problem 64 (a)

Internal problem ID [14769]
Internal ﬁle name [OUTPUT/14449_Monday_April_08_2024_06_25_15_AM_12591752/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: 64 (a).
ODE order: 3.
ODE degree: 1.

The type(s) of ODE detected by this program : "higher_order_ODE_non_constant_coeﬃcients_of_type_Euler"

\[ \boxed {x^{3} y^{\prime \prime \prime }+16 x^{2} y^{\prime \prime }+79 y^{\prime } x +125 y=0} \] 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 \[ x^{3} y^{\prime \prime \prime }+16 x^{2} y^{\prime \prime }+79 y^{\prime } x +125 y = 0 \] gives \[ 79 x \lambda \,x^{\lambda -1}+16 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+125 x^{\lambda } = 0 \] Which simpliﬁes to \[ 79 \lambda \,x^{\lambda }+16 \lambda \left (\lambda -1\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+125 x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes

\[ 79 \lambda +16 \lambda \left (\lambda -1\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right )+125 = 0 \] Simplifying gives the characteristic equation as \[ \lambda ^{3}+13 \lambda ^{2}+65 \lambda +125 = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= -5\\ \lambda _2 &= -4-3 i\\ \lambda _3 &= -4+3 i \end {align*}


root	multiplicity	type of root

\(-5\)	\(1\)	real root

\(-4 \pm 3 i\)	\(1\)	complex conjugate 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

\[ y = \frac {c_{1}}{x^{5}}+\frac {c_{2} \cos \left (3 \ln \left (x \right )\right )+c_{3} \sin \left (3 \ln \left (x \right )\right )}{x^{4}} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= \frac {1}{x^{5}}\\ y_2 &= \frac {\cos \left (3 \ln \left (x \right )\right )}{x^{4}}\\ y_3 &= \frac {\sin \left (3 \ln \left (x \right )\right )}{x^{4}} \end {align*}

Summary

The solution(s) found are the following \begin{align*} \tag{1} y &= \frac {c_{1}}{x^{5}}+\frac {c_{2} \cos \left (3 \ln \left (x \right )\right )+c_{3} \sin \left (3 \ln \left (x \right )\right )}{x^{4}} \\ \end{align*}

Veriﬁcation of solutions

\[ y = \frac {c_{1}}{x^{5}}+\frac {c_{2} \cos \left (3 \ln \left (x \right )\right )+c_{3} \sin \left (3 \ln \left (x \right )\right )}{x^{4}} \] Veriﬁed OK.

15.61.1 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime } x^{3}+16 x^{2} y^{\prime \prime }+79 y^{\prime } x +125 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 3 \\ {} & {} & y^{\prime \prime \prime } \\ \bullet & {} & \textrm {Isolate 3rd derivative}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime }=-\frac {125 y}{x^{3}}-\frac {16 y^{\prime \prime } x +79 y^{\prime }}{x^{2}} \\ \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 }+\frac {16 y^{\prime \prime }}{x}+\frac {79 y^{\prime }}{x^{2}}+\frac {125 y}{x^{3}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & y^{\prime \prime \prime } x^{3}+16 x^{2} y^{\prime \prime }+79 y^{\prime } x +125 y=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}} \\ & {} & \textrm {Substitute the change of variables back into the ODE}\hspace {3pt} \\ {} & {} & \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^{3}+16 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 )+79 \frac {d}{d t}y \left (t \right )+125 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{3}}{d t^{3}}y \left (t \right )+13 \frac {d^{2}}{d t^{2}}y \left (t \right )+65 \frac {d}{d t}y \left (t \right )+125 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 )=-13 y_{3}\left (t \right )-65 y_{2}\left (t \right )-125 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 )=-13 y_{3}\left (t \right )-65 y_{2}\left (t \right )-125 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 \\ -125 & -65 & -13 \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 \\ -125 & -65 & -13 \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 [-5, \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]\right ], \left [-4-3 \,\mathrm {I}, \left [\begin {array}{c} \frac {7}{625}-\frac {24 \,\mathrm {I}}{625} \\ -\frac {4}{25}+\frac {3 \,\mathrm {I}}{25} \\ 1 \end {array}\right ]\right ], \left [-4+3 \,\mathrm {I}, \left [\begin {array}{c} \frac {7}{625}+\frac {24 \,\mathrm {I}}{625} \\ -\frac {4}{25}-\frac {3 \,\mathrm {I}}{25} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-5, \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-5 t}\cdot \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-4-3 \,\mathrm {I}, \left [\begin {array}{c} \frac {7}{625}-\frac {24 \,\mathrm {I}}{625} \\ -\frac {4}{25}+\frac {3 \,\mathrm {I}}{25} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-4-3 \,\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} \frac {7}{625}-\frac {24 \,\mathrm {I}}{625} \\ -\frac {4}{25}+\frac {3 \,\mathrm {I}}{25} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Use Euler identity to write solution in terms of}\hspace {3pt} \sin \hspace {3pt}\textrm {and}\hspace {3pt} \cos \\ {} & {} & {\mathrm e}^{-4 t}\cdot \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right )\cdot \left [\begin {array}{c} \frac {7}{625}-\frac {24 \,\mathrm {I}}{625} \\ -\frac {4}{25}+\frac {3 \,\mathrm {I}}{25} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-4 t}\cdot \left [\begin {array}{c} \left (\frac {7}{625}-\frac {24 \,\mathrm {I}}{625}\right ) \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right ) \\ \left (-\frac {4}{25}+\frac {3 \,\mathrm {I}}{25}\right ) \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right ) \\ \cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{2}\left (t \right )={\mathrm e}^{-4 t}\cdot \left [\begin {array}{c} \frac {7 \cos \left (3 t \right )}{625}-\frac {24 \sin \left (3 t \right )}{625} \\ -\frac {4 \cos \left (3 t \right )}{25}+\frac {3 \sin \left (3 t \right )}{25} \\ \cos \left (3 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{-4 t}\cdot \left [\begin {array}{c} -\frac {7 \sin \left (3 t \right )}{625}-\frac {24 \cos \left (3 t \right )}{625} \\ \frac {4 \sin \left (3 t \right )}{25}+\frac {3 \cos \left (3 t \right )}{25} \\ -\sin \left (3 t \right ) \end {array}\right ]\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}\left (t \right )+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-5 t}\cdot \left [\begin {array}{c} \frac {1}{25} \\ -\frac {1}{5} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-4 t}\cdot \left [\begin {array}{c} \frac {7 \cos \left (3 t \right )}{625}-\frac {24 \sin \left (3 t \right )}{625} \\ -\frac {4 \cos \left (3 t \right )}{25}+\frac {3 \sin \left (3 t \right )}{25} \\ \cos \left (3 t \right ) \end {array}\right ]+c_{3} {\mathrm e}^{-4 t}\cdot \left [\begin {array}{c} -\frac {7 \sin \left (3 t \right )}{625}-\frac {24 \cos \left (3 t \right )}{625} \\ \frac {4 \sin \left (3 t \right )}{25}+\frac {3 \cos \left (3 t \right )}{25} \\ -\sin \left (3 t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {First component of the vector is the solution to the ODE}\hspace {3pt} \\ {} & {} & y \left (t \right )=\frac {\left (\left (7 c_{2} -24 c_{3} \right ) \cos \left (3 t \right )-24 \sin \left (3 t \right ) \left (c_{2} +\frac {7 c_{3}}{24}\right )\right ) {\mathrm e}^{-4 t}}{625}+\frac {c_{1} {\mathrm e}^{-5 t}}{25} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=\frac {\left (7 c_{2} -24 c_{3} \right ) \cos \left (3 \ln \left (x \right )\right )-24 \sin \left (3 \ln \left (x \right )\right ) \left (c_{2} +\frac {7 c_{3}}{24}\right )}{625 x^{4}}+\frac {c_{1}}{25 x^{5}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=\frac {7 \cos \left (3 \ln \left (x \right )\right ) c_{2}}{625 x^{4}}-\frac {24 \cos \left (3 \ln \left (x \right )\right ) c_{3}}{625 x^{4}}-\frac {24 \sin \left (3 \ln \left (x \right )\right ) c_{2}}{625 x^{4}}-\frac {7 \sin \left (3 \ln \left (x \right )\right ) c_{3}}{625 x^{4}}+\frac {c_{1}}{25 x^{5}} \end {array} \]

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`

\[ y \left (x \right ) = \frac {c_{2} \sin \left (3 \ln \left (x \right )\right ) x +c_{3} \cos \left (3 \ln \left (x \right )\right ) x +c_{1}}{x^{5}} \]

15.61 problem 64 (a)

15.61.1 Maple step by step solution