ y_{4}\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}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \end {array}\right ]\cdot {\moverset {\rightarrow }{y}}\left (t \right ) \\ \bullet & {} & \textrm {Define the coefficient matrix}\hspace {3pt} \\ {} & {} & A =\left [\begin {array}{cccc} 0 & 1 & 0 & 0 \\ 0 & 0 & 1 & 0 \\ 0 & 0 & 0 & 1 \\ 0 & 0 & 0 & 0 \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 [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ], \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}=\left [\begin {array}{c} 1 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{2}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{3}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [0, \left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{4}=\left [\begin {array}{c} 0 \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {General solution to the system of ODEs}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}+c_{4} {\moverset {\rightarrow }{y}}_{4} \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\left [\begin {array}{c} c_{1} \\ 0 \\ 0 \\ 0 \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=c_{1} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=c_{1} \end {array} \]

Maple trace

`Methods for high 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`

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

\[ y(x)\to \frac {1}{3} c_3 \log ^3(x)+\frac {1}{2} c_2 \log ^2(x)+c_1 \log (x)+c_4 \]

15.20 problem 20

15.20.1 Maple step by step solution