Internal problem ID [14772]
Internal file name [OUTPUT/14452_Monday_April_08_2024_06_25_16_AM_42494436/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 (d).
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, _with_linear_symmetries]]
\[ \boxed {x^{4} y^{\prime \prime \prime \prime }+8 x^{3} y^{\prime \prime \prime }+27 x^{2} y^{\prime \prime }+35 y^{\prime } x +45 y=0} \] 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 \[ x^{4} y^{\prime \prime \prime \prime }+8 x^{3} y^{\prime \prime \prime }+27 x^{2} y^{\prime \prime }+35 y^{\prime } x +45 y = 0 \] gives \[ 35 x \lambda \,x^{\lambda -1}+27 x^{2} \lambda \left (\lambda -1\right ) x^{\lambda -2}+8 x^{3} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda -3}+x^{4} \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda -4}+45 x^{\lambda } = 0 \] Which simplifies to \[ 35 \lambda \,x^{\lambda }+27 \lambda \left (\lambda -1\right ) x^{\lambda }+8 \lambda \left (\lambda -1\right ) \left (\lambda -2\right ) x^{\lambda }+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right ) x^{\lambda }+45 x^{\lambda } = 0 \] And since \(x^{\lambda }\neq 0\) then dividing through by \(x^{\lambda }\), the above becomes
\[ 35 \lambda +27 \lambda \left (\lambda -1\right )+8 \lambda \left (\lambda -1\right ) \left (\lambda -2\right )+\lambda \left (\lambda -1\right ) \left (\lambda -2\right ) \left (\lambda -3\right )+45 = 0 \] Simplifying gives the characteristic equation as \[ \left (\lambda ^{2}+2 \lambda +5\right ) \left (\lambda ^{2}+9\right ) = 0 \] Solving the above gives the following roots \begin {align*} \lambda _1 &= 3 i\\ \lambda _2 &= -3 i\\ \lambda _3 &= -1+2 i\\ \lambda _4 &= -1-2 i \end {align*}
This table summarises the result
| root | multiplicity | type of root |
| \( \pm 3 i\) | \(1\) | complex conjugate root |
| \(-1 \pm 2 i\) | \(1\) | 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
\[ y = c_{1} \cos \left (3 \ln \left (x \right )\right )+c_{2} \sin \left (3 \ln \left (x \right )\right )+\frac {c_{3} \cos \left (2 \ln \left (x \right )\right )+c_{4} \sin \left (2 \ln \left (x \right )\right )}{x} \] The fundamental set of solutions for the homogeneous solution are the following \begin {align*} y_1 &= \cos \left (3 \ln \left (x \right )\right )\\ y_2 &= \sin \left (3 \ln \left (x \right )\right )\\ y_3 &= \frac {\cos \left (2 \ln \left (x \right )\right )}{x}\\ y_4 &= \frac {\sin \left (2 \ln \left (x \right )\right )}{x} \end {align*}
Summary
The solution(s) found are the following \begin{align*} \tag{1} y &= c_{1} \cos \left (3 \ln \left (x \right )\right )+c_{2} \sin \left (3 \ln \left (x \right )\right )+\frac {c_{3} \cos \left (2 \ln \left (x \right )\right )+c_{4} \sin \left (2 \ln \left (x \right )\right )}{x} \\ \end{align*}
Verification of solutions
\[ y = c_{1} \cos \left (3 \ln \left (x \right )\right )+c_{2} \sin \left (3 \ln \left (x \right )\right )+\frac {c_{3} \cos \left (2 \ln \left (x \right )\right )+c_{4} \sin \left (2 \ln \left (x \right )\right )}{x} \] Verified OK.
\[ \begin {array}{lll} & {} & \textrm {Let's solve}\hspace {3pt} \\ {} & {} & x^{4} y^{\prime \prime \prime \prime }+8 y^{\prime \prime \prime } x^{3}+27 x^{2} y^{\prime \prime }+35 y^{\prime } x +45 y=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 {45 y}{x^{4}}-\frac {8 x^{2} y^{\prime \prime \prime }+27 y^{\prime \prime } x +35 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} \\ {} & {} & y^{\prime \prime \prime \prime }+\frac {8 y^{\prime \prime \prime }}{x}+\frac {27 y^{\prime \prime }}{x^{2}}+\frac {35 y^{\prime }}{x^{3}}+\frac {45 y}{x^{4}}=0 \\ \bullet & {} & \textrm {Multiply by denominators of the ODE}\hspace {3pt} \\ {} & {} & x^{4} y^{\prime \prime \prime \prime }+8 y^{\prime \prime \prime } x^{3}+27 x^{2} y^{\prime \prime }+35 y^{\prime } x +45 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}} \\ {} & \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^{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 )+8 \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}+27 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 )+35 \frac {d}{d t}y \left (t \right )+45 y \left (t \right )=0 \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & \frac {d^{4}}{d t^{4}}y \left (t \right )+2 \frac {d^{3}}{d t^{3}}y \left (t \right )+14 \frac {d^{2}}{d t^{2}}y \left (t \right )+18 \frac {d}{d t}y \left (t \right )+45 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 )=-2 y_{4}\left (t \right )-14 y_{3}\left (t \right )-18 y_{2}\left (t \right )-45 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 )=-2 y_{4}\left (t \right )-14 y_{3}\left (t \right )-18 y_{2}\left (t \right )-45 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 \\ -45 & -18 & -14 & -2 \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 \\ -45 & -18 & -14 & -2 \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 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ \frac {\mathrm {I}}{3} \\ 1 \end {array}\right ]\right ], \left [3 \,\mathrm {I}, \left [\begin {array}{c} \frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ -\frac {\mathrm {I}}{3} \\ 1 \end {array}\right ]\right ], \left [-1-2 \,\mathrm {I}, \left [\begin {array}{c} \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ], \left [-1+2 \,\mathrm {I}, \left [\begin {array}{c} \frac {11}{125}+\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}+\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}-\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-3 \,\mathrm {I}, \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ \frac {\mathrm {I}}{3} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-3 \,\mathrm {I} t}\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ \frac {\mathrm {I}}{3} \\ 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 \\ {} & {} & \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right )\cdot \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \\ -\frac {1}{9} \\ \frac {\mathrm {I}}{3} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & \left [\begin {array}{c} -\frac {\mathrm {I}}{27} \left (\cos \left (3 t \right )-\mathrm {I} \sin \left (3 t \right )\right ) \\ -\frac {\cos \left (3 t \right )}{9}+\frac {\mathrm {I} \sin \left (3 t \right )}{9} \\ \frac {\mathrm {I}}{3} \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}}_{1}\left (t \right )=\left [\begin {array}{c} -\frac {\sin \left (3 t \right )}{27} \\ -\frac {\cos \left (3 t \right )}{9} \\ \frac {\sin \left (3 t \right )}{3} \\ \cos \left (3 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{2}\left (t \right )=\left [\begin {array}{c} -\frac {\cos \left (3 t \right )}{27} \\ \frac {\sin \left (3 t \right )}{9} \\ \frac {\cos \left (3 t \right )}{3} \\ -\sin \left (3 t \right ) \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Consider complex eigenpair, complex conjugate eigenvalue can be ignored}\hspace {3pt} \\ {} & {} & \left [-1-2 \,\mathrm {I}, \left [\begin {array}{c} \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ]\right ] \\ \bullet & {} & \textrm {Solution from eigenpair}\hspace {3pt} \\ {} & {} & {\mathrm e}^{\left (-1-2 \,\mathrm {I}\right ) t}\cdot \left [\begin {array}{c} \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\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}^{-t}\cdot \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right )\cdot \left [\begin {array}{c} \frac {11}{125}-\frac {2 \,\mathrm {I}}{125} \\ -\frac {3}{25}-\frac {4 \,\mathrm {I}}{25} \\ -\frac {1}{5}+\frac {2 \,\mathrm {I}}{5} \\ 1 \end {array}\right ] \\ \bullet & {} & \textrm {Simplify expression}\hspace {3pt} \\ {} & {} & {\mathrm e}^{-t}\cdot \left [\begin {array}{c} \left (\frac {11}{125}-\frac {2 \,\mathrm {I}}{125}\right ) \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \left (-\frac {3}{25}-\frac {4 \,\mathrm {I}}{25}\right ) \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \left (-\frac {1}{5}+\frac {2 \,\mathrm {I}}{5}\right ) \left (\cos \left (2 t \right )-\mathrm {I} \sin \left (2 t \right )\right ) \\ \cos \left (2 t \right )-\mathrm {I} \sin \left (2 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}^{-t}\cdot \left [\begin {array}{c} \frac {11 \cos \left (2 t \right )}{125}-\frac {2 \sin \left (2 t \right )}{125} \\ -\frac {3 \cos \left (2 t \right )}{25}-\frac {4 \sin \left (2 t \right )}{25} \\ -\frac {\cos \left (2 t \right )}{5}+\frac {2 \sin \left (2 t \right )}{5} \\ \cos \left (2 t \right ) \end {array}\right ], {\moverset {\rightarrow }{y}}_{4}\left (t \right )={\mathrm e}^{-t}\cdot \left [\begin {array}{c} -\frac {11 \sin \left (2 t \right )}{125}-\frac {2 \cos \left (2 t \right )}{125} \\ \frac {3 \sin \left (2 t \right )}{25}-\frac {4 \cos \left (2 t \right )}{25} \\ \frac {\sin \left (2 t \right )}{5}+\frac {2 \cos \left (2 t \right )}{5} \\ -\sin \left (2 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}\left (t \right )+c_{2} {\moverset {\rightarrow }{y}}_{2}\left (t \right )+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_{3} {\mathrm e}^{-t}\cdot \left [\begin {array}{c} \frac {11 \cos \left (2 t \right )}{125}-\frac {2 \sin \left (2 t \right )}{125} \\ -\frac {3 \cos \left (2 t \right )}{25}-\frac {4 \sin \left (2 t \right )}{25} \\ -\frac {\cos \left (2 t \right )}{5}+\frac {2 \sin \left (2 t \right )}{5} \\ \cos \left (2 t \right ) \end {array}\right ]+c_{4} {\mathrm e}^{-t}\cdot \left [\begin {array}{c} -\frac {11 \sin \left (2 t \right )}{125}-\frac {2 \cos \left (2 t \right )}{125} \\ \frac {3 \sin \left (2 t \right )}{25}-\frac {4 \cos \left (2 t \right )}{25} \\ \frac {\sin \left (2 t \right )}{5}+\frac {2 \cos \left (2 t \right )}{5} \\ -\sin \left (2 t \right ) \end {array}\right ]+\left [\begin {array}{c} -\frac {c_{1} \sin \left (3 t \right )}{27}-\frac {c_{2} \cos \left (3 t \right )}{27} \\ -\frac {c_{1} \cos \left (3 t \right )}{9}+\frac {c_{2} \sin \left (3 t \right )}{9} \\ \frac {c_{1} \sin \left (3 t \right )}{3}+\frac {c_{2} \cos \left (3 t \right )}{3} \\ c_{1} \cos \left (3 t \right )-c_{2} \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 (11 c_{3} -2 c_{4} \right ) \cos \left (2 t \right )-2 \sin \left (2 t \right ) \left (c_{3} +\frac {11 c_{4}}{2}\right )\right ) {\mathrm e}^{-t}}{125}-\frac {c_{1} \sin \left (3 t \right )}{27}-\frac {c_{2} \cos \left (3 t \right )}{27} \\ \bullet & {} & \textrm {Change variables back using}\hspace {3pt} t =\ln \left (x \right ) \\ {} & {} & y=\frac {\left (11 c_{3} -2 c_{4} \right ) \cos \left (2 \ln \left (x \right )\right )-2 \sin \left (2 \ln \left (x \right )\right ) \left (c_{3} +\frac {11 c_{4}}{2}\right )}{125 x}-\frac {c_{1} \sin \left (3 \ln \left (x \right )\right )}{27}-\frac {c_{2} \cos \left (3 \ln \left (x \right )\right )}{27} \\ \bullet & {} & \textrm {Simplify}\hspace {3pt} \\ {} & {} & y=-\frac {c_{1} \sin \left (3 \ln \left (x \right )\right )}{27}-\frac {c_{2} \cos \left (3 \ln \left (x \right )\right )}{27}+\frac {11 c_{3} \cos \left (2 \ln \left (x \right )\right )}{125 x}-\frac {2 c_{4} \cos \left (2 \ln \left (x \right )\right )}{125 x}-\frac {2 c_{3} \sin \left (2 \ln \left (x \right )\right )}{125 x}-\frac {11 c_{4} \sin \left (2 \ln \left (x \right )\right )}{125 x} \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: 39
dsolve(x^4*diff(y(x),x$4)+8*x^3*diff(y(x),x$3)+27*x^2*diff(y(x),x$2)+35*x*diff(y(x),x)+45*y(x)=0,y(x), singsol=all)
\[ y \left (x \right ) = \frac {c_{1} \sin \left (2 \ln \left (x \right )\right )}{x}+\frac {c_{2} \cos \left (2 \ln \left (x \right )\right )}{x}+c_{3} \sin \left (3 \ln \left (x \right )\right )+c_{4} \cos \left (3 \ln \left (x \right )\right ) \]
✓ Solution by Mathematica
Time used: 0.004 (sec). Leaf size: 44
DSolve[x^4*y''''[x]+8*x^3*y'''[x]+27*x^2*y''[x]+35*x*y'[x]+45*y[x]==0,y[x],x,IncludeSingularSolutions -> True]
\[ y(x)\to \frac {c_2 \cos (2 \log (x))+c_3 x \cos (3 \log (x))+c_1 \sin (2 \log (x))+c_4 x \sin (3 \log (x))}{x} \]