15.63 problem 64 (c)

Internal problem ID [15410]
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 (c)
Date solved : Friday, October 18, 2024 at 01:36:43 PM
CAS classification : [[_high_order, _exact, _linear, _homogeneous]]

Solve

\begin{align*} x^{4} y^{\prime \prime \prime \prime }+14 x^{3} y^{\prime \prime \prime }+55 x^{2} y^{\prime \prime }+65 x y^{\prime }+15 y&=0 \end{align*}

15.63.1 Solved as higher order Euler type ode

Time used: 0.125 (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}\\ y^{\prime \prime \prime \prime } &= \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda -4} \end{align*}

Substituting these back into

\[ x^{4} y^{\prime \prime \prime \prime }+14 x^{3} y^{\prime \prime \prime }+55 x^{2} y^{\prime \prime }+65 x y^{\prime }+15 y = 0 \]

gives

\[ 65 x \lambda \,x^{\lambda -1}+55 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+14 x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+x^{4} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda -4}+15 x^{\lambda } = 0 \]

Which simplifies to

\[ 65 \lambda \,x^{\lambda }+55 \lambda \left (\lambda -1\right ) x^{\lambda }+14 \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda }+15 x^{\lambda } = 0 \]

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

\[ 65 \lambda +55 \lambda \left (\lambda -1\right )+14 \lambda \left (\lambda -1\right ) \left (\lambda -2\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right )+15 = 0 \]

Simplifying gives the characteristic equation as

\[ \lambda ^{4}+8 \lambda ^{3}+24 \lambda ^{2}+32 \lambda +15 = 0 \]

Solving the above gives the following roots

\begin{align*} \lambda _1 &= -1\\ \lambda _2 &= -3\\ \lambda _3 &= -2+i\\ \lambda _4 &= -2-i \end{align*}

This table summarises the result

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

\[ y = \frac {c_1}{x}+\frac {c_2}{x^{3}}+\frac {c_3 \cos \left (\ln \left (x \right )\right )+c_4 \sin \left (\ln \left (x \right )\right )}{x^{2}} \]

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

\begin{align*} y_1 &= \frac {1}{x} \\ y_2 &= \frac {1}{x^{3}} \\ y_3 &= \frac {\cos \left (\ln \left (x \right )\right )}{x^{2}} \\ y_4 &= \frac {\sin \left (\ln \left (x \right )\right )}{x^{2}} \\ \end{align*}

15.63.2 Maple step by step solution

\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{4} \left (\frac {d^{4}}{d x^{4}}y \left (x \right )\right )+14 x^{3} \left (\frac {d^{3}}{d x^{3}}y \left (x \right )\right )+55 x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+65 x \left (\frac {d}{d x}y \left (x \right )\right )+15 y \left (x \right )=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right ) \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right )=-\frac {15 y \left (x \right )}{x^{4}}-\frac {14 x^{2} \left (\frac {d^{3}}{d x^{3}}y \left (x \right )\right )+55 x \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+65 \frac {d}{d x}y \left (x \right )}{x^{3}} \\ \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^{4}}{d x^{4}}y \left (x \right )+\frac {14 \left (\frac {d^{3}}{d x^{3}}y \left (x \right )\right )}{x}+\frac {55 \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )}{x^{2}}+\frac {65 \left (\frac {d}{d x}y \left (x \right )\right )}{x^{3}}+\frac {15 y \left (x \right )}{x^{4}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{4} \left (\frac {d^{4}}{d x^{4}}y \left (x \right )\right )+14 x^{3} \left (\frac {d^{3}}{d x^{3}}y \left (x \right )\right )+55 x^{2} \left (\frac {d^{2}}{d x^{2}}y \left (x \right )\right )+65 x \left (\frac {d}{d x}y \left (x \right )\right )+15 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}} \\ {} & \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} \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right )=\left (\frac {d^{4}}{d t^{4}}y \left (t \right )\right ) \left (\frac {d}{d x}t \left (x \right )\right )^{4}+3 \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^{3}}{d t^{3}}y \left (t \right )\right )+3 \left (\frac {d^{2}}{d x^{2}}t \left (x \right )\right )^{2} \left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+3 \left (\left (\frac {d^{3}}{d x^{3}}t \left (x \right )\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 ) \left (\frac {d}{d x}t \left (x \right )\right ) \left (\frac {d^{2}}{d x^{2}}t \left (x \right )\right )\right ) \left (\frac {d}{d x}t \left (x \right )\right )+\left (\frac {d^{4}}{d x^{4}}t \left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right )+\left (\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) \left (\frac {d}{d x}t \left (x \right )\right ) \left (\frac {d^{3}}{d x^{3}}t \left (x \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d x^{4}}y \left (x \right )=\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^{4} \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 )+14 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 )+55 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 )+65 \frac {d}{d t}y \left (t \right )+15 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d t^{4}}y \left (t \right )+8 \frac {d^{3}}{d t^{3}}y \left (t \right )+24 \frac {d^{2}}{d t^{2}}y \left (t \right )+32 \frac {d}{d t}y \left (t \right )+15 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 )=-8 y_{4}\left (t \right )-24 y_{3}\left (t \right )-32 y_{2}\left (t \right )-15 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 ), y_{4}\left (t \right )=\frac {d}{d t}y_{3}\left (t \right ), \frac {d}{d t}y_{4}\left (t \right )=-8 y_{4}\left (t \right )-24 y_{3}\left (t \right )-32 y_{2}\left (t \right )-15 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 ) \\ 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 \\ -15 & -32 & -24 & -8 \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 \\ -15 & -32 & -24 & -8 \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 [-3, \left [\begin {array}{c} -\frac {1}{27} \\ \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ]\right ], \left [-1, \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]\right ], \left [-2-\mathrm {I}, \left [\begin {array}{c} -\frac {2}{125}+\frac {11 \,\mathrm {I}}{125} \\ \frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}+\frac {\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ], \left [-2+\mathrm {I}, \left [\begin {array}{c} -\frac {2}{125}-\frac {11 \,\mathrm {I}}{125} \\ \frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}-\frac {\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-3, \left [\begin {array}{c} -\frac {1}{27} \\ \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution to homogeneous system from eigenpair}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}_{1}={\mathrm e}^{-3 t}\cdot \left [\begin {array}{c} -\frac {1}{27} \\ \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider eigenpair}\hspace {3pt} \\ {} & {} & \left [-1, \left [\begin {array}{c} -1 \\ 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 \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-2-\mathrm {I}, \left [\begin {array}{c} -\frac {2}{125}+\frac {11 \,\mathrm {I}}{125} \\ \frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}+\frac {\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-2-\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} -\frac {2}{125}+\frac {11 \,\mathrm {I}}{125} \\ \frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}+\frac {\mathrm {I}}{5} \\ 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}^{-2 t}\cdot \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right )\cdot \left [\begin {array}{c} -\frac {2}{125}+\frac {11 \,\mathrm {I}}{125} \\ \frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {2}{5}+\frac {\mathrm {I}}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \left (-\frac {2}{125}+\frac {11 \,\mathrm {I}}{125}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \left (\frac {3}{25}-\frac {4 \,\mathrm {I}}{25}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \left (-\frac {2}{5}+\frac {\mathrm {I}}{5}\right ) \left (\cos \left (t \right )-\mathrm {I} \sin \left (t \right )\right ) \\ \cos \left (t \right )-\mathrm {I} \sin \left (t \right ) \end {array}\right ] \\ \bullet & {} & \textrm {Both real and imaginary parts are solutions to the homogeneous system}\hspace {3pt} \\ {} & {} & \left [{\moverset {\rightarrow }{y}}_{3}\left (t \right )={\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {2 \cos \left (t \right )}{125}+\frac {11 \sin \left (t \right )}{125} \\ \frac {3 \cos \left (t \right )}{25}-\frac {4 \sin \left (t \right )}{25} \\ -\frac {2 \cos \left (t \right )}{5}+\frac {\sin \left (t \right )}{5} \\ \cos \left (t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (t \right )={\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \frac {2 \sin \left (t \right )}{125}+\frac {11 \cos \left (t \right )}{125} \\ -\frac {3 \sin \left (t \right )}{25}-\frac {4 \cos \left (t \right )}{25} \\ \frac {2 \sin \left (t \right )}{5}+\frac {\cos \left (t \right )}{5} \\ -\sin \left (t \right ) \end {array}\right ]\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}\left (t \right )+\mathit {C4} {\moverset {\rightarrow }{y}}_{4}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=\mathit {C1} \,{\mathrm e}^{-3 t}\cdot \left [\begin {array}{c} -\frac {1}{27} \\ \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ]+\mathit {C2} \,{\mathrm e}^{-t}\cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+\mathit {C3} \,{\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} -\frac {2 \cos \left (t \right )}{125}+\frac {11 \sin \left (t \right )}{125} \\ \frac {3 \cos \left (t \right )}{25}-\frac {4 \sin \left (t \right )}{25} \\ -\frac {2 \cos \left (t \right )}{5}+\frac {\sin \left (t \right )}{5} \\ \cos \left (t \right ) \end {array}\right ]+\mathit {C4} \,{\mathrm e}^{-2 t}\cdot \left [\begin {array}{c} \frac {2 \sin \left (t \right )}{125}+\frac {11 \cos \left (t \right )}{125} \\ -\frac {3 \sin \left (t \right )}{25}-\frac {4 \cos \left (t \right )}{25} \\ \frac {2 \sin \left (t \right )}{5}+\frac {\cos \left (t \right )}{5} \\ -\sin \left (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 (-2 \mathit {C3} +11 \mathit {C4} \right ) \cos \left (t \right )+11 \sin \left (t \right ) \left (\mathit {C3} +\frac {2 \mathit {C4}}{11}\right )\right ) {\mathrm e}^{-2 t}}{125}-\mathit {C2} \,{\mathrm e}^{-t}-\frac {\mathit {C1} \,{\mathrm e}^{-3 t}}{27} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y \left (x \right )=\frac {\left (-2 \mathit {C3} +11 \mathit {C4} \right ) \cos \left (\ln \left (x \right )\right )+11 \sin \left (\ln \left (x \right )\right ) \left (\mathit {C3} +\frac {2 \mathit {C4}}{11}\right )}{125 x^{2}}-\frac {\mathit {C2}}{x}-\frac {\mathit {C1}}{27 x^{3}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y \left (x \right )=-\frac {2 \cos \left (\ln \left (x \right )\right ) \mathit {C3}}{125 x^{2}}+\frac {11 \cos \left (\ln \left (x \right )\right ) \mathit {C4}}{125 x^{2}}+\frac {11 \sin \left (\ln \left (x \right )\right ) \mathit {C3}}{125 x^{2}}+\frac {2 \sin \left (\ln \left (x \right )\right ) \mathit {C4}}{125 x^{2}}-\frac {\mathit {C2}}{x}-\frac {\mathit {C1}}{27 x^{3}} \end {array} \]

15.63.3 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`
 

15.63.4 Maple dsolve solution

Solving time : 0.005 (sec)
Leaf size : 27

dsolve(x^4*diff(diff(diff(diff(y(x),x),x),x),x)+14*x^3*diff(diff(diff(y(x),x),x),x)+55*x^2*diff(diff(y(x),x),x)+65*x*diff(y(x),x)+15*y(x) = 0, 
       y(x),singsol=all)
 

\[ y \left (x \right ) = \frac {c_{3} x \sin \left (\ln \left (x \right )\right )+c_4 \cos \left (\ln \left (x \right )\right ) x +c_{1} x^{2}+c_{2}}{x^{3}} \]

15.63.5 Mathematica DSolve solution

Solving time : 0.005 (sec)
Leaf size : 32

DSolve[{x^4*D[y[x],{x,4}]+14*x^3*D[y[x],{x,3}]+55*x^2*D[y[x],{x,2}]+65*x*D[y[x],x]+15*y[x]==0,{}}, 
       y[x],x,IncludeSingularSolutions->True]
 

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