Internal problem ID [14771]
Internal file name [OUTPUT/14451_Monday_April_08_2024_06_25_16_AM_3263092/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 (c).
ODE order: 4.
ODE degree: 1.
The type(s) of ODE detected by this program : "higher_order_ODE_non_constant_coefficients_of_type_Euler"
Maple gives the following as the ode type
[[_high_order, _exact, _linear, _homogeneous]]
\[ \boxed {x^{4} y^{\prime \prime \prime \prime }+14 x^{3} y^{\prime \prime \prime }+55 x^{2} y^{\prime \prime }+65 y^{\prime } x +15 y=0} \] 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}\\ 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 \[ x^{4} y^{\prime \prime \prime \prime }+14 x^{3} y^{\prime \prime \prime }+55 x^{2} y^{\prime \prime }+65 y^{\prime } x +15 y = 0 \] 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*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} 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}} \\ \end{align*}
Verification of solutions
\[ y = \frac {c_{1}}{x}+\frac {c_{2}}{x^{3}}+\frac {c_{3} \cos \left (\ln \left (x \right )\right )+c_{4} \sin \left (\ln \left (x \right )\right )}{x^{2}} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{4} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )+14 x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )+55 \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+65 y^{\prime } x +15 y=0 \\ \bullet & {} & \textrm {Highest derivative means the order of the ODE is}\hspace {3pt} 4 \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime } \\ \bullet & {} & \textrm {Isolate 4th derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }=-\frac {15 y}{x^{4}}-\frac {14 \left (\frac {d}{d x}y^{\prime \prime }\right ) x^{2}+55 \left (\frac {d}{d x}y^{\prime }\right ) x +65 y^{\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} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }+\frac {14 \left (\frac {d}{d x}y^{\prime \prime }\right )}{x}+\frac {55 \left (\frac {d}{d x}y^{\prime }\right )}{x^{2}}+\frac {65 y^{\prime }}{x^{3}}+\frac {15 y}{x^{4}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{4} \left (\frac {d}{d x}\frac {d}{d x}y^{\prime \prime }\right )+14 x^{3} \left (\frac {d}{d x}y^{\prime \prime }\right )+55 \left (\frac {d}{d x}y^{\prime }\right ) x^{2}+65 y^{\prime } x +15 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} \\ {} & {} & \frac {d}{d x}y^{\prime }=\left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{2}+\left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime }=\frac {\frac {d}{d t}\frac {d}{d t}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}{d x}y^{\prime \prime }=\left (\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{3}+3 t^{\prime }\left (x \right ) \left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )+\left (\frac {d}{d x}t^{\prime \prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}y^{\prime \prime }=\frac {\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d}{d t}\frac {d}{d t}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}{d x}\frac {d}{d x}y^{\prime \prime }=\left (\frac {d}{d t}\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) {t^{\prime }\left (x \right )}^{4}+3 {t^{\prime }\left (x \right )}^{2} \left (\frac {d}{d x}t^{\prime }\left (x \right )\right ) \left (\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )\right )+3 \left (\frac {d}{d x}t^{\prime }\left (x \right )\right )^{2} \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )+3 \left (\left (\frac {d}{d x}t^{\prime \prime }\left (x \right )\right ) \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )+\left (\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )\right ) t^{\prime }\left (x \right ) \left (\frac {d}{d x}t^{\prime }\left (x \right )\right )\right ) t^{\prime }\left (x \right )+\left (\frac {d}{d x}\frac {d}{d x}t^{\prime \prime }\left (x \right )\right ) \left (\frac {d}{d t}y \left (t \right )\right )+\left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right ) t^{\prime }\left (x \right ) \left (\frac {d}{d x}t^{\prime \prime }\left (x \right )\right ) \\ {} & \circ & \textrm {Compute derivative}\hspace {3pt} \\ {} & {} & \frac {d}{d x}\frac {d}{d x}y^{\prime \prime }=\frac {\frac {d}{d t}\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{4}}-\frac {3 \left (\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{4}}+\frac {5 \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )}{x^{4}}+\frac {3 \left (\frac {2 \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )}{x^{3}}-\frac {\frac {d}{d t}\frac {d^{2}}{d t^{2}}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}{d t}\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{4}}-\frac {3 \left (\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )\right )}{x^{4}}+\frac {5 \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )}{x^{4}}+\frac {3 \left (\frac {2 \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )}{x^{3}}-\frac {\frac {d}{d t}\frac {d^{2}}{d t^{2}}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}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )}{x^{3}}-\frac {3 \left (\frac {d}{d t}\frac {d}{d t}y \left (t \right )\right )}{x^{3}}+\frac {2 \left (\frac {d}{d t}y \left (t \right )\right )}{x^{3}}\right )+55 \left (\frac {\frac {d}{d t}\frac {d}{d t}y \left (t \right )}{x^{2}}-\frac {\frac {d}{d t}y \left (t \right )}{x^{2}}\right ) x^{2}+65 \frac {d}{d t}y \left (t \right )+15 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d}{d t}\frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )+8 \frac {d}{d t}\frac {d^{2}}{d t^{2}}y \left (t \right )+24 \frac {d}{d t}\frac {d}{d t}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}{d t}\frac {d}{d t}y \left (t \right ) \\ {} & \circ & \textrm {Define new variable}\hspace {3pt} y_{4}\left (t \right ) \\ {} & {} & y_{4}\left (t \right )=\frac {d}{d t}\frac {d^{2}}{d t^{2}}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}}=c_{1} {\moverset {\rightarrow }{y}}_{1}+c_{2} {\moverset {\rightarrow }{y}}_{2}+c_{3} {\moverset {\rightarrow }{y}}_{3}\left (t \right )+c_{4} {\moverset {\rightarrow }{y}}_{4}\left (t \right ) \\ \bullet & {} & \textrm {Substitute solutions into the general solution}\hspace {3pt} \\ {} & {} & {\moverset {\rightarrow }{y}}=c_{1} {\mathrm e}^{-3 t}\cdot \left [\begin {array}{c} -\frac {1}{27} \\ \frac {1}{9} \\ -\frac {1}{3} \\ 1 \end {array}\right ]+c_{2} {\mathrm e}^{-t}\cdot \left [\begin {array}{c} -1 \\ 1 \\ -1 \\ 1 \end {array}\right ]+c_{3} {\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 ]+c_{4} {\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 c_{3} +11 c_{4} \right ) \cos \left (t \right )+11 \sin \left (t \right ) \left (c_{3} +\frac {2 c_{4}}{11}\right )\right ) {\mathrm e}^{-2 t}}{125}-c_{2} {\mathrm e}^{-t}-\frac {c_{1} {\mathrm e}^{-3 t}}{27} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=\frac {\left (-2 c_{3} +11 c_{4} \right ) \cos \left (\ln \left (x \right )\right )+11 \sin \left (\ln \left (x \right )\right ) \left (c_{3} +\frac {2 c_{4}}{11}\right )}{125 x^{2}}-\frac {c_{2}}{x}-\frac {c_{1}}{27 x^{3}} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=-\frac {2 c_{3} \cos \left (\ln \left (x \right )\right )}{125 x^{2}}+\frac {11 c_{4} \cos \left (\ln \left (x \right )\right )}{125 x^{2}}+\frac {11 c_{3} \sin \left (\ln \left (x \right )\right )}{125 x^{2}}+\frac {2 c_{4} \sin \left (\ln \left (x \right )\right )}{125 x^{2}}-\frac {c_{2}}{x}-\frac {c_{1}}{27 x^{3}} \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`
✓ Solution by Maple
Time used: 0.0 (sec). Leaf size: 27
dsolve(x^4*diff(y(x),x$4)+14*x^3*diff(y(x),x$3)+55*x^2*diff(y(x),x$2)+65*x*diff(y(x),x)+15*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{3} \sin \left (\ln \left (x \right )\right ) x +c_{4} \cos \left (\ln \left (x \right )\right ) x +c_{2} x^{2}+c_{1}}{x^{3}} \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 32
DSolve[x^4*y''''[x]+14*x^3*y'''[x]+55*x^2*y''[x]+65*x*y'[x]+15*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_4 x^2+c_2 x \cos (\log (x))+c_1 x \sin (\log (x))+c_3}{x^3} \]